{"id":241,"date":"2017-11-14T13:47:26","date_gmt":"2017-11-14T13:47:26","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/chapter\/newtons-second-law\/"},"modified":"2017-11-18T01:29:24","modified_gmt":"2017-11-18T01:29:24","slug":"newtons-second-law","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/chapter\/newtons-second-law\/","title":{"raw":"Newton&#8217;s Second Law","rendered":"Newton&#8217;s Second Law"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of the section, you will be able to:<\/p>\n<ul>\n<li>Distinguish between external and internal forces<\/li>\n<li>Describe Newton's second law of motion<\/li>\n<li>Explain the dependence of acceleration on net force and mass<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165039275238\">Newton\u2019s second law is closely related to his first law. It mathematically gives the cause-and-effect relationship between force and changes in motion. Newton\u2019s second law is quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton\u2019s second law as a simple equation that gives the exact relationship of force, mass, and acceleration, we need to sharpen some ideas we mentioned earlier.<\/p>\n<div class=\"bc-section section\">\n<h3>Force and Acceleration<\/h3>\n<p id=\"fs-id1165039092833\">First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is acceleration. Newton\u2019s first law says that a net external force causes a change in motion; thus, we see that a <em>net external force causes nonzero acceleration<\/em>.<\/p>\n<p id=\"fs-id1165039083134\">We defined external force in <a href=\"\/contents\/db08e30b-406e-4711-935c-60e064cb2271\" class=\"target-chapter\">Forces<\/a> as force acting on an object or system that originates outside of the object or system. Let\u2019s consider this concept further. An intuitive notion of <em>external<\/em> is correct\u2014it is outside the system of interest. For example, in <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>(a), the system of interest is the car plus the person within it. The two forces exerted by the two students are external forces. In contrast, an internal force acts between elements of the system. Thus, the force the person in the car exerts to hang on to the steering wheel is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton\u2019s first law. (The internal forces cancel each other out, as explained in the next section.) Therefore, we must define the boundaries of the system before we can determine which forces are external. Sometimes, the system is obvious, whereas at other times, identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton\u2019s laws. This concept is revisited many times in the study of physics.<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_ExtForce\">\n<div class=\"bc-figcaption figcaption\">Different forces exerted on the same mass produce different accelerations. (a) Two students push a stalled car. All external forces acting on the car are shown. (b) The forces acting on the car are transferred to a coordinate plane (free-body diagram) for simpler analysis. (c) The tow truck can produce greater external force on the same mass, and thus greater acceleration.<\/div>\n<p><span id=\"fs-id1165039264684\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_ExtForce.jpg\" alt=\"Figure a shows two people pushing a car with forces F1 and F2 in the right direction. Acceleration a is also in the same direction. Frictional force f is shown near the tire in the opposite direction, left. Upward force N and downward force W are equal in magnitude and are shown near the ground. Figure b puts all the forces of figure a together and shows a net force F net. These forces are also shown in a free body diagram. Figure c shows the car being towed by a tow-truck. Here, the forces N, W and f are the same as those in figure a. F subscript tow truck has a greater magnitude than F1 or F2. Acceleration a prime has a greater magnitude than a. All forces of this system are also shown in a free body diagram.\"><\/span><\/p><\/div>\n<p id=\"fs-id1165039092504\">From this example, you can see that different forces exerted on the same mass produce different accelerations. In <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>(a), the two students push a car with a driver in it. Arrows representing all external forces are shown. The system of interest is the car and its driver. The weight [latex]\\stackrel{\\to }{w}[\/latex] of the system and the support of the ground [latex]\\stackrel{\\to }{N}[\/latex] are also shown for completeness and are assumed to cancel (because there was no vertical motion and no imbalance of forces in the vertical direction to create a change in motion). The vector [latex]\\stackrel{\\to }{f}[\/latex] represents the friction acting on the car, and it acts to the left, opposing the motion of the car. (We discuss friction in more detail in the next chapter.) In <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>(b), all external forces acting on the system add together to produce the net force [latex]{\\stackrel{\\to }{F}}_{\\text{net}}.[\/latex] The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from this dot. Because there are two forces acting to the right, the vectors are shown collinearly. Finally, in <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>(c), a larger net external force produces a larger acceleration [latex]\\left(\\stackrel{\\to }{{a}^{\\prime }}&gt;\\stackrel{\\to }{a}\\right)[\/latex] when the tow truck pulls the car.<\/p>\n<p>It seems reasonable that acceleration would be directly proportional to and in the same direction as the net external force acting on a system. This assumption has been verified experimentally and is illustrated in <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>. To obtain an equation for Newton\u2019s second law, we first write the relationship of acceleration [latex]\\stackrel{\\to }{a}[\/latex] and net external force [latex]{\\stackrel{\\to }{F}}_{\\text{net}}[\/latex] as the proportionality<\/p>\n<div id=\"fs-id1165039046494\" class=\"unnumbered\">[latex]\\stackrel{\\to }{a}\\propto {\\stackrel{\\to }{F}}_{\\text{net}}[\/latex]<\/div>\n<p id=\"fs-id1165039109910\">where the symbol [latex]\\propto [\/latex] means \u201cproportional to.\u201d (Recall from <a href=\"\/contents\/db08e30b-406e-4711-935c-60e064cb2271\" class=\"target-chapter\">Forces<\/a> that the net external force is the vector sum of all external forces and is sometimes indicated as [latex]\\sum \\stackrel{\\to }{F}.[\/latex]) This proportionality shows what we have said in words\u2014acceleration is directly proportional to net external force. Once the system of interest is chosen, identify the external forces and ignore the internal ones. It is a tremendous simplification to disregard the numerous internal forces acting between objects within the system, such as muscular forces within the students\u2019 bodies, let alone the myriad forces between the atoms in the objects. Still, this simplification helps us solve some complex problems.<\/p>\n<p id=\"fs-id1165039188181\">It also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. As illustrated in <a href=\"#CNX_UPhysics_05_03_Basketball\" class=\"autogenerated-content\">(Figure)<\/a>, the same net external force applied to a basketball produces a much smaller acceleration when it is applied to an SUV. The proportionality is written as<\/p>\n<div id=\"fs-id1165039088901\" class=\"unnumbered\">[latex]a\\propto \\frac{1}{m},[\/latex]<\/div>\n<p id=\"fs-id1165039337544\">where <em>m<\/em> is the mass of the system and <em>a<\/em> is the magnitude of the acceleration. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is directly proportional to net external force.<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_Basketball\">\n<div class=\"bc-figcaption figcaption\">The same force exerted on systems of different masses produces different accelerations. (a) A basketball player pushes on a basketball to make a pass. (Ignore the effect of gravity on the ball.) (b) The same player exerts an identical force on a stalled SUV and produces far less acceleration. (c) The free-body diagrams are identical, permitting direct comparison of the two situations. A series of patterns for free-body diagrams will emerge as you do more problems and learn how to draw them in <a href=\"\/contents\/595efb44-2031-41fd-b3a2-f08f0e2ab46a\" class=\"target-chapter\">Drawing Free-Body Diagrams<\/a>.<\/div>\n<p><span id=\"fs-id1165039052516\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_Basketball.jpg\" alt=\"Figure a shows a person exerting force F on a basketball with mass m1. The ball is shown to move to the rigth with an acceleration a1. Figure b shows the person exerting the same amount of force, F on an SUV with mass m2. The acceleration is a2, which is much smaller than a1. Figure c shows the free body diagrams of both systems shown in figure a and figure b. Both show the force F having the same magnitude and direction. The label reads: the free-body diagrams of both objects are the same.\"><\/span><\/p><\/div>\n<p id=\"fs-id1165039347931\">It has been found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields <span>Newton\u2019s second law<\/span>.<\/p>\n<div id=\"fs-id1165039096883\">\n<div>Newton\u2019s Second Law of Motion<\/div>\n<p id=\"fs-id1165038999240\">The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system and is inversely proportion to its mass. In equation form, Newton\u2019s second law is<\/p>\n<div id=\"fs-id1165039299982\" class=\"unnumbered\">[latex]\\stackrel{\\to }{a}=\\frac{{\\stackrel{\\to }{F}}_{\\text{net}}}{m},[\/latex]<\/div>\n<p id=\"fs-id1165039257622\">where [latex]\\stackrel{\\to }{a}[\/latex] is the acceleration, [latex]{\\stackrel{\\to }{F}}_{\\text{net}}[\/latex] is the net force, and <em>m<\/em> is the mass. This is often written in the more familiar form<\/p>\n<div id=\"fs-id1165039079819\">[latex]{\\stackrel{\\to }{F}}_{\\text{net}}=\\sum \\stackrel{\\to }{F}=m\\stackrel{\\to }{a},[\/latex]<\/div>\n<p id=\"fs-id1165039083076\">but the first equation gives more insight into what Newton\u2019s second law means. When only the magnitude of force and acceleration are considered, this equation can be written in the simpler scalar form:<\/p>\n<div id=\"fs-id1165039009387\">[latex]{F}_{\\text{net}}=ma.[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1165039401832\">The law is a cause-and-effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is based on experimental verification. The free-body diagram, which you will learn to draw in <a href=\"\/contents\/595efb44-2031-41fd-b3a2-f08f0e2ab46a\" class=\"target-chapter\">Drawing Free-Body Diagrams<\/a>, is the basis for writing Newton\u2019s second law.<\/p>\n<div id=\"fs-id1165039247388\" class=\"textbox examples\">\n<p id=\"fs-id1165039088471\"><span>What Acceleration Can a Person Produce When Pushing a Lawn Mower?<\/span><br>\nSuppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb.) parallel to the ground (<a href=\"#CNX_UPhysics_05_03_Mower\" class=\"autogenerated-content\">(Figure)<\/a>). The mass of the mower is 24 kg. What is its acceleration?<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_Mower\">\n<div class=\"bc-figcaption figcaption\">(a) The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right? (b) The free-body diagram for this problem is shown.<\/div>\n<p><span id=\"fs-id1165039252210\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_Mower.jpg\" alt=\"Figure a shows a person using a lawn mower on a lawn. Force F net points right, from the person\u2019s hands. Figure b shows the force F net along the positive x axis.\"><\/span><\/p><\/div>\n<p id=\"fs-id1165038978303\"><span>Strategy<\/span><br>\nThis problem involves only motion in the horizontal direction; we are also given the net force, indicated by the single vector, but we can suppress the vector nature and concentrate on applying Newton\u2019s second law. Since [latex]{F}_{\\text{net}}[\/latex] and <em>m<\/em> are given, the acceleration can be calculated directly from Newton\u2019s second law as [latex]{F}_{\\text{net}}=ma.[\/latex]<\/p>\n<p id=\"fs-id1165038974688\"><span>Solution<\/span><br>\nThe magnitude of the acceleration <em>a<\/em> is [latex]a={F}_{\\text{net}}\\text{\/}m[\/latex]. Entering known values gives<\/p>\n<div id=\"fs-id1165039334536\" class=\"unnumbered\">[latex]a=\\frac{51\\phantom{\\rule{0.2em}{0ex}}\\text{N}}{24\\phantom{\\rule{0.2em}{0ex}}\\text{kg}}.[\/latex]<\/div>\n<p id=\"fs-id1165039253539\">Substituting the unit of kilograms times meters per square second for newtons yields<\/p>\n<div id=\"fs-id1165039123852\" class=\"unnumbered\">[latex]a=\\frac{51\\phantom{\\rule{0.2em}{0ex}}\\text{kg}\u00b7{\\text{m\/s}}^{2}}{24\\phantom{\\rule{0.2em}{0ex}}\\text{kg}}=2.1\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}.[\/latex]<\/div>\n<p id=\"fs-id1165038998097\"><span>Significance<\/span><br>\nThe direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. This is a result of the vector relationship expressed in Newton\u2019s second law, that is, the vector representing net force is the scalar multiple of the acceleration vector. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moved forward), and the vertical forces must cancel because no acceleration occurs in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long, because the person\u2019s top speed would soon be reached.<\/p>\n<\/div>\n<div id=\"fs-id1165039079950\" class=\"check-understanding\">\n<div id=\"fs-id1165039418564\">\n<div id=\"fs-id1165039010637\">\n<p id=\"fs-id1165039295994\"><strong>Check Your Understanding<\/strong> At the time of its launch, the HMS <em>Titanic<\/em> was the most massive mobile object ever built, with a mass of [latex]6.0\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{7}\\phantom{\\rule{0.2em}{0ex}}\\text{kg}[\/latex]. If a force of 6 MN [latex]\\left(6\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{6}\\phantom{\\rule{0.2em}{0ex}}\\text{N}\\right)[\/latex] was applied to the ship, what acceleration would it experience?<\/p>\n<\/div>\n<div id=\"fs-id1165039093866\">\n[latex]0.1\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165039028823\">In the preceding example, we dealt with net force only for simplicity. However, several forces act on the lawn mower. The weight [latex]\\stackrel{\\to }{w}[\/latex] (discussed in detail in <a href=\"\/contents\/68003722-b144-499e-859f-615de26fd893\" class=\"target-chapter\">Mass and Weight<\/a>) pulls down on the mower, toward the center of Earth; this produces a contact force on the ground. The ground must exert an upward force on the lawn mower, known as the normal force [latex]\\stackrel{\\to }{N}[\/latex], which we define in <a href=\"\/contents\/7517c753-4d3a-4137-8cb8-4d9ad48844aa\" class=\"target-chapter\">Common Forces<\/a>. These forces are balanced and therefore do not produce vertical acceleration. In the next example, we show both of these forces. As you continue to solve problems using Newton\u2019s second law, be sure to show multiple forces.<\/p>\n<div id=\"fs-id1165038975256\" class=\"textbox examples\">\n<p id=\"fs-id1165039273309\"><span>Which Force Is Bigger?<\/span><br>\n(a) The car shown in <a href=\"#CNX_UPhysics_05_03_RedCar\" class=\"autogenerated-content\">(Figure)<\/a> is moving at a constant speed. Which force is bigger, [latex]{\\stackrel{\\to }{F}}_{\\text{engine}}[\/latex] or [latex]{\\stackrel{\\to }{F}}_{\\text{friction}}[\/latex]? Explain.<\/p>\n<p>(b) The same car is now accelerating to the right. Which force is bigger, [latex]{\\stackrel{\\to }{F}}_{\\text{engine}}[\/latex] or [latex]{\\stackrel{\\to }{F}}_{\\text{friction}}?[\/latex] Explain.<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_RedCar\">\n<div class=\"bc-figcaption figcaption\">A car is shown (a) moving at constant speed and (b) accelerating. How do the forces acting on the car compare in each case? (a) What does the knowledge that the car is moving at constant velocity tell us about the net horizontal force on the car compared to the friction force? (b) What does the knowledge that the car is accelerating tell us about the horizontal force on the car compared to the friction force?<\/div>\n<p><span id=\"fs-id1165039085891\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_RedCar_img.jpg\" alt=\"Figure a shows a car with velocity 10 meters per second, moving right. F subscript engine right and F subscript friction points left. Figure b shows the car moving with an acceleration of 10 meters per second squared, towards the right. Forces F subscript engine and F subscript friction are the same as those in figure a.\"><\/span><\/p><\/div>\n<p id=\"fs-id1165039333800\"><span>Strategy<\/span><br>\nWe must consider Newton\u2019s first and second laws to analyze the situation. We need to decide which law applies; this, in turn, will tell us about the relationship between the forces.<\/p>\n<p id=\"fs-id1165039286658\"><span>Solution<\/span><\/p>\n<ol id=\"fs-id1165039315375\" type=\"a\">\n<li>The forces are equal. According to Newton\u2019s first law, if the net force is zero, the velocity is constant.<\/li>\n<li>In this case, [latex]{\\stackrel{\\to }{F}}_{\\text{engine}}[\/latex] must be larger than [latex]{\\stackrel{\\to }{F}}_{\\text{friction}}.[\/latex] According to Newton\u2019s second law, a net force is required to cause acceleration.<\/li>\n<\/ol>\n<p><span>Significance<\/span><br>\nThese questions may seem trivial, but they are commonly answered incorrectly. For a car or any other object to move, it must be accelerated from rest to the desired speed; this requires that the engine force be greater than the friction force. Once the car is moving at constant velocity, the net force must be zero; otherwise, the car will accelerate (gain speed). To solve problems involving Newton\u2019s laws, we must understand whether to apply Newton\u2019s first law (where [latex]\\sum \\stackrel{\\to }{F}=\\stackrel{\\to }{0}[\/latex]) or Newton\u2019s second law (where [latex]\\sum \\stackrel{\\to }{F}[\/latex] is not zero). This will be apparent as you see more examples and attempt to solve problems on your own.<\/p>\n<\/div>\n<div id=\"fs-id1165039286668\" class=\"textbox examples\">\n<p id=\"fs-id1165038973771\"><span>What Rocket Thrust Accelerates This Sled?<\/span><br>\nBefore manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets.<\/p>\n<p id=\"fs-id1165039110006\">Calculate the magnitude of force exerted by each rocket, called its thrust <em>T<\/em>, for the four-rocket propulsion system shown in <a href=\"#CNX_UPhysics_05_03_RocketSled\" class=\"autogenerated-content\">(Figure)<\/a>. The sled\u2019s initial acceleration is [latex]49\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex], the mass of the system is 2100 kg, and the force of friction opposing the motion is 650 N.<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_RocketSled\">\n<div class=\"bc-figcaption figcaption\">A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust <em>T<\/em>. The system here is the sled, its rockets, and its rider, so none of the forces between these objects are considered. The arrow representing friction [latex]\\left(\\stackrel{\\to }{f}\\right)[\/latex] is drawn larger than scale.<\/div>\n<p><span id=\"fs-id1165039065298\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_RocketSled.jpg\" alt=\"Figure shows a sled going right. It has four rockets at the back, with each thrust vector having the same magnitude and pointing right. Friction f points left. The upward normal force N and downward weight, are both equal in magnitude. Acceleration a is towards the right. All these forces are also shown in a free body diagram.\"><\/span><\/p><\/div>\n<p id=\"fs-id1165039065549\"><span>Strategy<\/span><br>\nAlthough forces are acting both vertically and horizontally, we assume the vertical forces cancel because there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in <a href=\"#CNX_UPhysics_05_03_RocketSled\" class=\"autogenerated-content\">(Figure)<\/a>.<\/p>\n<p id=\"fs-id1165039303979\"><span>Solution<\/span><br>\nSince acceleration, mass, and the force of friction are given, we start with Newton\u2019s second law and look for ways to find the thrust of the engines. We have defined the direction of the force and acceleration as acting \u201cto the right,\u201d so we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with<\/p>\n<div id=\"fs-id1165039095877\" class=\"unnumbered\">[latex]{F}_{\\text{net}}=ma[\/latex]<\/div>\n<p id=\"fs-id1165039123524\">where [latex]{F}_{\\text{net}}[\/latex] is the net force along the horizontal direction. We can see from the figure that the engine thrusts add, whereas friction opposes the thrust. In equation form, the net external force is<\/p>\n<div id=\"fs-id1165039115293\" class=\"unnumbered\">[latex]{F}_{\\text{net}}=4T-f.[\/latex]<\/div>\n<p id=\"fs-id1165039300255\">Substituting this into Newton\u2019s second law gives us<\/p>\n<div id=\"fs-id1165039084866\" class=\"unnumbered\">[latex]{F}_{\\text{net}}=ma=4T-f.[\/latex]<\/div>\n<p id=\"fs-id1165038968653\">Using a little algebra, we solve for the total thrust 4<em>T<\/em>:<\/p>\n<div class=\"unnumbered\">[latex]4T=ma+f.[\/latex]<\/div>\n<p id=\"fs-id1165039109216\">Substituting known values yields<\/p>\n<div id=\"fs-id1165039098409\" class=\"unnumbered\">[latex]4T=ma+f=\\left(2100\\phantom{\\rule{0.2em}{0ex}}\\text{kg}\\right)\\left(49\\phantom{\\rule{0.2em}{0ex}}{\\text{m}\\text{\/}\\text{s}}^{2}\\right)+650\\phantom{\\rule{0.2em}{0ex}}\\text{N}.[\/latex]<\/div>\n<p id=\"fs-id1165039122508\">Therefore, the total thrust is<\/p>\n<div id=\"fs-id1165039044869\" class=\"unnumbered\">[latex]4T=1.0\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{5}\\phantom{\\rule{0.2em}{0ex}}\\text{N},[\/latex]<\/div>\n<p id=\"fs-id1165039110278\">and the individual thrusts are<\/p>\n<div id=\"fs-id1165039053608\" class=\"unnumbered\">[latex]T=\\frac{1.0\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{5}\\phantom{\\rule{0.2em}{0ex}}\\text{N}}{4}=2.5\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{4}\\phantom{\\rule{0.2em}{0ex}}\\text{N}.[\/latex]<\/div>\n<p id=\"fs-id1165038998398\"><span>Significance<\/span><br>\nThe numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance, and the setup was designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km\/h were obtained, with accelerations of 45 <em>g<\/em>\u2019s. (Recall that <em>g<\/em>, acceleration due to gravity, is [latex]9.80\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex]. When we say that acceleration is 45 <em>g<\/em>\u2019s, it is [latex]45\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}9.8\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2},[\/latex] which is approximately [latex]440\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex].) Although living subjects are not used anymore, land speeds of 10,000 km\/h have been obtained with a rocket sled.<\/p>\n<p id=\"fs-id1165039114554\">In this example, as in the preceding one, the system of interest is obvious. We see in later examples that choosing the system of interest is crucial\u2014and the choice is not always obvious.<\/p>\n<p id=\"fs-id1165039247454\">Newton\u2019s second law is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature.<\/p>\n<\/div>\n<div id=\"fs-id1165039048070\" class=\"check-understanding\">\n<div id=\"fs-id1165039069953\">\n<div id=\"fs-id1165039235667\">\n<p id=\"fs-id1165038988394\"><strong>Check Your Understanding<\/strong> A 550-kg sports car collides with a 2200-kg truck, and during the collision, the net force on each vehicle is the force exerted by the other. If the magnitude of the truck\u2019s acceleration is [latex]10\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2},[\/latex] what is the magnitude of the sports car\u2019s acceleration?<\/p>\n<\/div>\n<div id=\"fs-id1165039454293\">\n<p id=\"fs-id1165039273067\">[latex]40\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\" id=\"fs-id1165039187319\">\n<h3>Component Form of Newton\u2019s Second Law<\/h3>\n<p id=\"fs-id1165039340543\">We have developed Newton\u2019s second law and presented it as a vector equation in <a href=\"#fs-id1165039079819\" class=\"autogenerated-content\">(Figure)<\/a>. This vector equation can be written as three component equations:<\/p>\n<div id=\"fs-id1165039122533\" class=\"equation-callout\">\n<div id=\"fs-id1165039419225\">[latex]\\sum {\\stackrel{\\to }{F}}_{x}=m{\\stackrel{\\to }{a}}_{x},\\phantom{\\rule{0.2em}{0ex}}\\sum {\\stackrel{\\to }{F}}_{y}=m{\\stackrel{\\to }{a}}_{y},\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}\\sum {\\stackrel{\\to }{F}}_{z}=m{\\stackrel{\\to }{a}}_{z}.[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1165038966002\">The second law is a description of how a body responds mechanically to its environment. The influence of the environment is the net force [latex]{\\stackrel{\\to }{F}}_{\\text{net}},[\/latex] the body\u2019s response is the acceleration [latex]\\stackrel{\\to }{a},[\/latex] and the strength of the response is inversely proportional to the mass <em>m<\/em>. The larger the mass of an object, the smaller its response (its acceleration) to the influence of the environment (a given net force). Therefore, a body\u2019s mass is a measure of its inertia, as we explained in <a href=\"\/contents\/e3652cf1-835a-4a39-a951-80c909f22251\" class=\"target-chapter\">Newton\u2019s First Law<\/a>.<\/p>\n<div id=\"fs-id1165039045618\" class=\"textbox examples\">\n<p id=\"fs-id1165039225277\"><span>Force on a Soccer Ball<\/span><br>\nA 0.400-kg soccer ball is kicked across the field by a player; it undergoes acceleration given by [latex]\\stackrel{\\to }{a}=3.00\\stackrel{^}{i}+7.00\\stackrel{^}{j}\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}.[\/latex] Find (a) the resultant force acting on the ball and (b) the magnitude and direction of the resultant force.<\/p>\n<p id=\"fs-id1165039123808\"><span>Strategy<\/span><br>\nThe vectors in [latex]\\stackrel{^}{i}[\/latex] and [latex]\\stackrel{^}{j}[\/latex] format, which indicate force direction along the <em>x<\/em>-axis and the <em>y<\/em>-axis, respectively, are involved, so we apply Newton\u2019s second law in vector form.<\/p>\n<p id=\"fs-id1165039269338\"><span>Solution<\/span><\/p>\n<ol id=\"fs-id1165039107084\" type=\"a\">\n<li>We apply Newton\u2019s second law:\n<div><\/div>\n<div id=\"fs-id1165038962202\" class=\"unnumbered\">[latex]{\\stackrel{\\to }{F}}_{\\text{net}}=m\\stackrel{\\to }{a}=\\left(0.400\\phantom{\\rule{0.2em}{0ex}}\\text{kg}\\right)\\left(3.00\\stackrel{^}{i}+7.00\\stackrel{^}{j}\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}\\right)=1.20\\stackrel{^}{i}+2.80\\stackrel{^}{j}\\phantom{\\rule{0.2em}{0ex}}\\text{N}.[\/latex]<\/div>\n<\/li>\n<li>Magnitude and direction are found using the components of [latex]{\\stackrel{\\to }{F}}_{\\text{net}}[\/latex]:\n<div><\/div>\n<div id=\"fs-id1165039293018\" class=\"unnumbered\">[latex]{F}_{\\text{net}}=\\sqrt{{\\left(1.20\\phantom{\\rule{0.2em}{0ex}}\\text{N}\\right)}^{2}+{\\left(2.80\\phantom{\\rule{0.2em}{0ex}}\\text{N}\\right)}^{2}}=3.05\\phantom{\\rule{0.2em}{0ex}}\\text{N}\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}\\theta ={\\text{tan}}^{-1}\\left(\\frac{2.80}{1.20}\\right)=66.8\\text{\u00b0}.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p id=\"fs-id1165039464375\"><span>Significance<\/span><br>\nWe must remember that Newton\u2019s second law is a vector equation. In (a), we are multiplying a vector by a scalar to determine the net force in vector form. While the vector form gives a compact representation of the force vector, it does not tell us how \u201cbig\u201d it is, or where it goes, in intuitive terms. In (b), we are determining the actual size (magnitude) of this force and the direction in which it travels.<\/p>\n<\/div>\n<div id=\"fs-id1165039443202\" class=\"textbox examples\">\n<p id=\"fs-id1165039475586\"><span>Mass of a Car<\/span><br>\nFind the mass of a car if a net force of [latex]-600.0\\stackrel{^}{j}\\phantom{\\rule{0.2em}{0ex}}\\text{N}[\/latex] produces an acceleration of [latex]-0.2\\stackrel{^}{j}\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex].<\/p>\n<p id=\"fs-id1165039311437\"><span>Strategy<\/span><br>\nVector division is not defined, so [latex]m={\\stackrel{\\to }{F}}_{\\text{net}}\\text{\/}\\stackrel{\\to }{a}[\/latex] cannot be performed. However, mass <em>m<\/em> is a scalar, so we can use the scalar form of Newton\u2019s second law, [latex]m={F}_{\\text{net}}\\text{\/}a[\/latex].<\/p>\n<p id=\"fs-id1165039295458\"><span>Solution<\/span><br>\nWe use [latex]m={F}_{\\text{net}}\\text{\/}a[\/latex] and substitute the magnitudes of the two vectors: [latex]{F}_{\\text{net}}=600.0\\phantom{\\rule{0.2em}{0ex}}\\text{N}[\/latex] and [latex]a=0.2\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}.[\/latex] Therefore,<\/p>\n<div id=\"fs-id1165039448064\" class=\"unnumbered\">[latex]m=\\frac{{F}_{\\text{net}}}{a}=\\frac{600.0\\phantom{\\rule{0.2em}{0ex}}\\text{N}}{0.2\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}}=3000\\phantom{\\rule{0.2em}{0ex}}\\text{kg}.[\/latex]<\/div>\n<p id=\"fs-id1165039345542\"><span>Significance<\/span><br>\nForce and acceleration were given in the [latex]\\stackrel{^}{i}[\/latex] and [latex]\\stackrel{^}{j}[\/latex] format, but the answer, mass <em>m<\/em>, is a scalar and thus is not given in [latex]\\stackrel{^}{i}[\/latex] and [latex]\\stackrel{^}{j}[\/latex] form.<\/p>\n<\/div>\n<div id=\"fs-id1165039464584\" class=\"textbox examples\">\n<p id=\"fs-id1165039464586\"><span>Several Forces on a Particle<\/span><br>\nA particle of mass [latex]m=4.0\\phantom{\\rule{0.2em}{0ex}}\\text{kg}[\/latex] is acted upon by four forces of magnitudes. [latex]{F}_{1}=10.0\\phantom{\\rule{0.2em}{0ex}}\\text{N},\\phantom{\\rule{0.2em}{0ex}}{F}_{2}=40.0\\phantom{\\rule{0.2em}{0ex}}\\text{N},\\phantom{\\rule{0.2em}{0ex}}{F}_{3}=5.0\\phantom{\\rule{0.2em}{0ex}}\\text{N},\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}{F}_{4}=2.0\\phantom{\\rule{0.2em}{0ex}}\\text{N}[\/latex], with the directions as shown in the free-body diagram in <a href=\"#CNX_UPhysics_05_03_FourForces\" class=\"autogenerated-content\">(Figure)<\/a>. What is the acceleration of the particle?<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_FourForces\">\n<div class=\"bc-figcaption figcaption\">Four forces in the <em>xy<\/em>-plane are applied to a 4.0-kg particle.<\/div>\n<p><span id=\"fs-id1165039374163\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_FourForces.jpg\" alt=\"A particle is shown in the xy plane. Force F1 is at an angle of 30 degrees with the positive x axis, force F2 is in the downward direction, force F3 points left and force F4 points upwards.\"><\/span><\/p><\/div>\n<p id=\"fs-id1165039296260\"><span>Strategy<\/span><br>\nBecause this is a two-dimensional problem, we must use a free-body diagram. First, [latex]{\\stackrel{\\to }{F}}_{1}[\/latex] must be resolved into <em>x<\/em>- and <em>y<\/em>-components. We can then apply the second law in each direction.<\/p>\n<p id=\"fs-id1165039464479\"><span>Solution<\/span><br>\nWe draw a free-body diagram as shown in <a href=\"#CNX_UPhysics_05_03_FourForces\" class=\"autogenerated-content\">(Figure)<\/a>. Now we apply Newton\u2019s second law. We consider all vectors resolved into <em>x<\/em>- and <em>y<\/em>-components:<\/p>\n<div id=\"fs-id1165039340892\" class=\"unnumbered\">[latex]\\begin{array}{cccc}\\sum {F}_{x}=m{a}_{x}\\hfill &amp; &amp; &amp; \\sum {F}_{y}=m{a}_{y}\\hfill \\\\ {F}_{1x}-{F}_{3x}=m{a}_{x}\\hfill &amp; &amp; &amp; {F}_{1y}+{F}_{4y}-{F}_{2y}=m{a}_{y}\\hfill \\\\ {F}_{1}\\phantom{\\rule{0.2em}{0ex}}\\text{cos}\\phantom{\\rule{0.2em}{0ex}}30\\text{\u00b0}-{F}_{3x}=m{a}_{x}\\hfill &amp; &amp; &amp; {F}_{1}\\text{sin}\\phantom{\\rule{0.2em}{0ex}}30\\text{\u00b0}+{F}_{4y}-{F}_{2y}=m{a}_{y}\\hfill \\\\ \\left(10.0\\phantom{\\rule{0.2em}{0ex}}\\text{N}\\right)\\left(\\text{cos}\\phantom{\\rule{0.2em}{0ex}}30\\text{\u00b0}\\right)-5.0\\phantom{\\rule{0.2em}{0ex}}\\text{N}=\\left(4.0\\phantom{\\rule{0.2em}{0ex}}\\text{kg}\\right){a}_{x}\\hfill &amp; &amp; &amp; \\left(10.0\\phantom{\\rule{0.2em}{0ex}}\\text{N}\\right)\\left(\\text{sin}\\phantom{\\rule{0.2em}{0ex}}30\\text{\u00b0}\\right)+2.0\\phantom{\\rule{0.2em}{0ex}}\\text{N}-40.0\\phantom{\\rule{0.2em}{0ex}}\\text{N}=\\left(4.0\\phantom{\\rule{0.2em}{0ex}}\\text{kg}\\right){a}_{y}\\hfill \\\\ {a}_{x}=0.92\\phantom{\\rule{0.2em}{0ex}}\\text{m}\\text{\/}{\\text{s}}^{2}.\\hfill &amp; &amp; &amp; {a}_{y}=-8.3\\phantom{\\rule{0.2em}{0ex}}{\\text{m}\\text{\/}\\text{s}}^{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165039196016\">Thus, the net acceleration is<\/p>\n<div id=\"fs-id1165039443012\" class=\"unnumbered\">[latex]\\stackrel{\\to }{a}=\\left(0.92\\stackrel{^}{i}-8.3\\stackrel{^}{j}\\right)\\phantom{\\rule{0.2em}{0ex}}\\text{m}\\text{\/}{\\text{s}}^{2},[\/latex]<\/div>\n<p id=\"fs-id1165039458298\">which is a vector of magnitude [latex]8.4\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex] directed at [latex]276\\text{\u00b0}[\/latex] to the positive <em>x<\/em>-axis.<\/p>\n<p id=\"fs-id1165039082704\"><span>Significance<\/span><br>\nNumerous examples in everyday life can be found that involve three or more forces acting on a single object, such as cables running from the Golden Gate Bridge or a football player being tackled by three defenders. We can see that the solution of this example is just an extension of what we have already done.<\/p>\n<\/div>\n<div id=\"fs-id1165039084354\" class=\"check-understanding\">\n<div id=\"fs-id1165039026512\">\n<div id=\"fs-id1165039026514\">\n<p id=\"fs-id1165039464613\"><strong>Check Your Understanding<\/strong> A car has forces acting on it, as shown below. The mass of the car is 1000.0 kg. The road is slick, so friction can be ignored. (a) What is the net force on the car? (b) What is the acceleration of the car?<\/p>\n<p><span id=\"fs-id1165039235873\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_RedCar3_img.jpg\" alt=\"The top view of a car is shown. Two force vectors originate from the car and point upwards and outwards. A force of 450 newtons makes an angle of 30 degrees with the straight line motion of the car, towards the right. Another force of 360 newtons makes an angle of 10 degrees with the straight line motion of the car, towards the left.\"><\/span><\/p><\/div>\n<div id=\"fs-id1165039463406\">\n<p id=\"fs-id1165039483330\">a. [latex]159.0\\stackrel{^}{i}+770.0\\stackrel{^}{j}\\phantom{\\rule{0.2em}{0ex}}\\text{N}[\/latex]; b. [latex]0.1590\\stackrel{^}{i}+0.7700\\stackrel{^}{j}\\phantom{\\rule{0.2em}{0ex}}\\text{N}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\" id=\"fs-id1165039485321\">\n<h3>Newton\u2019s Second Law and Momentum<\/h3>\n<p id=\"fs-id1165039420955\">Newton actually stated his second law in terms of momentum: \u201cThe instantaneous rate at which a body\u2019s momentum changes is equal to the net force acting on the body.\u201d (\u201cInstantaneous rate\u201d implies that the derivative is involved.) This can be given by the vector equation<\/p>\n<div id=\"fs-id1165039420958\" class=\"equation-callout\">\n<div id=\"fs-id1165039247034\">[latex]{\\stackrel{\\to }{F}}_{\\text{net}}=\\frac{d\\stackrel{\\to }{p}}{dt}.[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1165039440108\">This means that Newton\u2019s second law addresses the central question of motion: What causes a change in motion of an object? Momentum was described by Newton as \u201cquantity of motion,\u201d a way of combining both the velocity of an object and its mass. We devote <a href=\"\/contents\/647dc0b6-01e5-4e1c-893e-4ada7841ba77\" class=\"target-chapter\">Linear Momentum and Collisions<\/a> to the study of <span class=\"no-emphasis\">momentum<\/span>.<\/p>\n<p id=\"fs-id1165039484763\">For now, it is sufficient to define <em>momentum<\/em> [latex]\\stackrel{\\to }{p}[\/latex] as the product of the mass of the object <em>m<\/em> and its velocity [latex]\\stackrel{\\to }{v}[\/latex]:<\/p>\n<div id=\"fs-id1165039081523\">[latex]\\stackrel{\\to }{p}=m\\stackrel{\\to }{v}.[\/latex]<\/div>\n<p id=\"fs-id1165039341205\">Since velocity is a vector, so is momentum.<\/p>\n<p id=\"fs-id1165039341208\">It is easy to visualize momentum. A train moving at 10 m\/s has more momentum than one that moves at 2 m\/s. In everyday life, we speak of one sports team as \u201chaving momentum\u201d when they score points against the opposing team.<\/p>\n<p id=\"fs-id1165039485414\">If we substitute <a href=\"#fs-id1165039081523\" class=\"autogenerated-content\">(Figure)<\/a> into <a href=\"#fs-id1165039247034\" class=\"autogenerated-content\">(Figure)<\/a>, we obtain<\/p>\n<div id=\"fs-id1165039412912\" class=\"unnumbered\">[latex]{\\stackrel{\\to }{F}}_{\\text{net}}=\\frac{d\\stackrel{\\to }{p}}{dt}=\\frac{d\\left(m\\stackrel{\\to }{v}\\right)}{dt}.[\/latex]<\/div>\n<p id=\"fs-id1165039461971\">When <em>m<\/em> is constant, we have<\/p>\n<div id=\"fs-id1165039340576\" class=\"unnumbered\">[latex]{\\stackrel{\\to }{F}}_{\\text{net}}=m\\frac{d\\left(\\stackrel{\\to }{v}\\right)}{dt}=m\\stackrel{\\to }{a}.[\/latex]<\/div>\n<p id=\"fs-id1165039495422\">Thus, we see that the momentum form of Newton\u2019s second law reduces to the form given earlier in this section.<\/p>\n<div id=\"fs-id1165039495428\" class=\"media-2\">\n<p id=\"fs-id1165039209657\">Explore the <a href=\"https:\/\/openstaxcollege.org\/l\/21forcesatwork\">forces at work<\/a> when <a href=\"https:\/\/openstaxcollege.org\/l\/21pullacart\">pulling a cart<\/a> or pushing a refrigerator, crate, or person. Create an <a href=\"https:\/\/openstaxcollege.org\/l\/21forcemotion\">applied force<\/a> and see how it makes objects move. Put <a href=\"https:\/\/openstaxcollege.org\/l\/21ramp\">an object on a ramp<\/a> and see how it affects its motion.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\" id=\"fs-id1165039440646\">\n<h3>Summary<\/h3>\n<ul id=\"fs-id1165039440654\">\n<li>An external force acts on a system from outside the system, as opposed to internal forces, which act between components within the system.<\/li>\n<li>Newton\u2019s second law of motion says that the net external force on an object with a certain mass is directly proportional to and in the same direction as the acceleration of the object.<\/li>\n<li>Newton\u2019s second law can also describe net force as the instantaneous rate of change of momentum. Thus, a net external force causes nonzero acceleration.<\/li>\n<\/ul>\n<\/div>\n<div class=\"review-conceptual-questions\" id=\"fs-id1165039484961\">\n<h3>Conceptual Questions<\/h3>\n<div id=\"fs-id1165039484968\">\n<div id=\"fs-id1165039483720\">\n<p id=\"fs-id1165039483722\">Why can we neglect forces such as those holding a body together when we apply Newton\u2019s second law?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039479736\">\n<div id=\"fs-id1165039479738\">\n<p id=\"fs-id1165039479740\">A rock is thrown straight up. At the top of the trajectory, the velocity is momentarily zero. Does this imply that the force acting on the object is zero? Explain your answer.<\/p>\n<\/div>\n<div id=\"fs-id1165039351636\">\n<p id=\"fs-id1165039351638\">No. If the force were zero at this point, then there would be nothing to change the object\u2019s momentary zero velocity. Since we do not observe the object hanging motionless in the air, the force could not be zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"review-problems\" id=\"fs-id1165039475560\">\n<h3>Problems<\/h3>\n<div id=\"fs-id1165039477558\">\n<div id=\"fs-id1165039477560\">\n<p id=\"fs-id1165039477562\">Andrea, a 63.0-kg sprinter, starts a race with an acceleration of [latex]4.200\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex]. What is the net external force on her?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039507155\">\n<div id=\"fs-id1165039485048\">\n<p id=\"fs-id1165039485051\">If the sprinter from the previous problem accelerates at that rate for 20.00 m and then maintains that velocity for the remainder of a 100.00-m dash, what will her time be for the race?<\/p>\n<\/div>\n<div id=\"fs-id1165039485057\">\n<p id=\"fs-id1165039485059\">Running from rest, the sprinter attains a velocity of [latex]v=12.96\\phantom{\\rule{0.2em}{0ex}}\\text{m\/s}[\/latex], at end of acceleration. We find the time for acceleration using [latex]x=20.00\\phantom{\\rule{0.2em}{0ex}}\\text{m}=0+0.5a{t}_{1}{}^{2}[\/latex], or [latex]{t}_{1}=3.086\\phantom{\\rule{0.2em}{0ex}}\\text{s.}[\/latex] For maintained velocity, [latex]{x}_{2}=v{t}_{2}[\/latex], or [latex]{t}_{2}={x}_{2}\\text{\/}v=80.00\\phantom{\\rule{0.2em}{0ex}}\\text{m}\\text{\/}12.96\\phantom{\\rule{0.2em}{0ex}}\\text{m}\\text{\/}\\text{s}=6.173\\phantom{\\rule{0.2em}{0ex}}\\text{s}[\/latex]. [latex]\\text{Total time}=9.259\\phantom{\\rule{0.2em}{0ex}}\\text{s}[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039340333\">\n<div id=\"fs-id1165039340335\">\n<p id=\"fs-id1165039486251\">A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of his cart\u2019s acceleration.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039487882\">\n<div id=\"fs-id1165039487884\">\n<p id=\"fs-id1165039487886\">Astronauts in orbit are apparently weightless. This means that a clever method of measuring the mass of astronauts is needed to monitor their mass gains or losses, and adjust their diet. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted, and an astronaut\u2019s acceleration is measured to be [latex]0.893\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex]. (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which she orbits experiences an equal and opposite force. Use this knowledge to find an equation for the acceleration of the system (astronaut and spaceship) that would be measured by a nearby observer. (c) Discuss how this would affect the measurement of the astronaut\u2019s acceleration. Propose a method by which recoil of the vehicle is avoided.<\/p>\n<\/div>\n<div id=\"fs-id1165039234638\">\n<p id=\"fs-id1165039495414\">a. [latex]m=56.0\\phantom{\\rule{0.2em}{0ex}}\\text{kg}[\/latex]; b. [latex]{a}_{\\text{meas}}={a}_{\\text{astro}}+{a}_{\\text{ship}},\\phantom{\\rule{0.2em}{0ex}}\\text{where}\\phantom{\\rule{0.2em}{0ex}}{a}_{\\text{ship}}=\\frac{{m}_{\\text{astro}}{a}_{\\text{astro}}}{{m}_{\\text{ship}}}[\/latex]; c. If the force could be exerted on the astronaut by another source (other than the spaceship), then the spaceship would not experience a recoil.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039125323\">\n<div id=\"fs-id1165039125326\">\n<p id=\"fs-id1165039125328\">In <a href=\"#CNX_UPhysics_05_03_Mower\" class=\"autogenerated-content\">(Figure)<\/a>, the net external force on the 24-kg mower is given as 51 N. If the force of friction opposing the motion is 24 N, what force <em>F<\/em> (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m\/s when the force <em>F<\/em> is removed. How far will the mower go before stopping?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039485302\">\n<div id=\"fs-id1165039485304\">\n<p id=\"fs-id1165039485307\">The rocket sled shown below decelerates at a rate of [latex]196\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex]. What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is [latex]2.10\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{3}[\/latex] kg.<\/p>\n<p><span id=\"fs-id1165039484359\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_RocketSld1.jpg\" alt=\"Figure shows a rocket sled pointing right. Frictional force f points left. Upward force N and downward force w are equal in magnitude.\"><\/span><\/p><\/div>\n<div id=\"fs-id1165039498612\">\n<p id=\"fs-id1165039498614\">[latex]{F}_{\\text{net}}=4.12\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{5}\\phantom{\\rule{0.2em}{0ex}}\\text{N}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039477539\">\n<div id=\"fs-id1165039477541\">\n<p id=\"fs-id1165039485070\">If the rocket sled shown in the previous problem starts with only one rocket burning, what is the magnitude of this acceleration? Assume that the mass of the system is [latex]2.10\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{3}[\/latex] kg, the thrust <em>T<\/em> is [latex]2.40\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{4}\\phantom{\\rule{0.2em}{0ex}}\\text{N},[\/latex] and the force of friction opposing the motion is 650.0 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039470570\">\n<div id=\"fs-id1165039470572\">\n<p id=\"fs-id1165039470574\">What is the deceleration of the rocket sled if it comes to rest in 1.10 s from a speed of 1000.0 km\/h? (Such deceleration caused one test subject to black out and have temporary blindness.)<\/p>\n<\/div>\n<div id=\"fs-id1165039470658\">\n<p id=\"fs-id1165039470661\">[latex]a={253\\phantom{\\rule{0.2em}{0ex}}\\text{m\/s}}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039485447\">\n<div id=\"fs-id1165039485449\">\n<p id=\"fs-id1165039485452\">Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second exerts a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (See the free-body diagram.) (b) Calculate the acceleration. (c) What would the acceleration be if friction were 15.0 N?<\/p>\n<p><span id=\"fs-id1165039485459\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_FBDwagon_img.jpg\" alt=\"Figure shows a free body diagram. Force Fr points right, force N points upwards, forces Fl and f point left and force w points downwards.\"><\/span><\/p><\/div>\n<\/div>\n<div id=\"fs-id1165039352949\">\n<div id=\"fs-id1165039352951\">\n<p id=\"fs-id1165039352954\">A powerful motorcycle can produce an acceleration of [latex]3.50\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex] while traveling at 90.0 km\/h. At that speed, the forces resisting motion, including friction and air resistance, total 400.0 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force that motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?<\/p>\n<\/div>\n<div id=\"fs-id1165039486236\">\n<p id=\"fs-id1165039486238\">[latex]{F}_{\\text{net}}=F-f=ma\u21d2F=1.26\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{3}\\phantom{\\rule{0.2em}{0ex}}\\text{N}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039324241\">\n<div id=\"fs-id1165039324243\">\n<p id=\"fs-id1165039447938\">A car with a mass of 1000.0 kg accelerates from 0 to 90.0 km\/h in 10.0 s. (a) What is its acceleration? (b) What is the net force on the car?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039466306\">\n<div id=\"fs-id1165039466308\">\n<p id=\"fs-id1165039466310\">The driver in the previous problem applies the brakes when the car is moving at 90.0 km\/h, and the car comes to rest after traveling 40.0 m. What is the net force on the car during its deceleration?<\/p>\n<\/div>\n<div id=\"fs-id1165039466316\">\n<p id=\"fs-id1165039466318\">[latex]\\begin{array}{}\\\\ \\\\ {v}^{2}={v}_{0}^{2}+2ax\u21d2a=\\text{\u2212}7.80\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}\\hfill \\\\ {F}_{\\text{net}}=-7.80\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{3}\\phantom{\\rule{0.2em}{0ex}}\\text{N}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039462140\">\n<div id=\"fs-id1165039462142\">\n<p id=\"fs-id1165039462144\">An 80.0-kg passenger in an SUV traveling at [latex]1.00\\phantom{\\rule{0.2em}{0ex}}\u00d7\\phantom{\\rule{0.2em}{0ex}}{10}^{3}[\/latex] km\/h is wearing a seat belt. The driver slams on the brakes and the SUV stops in 45.0 m. Find the force of the seat belt on the passenger.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039461707\">\n<div id=\"fs-id1165039461709\">\n<p id=\"fs-id1165039461711\">A particle of mass 2.0 kg is acted on by a single force [latex]{\\stackrel{\\to }{F}}_{1}=18\\stackrel{^}{i}\\phantom{\\rule{0.2em}{0ex}}\\text{N}.[\/latex] (a) What is the particle\u2019s acceleration? (b) If the particle starts at rest, how far does it travel in the first 5.0 s?<\/p>\n<\/div>\n<div id=\"fs-id1165039483687\">\n<p id=\"fs-id1165039483689\">a. [latex]{\\stackrel{\\to }{F}}_{\\text{net}}=m\\stackrel{\\to }{a}\u21d2\\stackrel{\\to }{a}=9.0\\stackrel{^}{i}\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex]; b. The acceleration has magnitude [latex]9.0\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex], so [latex]x=110\\phantom{\\rule{0.2em}{0ex}}\\text{m}[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039477078\">\n<div id=\"fs-id1165039477080\">\n<p id=\"fs-id1165039477082\">Suppose that the particle of the previous problem also experiences forces [latex]{\\stackrel{\\to }{F}}_{2}=-15\\stackrel{^}{i}\\phantom{\\rule{0.2em}{0ex}}\\text{N}[\/latex] and [latex]{\\stackrel{\\to }{F}}_{3}=6.0\\stackrel{^}{j}\\phantom{\\rule{0.2em}{0ex}}\\text{N}.[\/latex] What is its acceleration in this case?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039478494\">\n<div id=\"fs-id1165039478496\">\n<p id=\"fs-id1165039478498\">Find the acceleration of the body of mass 5.0 kg shown below.<\/p>\n<p><span id=\"fs-id1165039478502\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_BodyMass1_img.jpg\" alt=\"Figure shows a circle labeled m in the xy plane. Three arrows originate from it. One points right and is labeled 10 i newtons. Another points left and is labeled -2 i newtons. The third points downwards and is labeled \u2013 4 j newtons.\"><\/span><\/p><\/div>\n<div id=\"fs-id1165039478514\">\n<p id=\"fs-id1165039478516\">[latex]1.6\\stackrel{^}{i}-0.8\\stackrel{^}{j}\\phantom{\\rule{0.2em}{0ex}}{\\text{m\/s}}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039484519\">\n<div id=\"fs-id1165039484521\">\n<p id=\"fs-id1165039484523\">In the following figure, the horizontal surface on which this block slides is frictionless. If the two forces acting on it each have magnitude [latex]F=30.0\\phantom{\\rule{0.2em}{0ex}}\\text{N}[\/latex] and [latex]M=10.0\\phantom{\\rule{0.2em}{0ex}}\\text{kg}[\/latex], what is the magnitude of the resulting acceleration of the block?<\/p>\n<p><span id=\"fs-id1165039484557\"><img src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_Prob5-12_img.jpg\" alt=\"Figure shows a box labeled M resting on a surface. An arrow forming an angle of minus 30 degrees with the horizontal is labeled F and points towards the box. Another arrow labeled F points right.\"><\/span><\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165039477315\">\n<dt>Newton\u2019s second law of motion<\/dt>\n<dd id=\"fs-id1165039477320\">acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system and is inversely proportional to its mass<\/dd>\n<\/dl>\n<\/div>\n\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of the section, you will be able to:<\/p>\n<ul>\n<li>Distinguish between external and internal forces<\/li>\n<li>Describe Newton&#8217;s second law of motion<\/li>\n<li>Explain the dependence of acceleration on net force and mass<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165039275238\">Newton\u2019s second law is closely related to his first law. It mathematically gives the cause-and-effect relationship between force and changes in motion. Newton\u2019s second law is quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton\u2019s second law as a simple equation that gives the exact relationship of force, mass, and acceleration, we need to sharpen some ideas we mentioned earlier.<\/p>\n<div class=\"bc-section section\">\n<h3>Force and Acceleration<\/h3>\n<p id=\"fs-id1165039092833\">First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is acceleration. Newton\u2019s first law says that a net external force causes a change in motion; thus, we see that a <em>net external force causes nonzero acceleration<\/em>.<\/p>\n<p id=\"fs-id1165039083134\">We defined external force in <a href=\"\/contents\/db08e30b-406e-4711-935c-60e064cb2271\" class=\"target-chapter\">Forces<\/a> as force acting on an object or system that originates outside of the object or system. Let\u2019s consider this concept further. An intuitive notion of <em>external<\/em> is correct\u2014it is outside the system of interest. For example, in <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>(a), the system of interest is the car plus the person within it. The two forces exerted by the two students are external forces. In contrast, an internal force acts between elements of the system. Thus, the force the person in the car exerts to hang on to the steering wheel is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton\u2019s first law. (The internal forces cancel each other out, as explained in the next section.) Therefore, we must define the boundaries of the system before we can determine which forces are external. Sometimes, the system is obvious, whereas at other times, identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton\u2019s laws. This concept is revisited many times in the study of physics.<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_ExtForce\">\n<div class=\"bc-figcaption figcaption\">Different forces exerted on the same mass produce different accelerations. (a) Two students push a stalled car. All external forces acting on the car are shown. (b) The forces acting on the car are transferred to a coordinate plane (free-body diagram) for simpler analysis. (c) The tow truck can produce greater external force on the same mass, and thus greater acceleration.<\/div>\n<p><span id=\"fs-id1165039264684\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_ExtForce.jpg\" alt=\"Figure a shows two people pushing a car with forces F1 and F2 in the right direction. Acceleration a is also in the same direction. Frictional force f is shown near the tire in the opposite direction, left. Upward force N and downward force W are equal in magnitude and are shown near the ground. Figure b puts all the forces of figure a together and shows a net force F net. These forces are also shown in a free body diagram. Figure c shows the car being towed by a tow-truck. Here, the forces N, W and f are the same as those in figure a. F subscript tow truck has a greater magnitude than F1 or F2. Acceleration a prime has a greater magnitude than a. All forces of this system are also shown in a free body diagram.\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-id1165039092504\">From this example, you can see that different forces exerted on the same mass produce different accelerations. In <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>(a), the two students push a car with a driver in it. Arrows representing all external forces are shown. The system of interest is the car and its driver. The weight <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-254a0f1c3515cc885417dcc5141e4ce3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#119;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"13\" style=\"vertical-align: 1px;\" \/> of the system and the support of the ground <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-06b0fb936557001a9c266b886bb1f77a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"16\" style=\"vertical-align: -2px;\" \/> are also shown for completeness and are assumed to cancel (because there was no vertical motion and no imbalance of forces in the vertical direction to create a change in motion). The vector <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-20dfc41e92b51968ea798fce6efbbd79_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#102;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"27\" width=\"13\" style=\"vertical-align: -4px;\" \/> represents the friction acting on the car, and it acts to the left, opposing the motion of the car. (We discuss friction in more detail in the next chapter.) In <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>(b), all external forces acting on the system add together to produce the net force <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-70fb686aa36bbe22d46b72cde600bd91_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"38\" style=\"vertical-align: -3px;\" \/> The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from this dot. Because there are two forces acting to the right, the vectors are shown collinearly. Finally, in <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>(c), a larger net external force produces a larger acceleration <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-7c6edcd8ad4b7ea65116f0624587603d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#123;&#97;&#125;&#94;&#123;&#92;&#112;&#114;&#105;&#109;&#101;&#32;&#125;&#125;&#62;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"62\" style=\"vertical-align: -17px;\" \/> when the tow truck pulls the car.<\/p>\n<p>It seems reasonable that acceleration would be directly proportional to and in the same direction as the net external force acting on a system. This assumption has been verified experimentally and is illustrated in <a href=\"#CNX_UPhysics_05_03_ExtForce\" class=\"autogenerated-content\">(Figure)<\/a>. To obtain an equation for Newton\u2019s second law, we first write the relationship of acceleration <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-9760d5d19a2fd88dc5619fe79b0348f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"13\" style=\"vertical-align: 1px;\" \/> and net external force <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4f25a14466fdda757c31c8d609203ba4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"33\" style=\"vertical-align: -3px;\" \/> as the proportionality<\/p>\n<div id=\"fs-id1165039046494\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-89d6fb0ea04ed7529421c153086c958b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#92;&#112;&#114;&#111;&#112;&#116;&#111;&#32;&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"65\" style=\"vertical-align: -3px;\" \/><\/div>\n<p id=\"fs-id1165039109910\">where the symbol <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-118c402462b54d8bd7e1748108d75195_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#114;&#111;&#112;&#116;&#111;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"12\" style=\"vertical-align: 0px;\" \/> means \u201cproportional to.\u201d (Recall from <a href=\"\/contents\/db08e30b-406e-4711-935c-60e064cb2271\" class=\"target-chapter\">Forces<\/a> that the net external force is the vector sum of all external forces and is sometimes indicated as <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-0e6bd8335e012d571cdd51e4ac837403_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#117;&#109;&#32;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"46\" style=\"vertical-align: -5px;\" \/>) This proportionality shows what we have said in words\u2014acceleration is directly proportional to net external force. Once the system of interest is chosen, identify the external forces and ignore the internal ones. It is a tremendous simplification to disregard the numerous internal forces acting between objects within the system, such as muscular forces within the students\u2019 bodies, let alone the myriad forces between the atoms in the objects. Still, this simplification helps us solve some complex problems.<\/p>\n<p id=\"fs-id1165039188181\">It also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. As illustrated in <a href=\"#CNX_UPhysics_05_03_Basketball\" class=\"autogenerated-content\">(Figure)<\/a>, the same net external force applied to a basketball produces a much smaller acceleration when it is applied to an SUV. The proportionality is written as<\/p>\n<div id=\"fs-id1165039088901\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-0d5a0cce80ca5241152fdaccb625c0fe_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#92;&#112;&#114;&#111;&#112;&#116;&#111;&#32;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#109;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"53\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"fs-id1165039337544\">where <em>m<\/em> is the mass of the system and <em>a<\/em> is the magnitude of the acceleration. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is directly proportional to net external force.<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_Basketball\">\n<div class=\"bc-figcaption figcaption\">The same force exerted on systems of different masses produces different accelerations. (a) A basketball player pushes on a basketball to make a pass. (Ignore the effect of gravity on the ball.) (b) The same player exerts an identical force on a stalled SUV and produces far less acceleration. (c) The free-body diagrams are identical, permitting direct comparison of the two situations. A series of patterns for free-body diagrams will emerge as you do more problems and learn how to draw them in <a href=\"\/contents\/595efb44-2031-41fd-b3a2-f08f0e2ab46a\" class=\"target-chapter\">Drawing Free-Body Diagrams<\/a>.<\/div>\n<p><span id=\"fs-id1165039052516\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_Basketball.jpg\" alt=\"Figure a shows a person exerting force F on a basketball with mass m1. The ball is shown to move to the rigth with an acceleration a1. Figure b shows the person exerting the same amount of force, F on an SUV with mass m2. The acceleration is a2, which is much smaller than a1. Figure c shows the free body diagrams of both systems shown in figure a and figure b. Both show the force F having the same magnitude and direction. The label reads: the free-body diagrams of both objects are the same.\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-id1165039347931\">It has been found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields <span>Newton\u2019s second law<\/span>.<\/p>\n<div id=\"fs-id1165039096883\">\n<div>Newton\u2019s Second Law of Motion<\/div>\n<p id=\"fs-id1165038999240\">The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system and is inversely proportion to its mass. In equation form, Newton\u2019s second law is<\/p>\n<div id=\"fs-id1165039299982\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-ccd7b48da00e818baad966aaf6662359_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#125;&#123;&#109;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"29\" width=\"70\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"fs-id1165039257622\">where <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-9760d5d19a2fd88dc5619fe79b0348f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"13\" style=\"vertical-align: 1px;\" \/> is the acceleration, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4f25a14466fdda757c31c8d609203ba4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"33\" style=\"vertical-align: -3px;\" \/> is the net force, and <em>m<\/em> is the mass. This is often written in the more familiar form<\/p>\n<div id=\"fs-id1165039079819\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-c769c2040458e09bac3cd85775f06a94_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#92;&#115;&#117;&#109;&#32;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#61;&#109;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"152\" style=\"vertical-align: -5px;\" \/><\/div>\n<p id=\"fs-id1165039083076\">but the first equation gives more insight into what Newton\u2019s second law means. When only the magnitude of force and acceleration are considered, this equation can be written in the simpler scalar form:<\/p>\n<div id=\"fs-id1165039009387\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-59eb7c24b65b6de807a6de51bb1b004a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#109;&#97;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"84\" style=\"vertical-align: -3px;\" \/><\/div>\n<\/div>\n<p id=\"fs-id1165039401832\">The law is a cause-and-effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is based on experimental verification. The free-body diagram, which you will learn to draw in <a href=\"\/contents\/595efb44-2031-41fd-b3a2-f08f0e2ab46a\" class=\"target-chapter\">Drawing Free-Body Diagrams<\/a>, is the basis for writing Newton\u2019s second law.<\/p>\n<div id=\"fs-id1165039247388\" class=\"textbox examples\">\n<p id=\"fs-id1165039088471\"><span>What Acceleration Can a Person Produce When Pushing a Lawn Mower?<\/span><br \/>\nSuppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb.) parallel to the ground (<a href=\"#CNX_UPhysics_05_03_Mower\" class=\"autogenerated-content\">(Figure)<\/a>). The mass of the mower is 24 kg. What is its acceleration?<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_Mower\">\n<div class=\"bc-figcaption figcaption\">(a) The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right? (b) The free-body diagram for this problem is shown.<\/div>\n<p><span id=\"fs-id1165039252210\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_Mower.jpg\" alt=\"Figure a shows a person using a lawn mower on a lawn. Force F net points right, from the person\u2019s hands. Figure b shows the force F net along the positive x axis.\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-id1165038978303\"><span>Strategy<\/span><br \/>\nThis problem involves only motion in the horizontal direction; we are also given the net force, indicated by the single vector, but we can suppress the vector nature and concentrate on applying Newton\u2019s second law. Since <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-60c38ca39cd2a768c7cd00182c2eaf03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"30\" style=\"vertical-align: -3px;\" \/> and <em>m<\/em> are given, the acceleration can be calculated directly from Newton\u2019s second law as <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-59eb7c24b65b6de807a6de51bb1b004a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#109;&#97;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"84\" style=\"vertical-align: -3px;\" \/><\/p>\n<p id=\"fs-id1165038974688\"><span>Solution<\/span><br \/>\nThe magnitude of the acceleration <em>a<\/em> is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4aa379f0335b23b384fd0887ce491d72_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#61;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#109;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"88\" style=\"vertical-align: -4px;\" \/>. Entering known values gives<\/p>\n<div id=\"fs-id1165039334536\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4f4b8cd5d2942df57cc7186e0efe45ed_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#53;&#49;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#125;&#123;&#50;&#52;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"72\" style=\"vertical-align: -9px;\" \/><\/div>\n<p id=\"fs-id1165039253539\">Substituting the unit of kilograms times meters per square second for newtons yields<\/p>\n<div id=\"fs-id1165039123852\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-9dea3144f510e6f7522517688133d4af_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#53;&#49;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;&middot;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#50;&#52;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;&#125;&#61;&#50;&#46;&#49;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"29\" width=\"191\" style=\"vertical-align: -9px;\" \/><\/div>\n<p id=\"fs-id1165038998097\"><span>Significance<\/span><br \/>\nThe direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. This is a result of the vector relationship expressed in Newton\u2019s second law, that is, the vector representing net force is the scalar multiple of the acceleration vector. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moved forward), and the vertical forces must cancel because no acceleration occurs in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long, because the person\u2019s top speed would soon be reached.<\/p>\n<\/div>\n<div id=\"fs-id1165039079950\" class=\"check-understanding\">\n<div id=\"fs-id1165039418564\">\n<div id=\"fs-id1165039010637\">\n<p id=\"fs-id1165039295994\"><strong>Check Your Understanding<\/strong> At the time of its launch, the HMS <em>Titanic<\/em> was the most massive mobile object ever built, with a mass of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-fef7e6f437e6dc88a5cf21bb665db4c8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#54;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#55;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"77\" style=\"vertical-align: -3px;\" \/>. If a force of 6 MN <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-32d7ab9a81b86ba28a2fef2576dbab91_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#54;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#54;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"70\" style=\"vertical-align: -7px;\" \/> was applied to the ship, what acceleration would it experience?<\/p>\n<\/div>\n<div id=\"fs-id1165039093866\">\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-f4e05fe2a498844353e54e49e48a7b12_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#48;&#46;&#49;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"64\" style=\"vertical-align: -4px;\" \/>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165039028823\">In the preceding example, we dealt with net force only for simplicity. However, several forces act on the lawn mower. The weight <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-254a0f1c3515cc885417dcc5141e4ce3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#119;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"13\" style=\"vertical-align: 1px;\" \/> (discussed in detail in <a href=\"\/contents\/68003722-b144-499e-859f-615de26fd893\" class=\"target-chapter\">Mass and Weight<\/a>) pulls down on the mower, toward the center of Earth; this produces a contact force on the ground. The ground must exert an upward force on the lawn mower, known as the normal force <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-06b0fb936557001a9c266b886bb1f77a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"16\" style=\"vertical-align: -2px;\" \/>, which we define in <a href=\"\/contents\/7517c753-4d3a-4137-8cb8-4d9ad48844aa\" class=\"target-chapter\">Common Forces<\/a>. These forces are balanced and therefore do not produce vertical acceleration. In the next example, we show both of these forces. As you continue to solve problems using Newton\u2019s second law, be sure to show multiple forces.<\/p>\n<div id=\"fs-id1165038975256\" class=\"textbox examples\">\n<p id=\"fs-id1165039273309\"><span>Which Force Is Bigger?<\/span><br \/>\n(a) The car shown in <a href=\"#CNX_UPhysics_05_03_RedCar\" class=\"autogenerated-content\">(Figure)<\/a> is moving at a constant speed. Which force is bigger, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-c53e0d2033a9609ecc894af72f17c8ad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#101;&#110;&#103;&#105;&#110;&#101;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"27\" width=\"53\" style=\"vertical-align: -6px;\" \/> or <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-c8897e06069c11c08680fb3d8d089b38_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#102;&#114;&#105;&#99;&#116;&#105;&#111;&#110;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"57\" style=\"vertical-align: -4px;\" \/>? Explain.<\/p>\n<p>(b) The same car is now accelerating to the right. Which force is bigger, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-c53e0d2033a9609ecc894af72f17c8ad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#101;&#110;&#103;&#105;&#110;&#101;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"27\" width=\"53\" style=\"vertical-align: -6px;\" \/> or <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-8da4ee9cf83ca7de25edbe0b4e508394_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#102;&#114;&#105;&#99;&#116;&#105;&#111;&#110;&#125;&#125;&#63;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"66\" style=\"vertical-align: -4px;\" \/> Explain.<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_RedCar\">\n<div class=\"bc-figcaption figcaption\">A car is shown (a) moving at constant speed and (b) accelerating. How do the forces acting on the car compare in each case? (a) What does the knowledge that the car is moving at constant velocity tell us about the net horizontal force on the car compared to the friction force? (b) What does the knowledge that the car is accelerating tell us about the horizontal force on the car compared to the friction force?<\/div>\n<p><span id=\"fs-id1165039085891\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_RedCar_img.jpg\" alt=\"Figure a shows a car with velocity 10 meters per second, moving right. F subscript engine right and F subscript friction points left. Figure b shows the car moving with an acceleration of 10 meters per second squared, towards the right. Forces F subscript engine and F subscript friction are the same as those in figure a.\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-id1165039333800\"><span>Strategy<\/span><br \/>\nWe must consider Newton\u2019s first and second laws to analyze the situation. We need to decide which law applies; this, in turn, will tell us about the relationship between the forces.<\/p>\n<p id=\"fs-id1165039286658\"><span>Solution<\/span><\/p>\n<ol id=\"fs-id1165039315375\" type=\"a\">\n<li>The forces are equal. According to Newton\u2019s first law, if the net force is zero, the velocity is constant.<\/li>\n<li>In this case, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-c53e0d2033a9609ecc894af72f17c8ad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#101;&#110;&#103;&#105;&#110;&#101;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"27\" width=\"53\" style=\"vertical-align: -6px;\" \/> must be larger than <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-cf951922b3cb9cf92784cce0c809a15a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#102;&#114;&#105;&#99;&#116;&#105;&#111;&#110;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"62\" style=\"vertical-align: -4px;\" \/> According to Newton\u2019s second law, a net force is required to cause acceleration.<\/li>\n<\/ol>\n<p><span>Significance<\/span><br \/>\nThese questions may seem trivial, but they are commonly answered incorrectly. For a car or any other object to move, it must be accelerated from rest to the desired speed; this requires that the engine force be greater than the friction force. Once the car is moving at constant velocity, the net force must be zero; otherwise, the car will accelerate (gain speed). To solve problems involving Newton\u2019s laws, we must understand whether to apply Newton\u2019s first law (where <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-49f9aef620c331b7aa02136c243a56ae_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#117;&#109;&#32;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#61;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#48;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"64\" style=\"vertical-align: -5px;\" \/>) or Newton\u2019s second law (where <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-ba8b7f9083070d29a37bbed47c7dcc5b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#117;&#109;&#32;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"38\" style=\"vertical-align: -5px;\" \/> is not zero). This will be apparent as you see more examples and attempt to solve problems on your own.<\/p>\n<\/div>\n<div id=\"fs-id1165039286668\" class=\"textbox examples\">\n<p id=\"fs-id1165038973771\"><span>What Rocket Thrust Accelerates This Sled?<\/span><br \/>\nBefore manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets.<\/p>\n<p id=\"fs-id1165039110006\">Calculate the magnitude of force exerted by each rocket, called its thrust <em>T<\/em>, for the four-rocket propulsion system shown in <a href=\"#CNX_UPhysics_05_03_RocketSled\" class=\"autogenerated-content\">(Figure)<\/a>. The sled\u2019s initial acceleration is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-5546570cca88b96dc63d1a4fc21bbdfa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#57;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"59\" style=\"vertical-align: -4px;\" \/>, the mass of the system is 2100 kg, and the force of friction opposing the motion is 650 N.<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_RocketSled\">\n<div class=\"bc-figcaption figcaption\">A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust <em>T<\/em>. The system here is the sled, its rockets, and its rider, so none of the forces between these objects are considered. The arrow representing friction <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-935a9527753dae4de52921fa81a8bccb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#102;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"34\" style=\"vertical-align: -17px;\" \/> is drawn larger than scale.<\/div>\n<p><span id=\"fs-id1165039065298\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_RocketSled.jpg\" alt=\"Figure shows a sled going right. It has four rockets at the back, with each thrust vector having the same magnitude and pointing right. Friction f points left. The upward normal force N and downward weight, are both equal in magnitude. Acceleration a is towards the right. All these forces are also shown in a free body diagram.\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-id1165039065549\"><span>Strategy<\/span><br \/>\nAlthough forces are acting both vertically and horizontally, we assume the vertical forces cancel because there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in <a href=\"#CNX_UPhysics_05_03_RocketSled\" class=\"autogenerated-content\">(Figure)<\/a>.<\/p>\n<p id=\"fs-id1165039303979\"><span>Solution<\/span><br \/>\nSince acceleration, mass, and the force of friction are given, we start with Newton\u2019s second law and look for ways to find the thrust of the engines. We have defined the direction of the force and acceleration as acting \u201cto the right,\u201d so we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with<\/p>\n<div id=\"fs-id1165039095877\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-436d46f92240967ca859819ece16adeb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#109;&#97;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"80\" style=\"vertical-align: -3px;\" \/><\/div>\n<p id=\"fs-id1165039123524\">where <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-60c38ca39cd2a768c7cd00182c2eaf03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"30\" style=\"vertical-align: -3px;\" \/> is the net force along the horizontal direction. We can see from the figure that the engine thrusts add, whereas friction opposes the thrust. In equation form, the net external force is<\/p>\n<div id=\"fs-id1165039115293\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-d47aedf0822495475c2fc7e8db42616e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#52;&#84;&#45;&#102;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"112\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"fs-id1165039300255\">Substituting this into Newton\u2019s second law gives us<\/p>\n<div id=\"fs-id1165039084866\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-f6bb4c79b2e96eb1126c3309ba9e65e9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#109;&#97;&#61;&#52;&#84;&#45;&#102;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"160\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"fs-id1165038968653\">Using a little algebra, we solve for the total thrust 4<em>T<\/em>:<\/p>\n<div class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-032b2b81084de48406ad85e0e88c2dae_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#84;&#61;&#109;&#97;&#43;&#102;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"106\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"fs-id1165039109216\">Substituting known values yields<\/p>\n<div id=\"fs-id1165039098409\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4642a4859b622c76c214e97928d7e51b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#84;&#61;&#109;&#97;&#43;&#102;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#50;&#49;&#48;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#57;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#43;&#54;&#53;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"33\" width=\"350\" style=\"vertical-align: -12px;\" \/><\/div>\n<p id=\"fs-id1165039122508\">Therefore, the total thrust is<\/p>\n<div id=\"fs-id1165039044869\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-1bb32d7258ca8850e3c4970d987696e2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#84;&#61;&#49;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#53;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"121\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"fs-id1165039110278\">and the individual thrusts are<\/p>\n<div id=\"fs-id1165039053608\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-6555ad1678973b28ea14e18ea5cdb6a8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#84;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#53;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#125;&#123;&#52;&#125;&#61;&#50;&#46;&#53;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#52;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"199\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"fs-id1165038998398\"><span>Significance<\/span><br \/>\nThe numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance, and the setup was designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km\/h were obtained, with accelerations of 45 <em>g<\/em>\u2019s. (Recall that <em>g<\/em>, acceleration due to gravity, is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-9532d26a5648e0ee09028f1614b6b3a6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#57;&#46;&#56;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"73\" style=\"vertical-align: -4px;\" \/>. When we say that acceleration is 45 <em>g<\/em>\u2019s, it is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-f3478fd060c6b7e22e26bcd83a9bb2cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#53;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#57;&#46;&#56;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"93\" style=\"vertical-align: -4px;\" \/> which is approximately <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-962a71dce3c226261bd4dec02912eb62_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#52;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"68\" style=\"vertical-align: -4px;\" \/>.) Although living subjects are not used anymore, land speeds of 10,000 km\/h have been obtained with a rocket sled.<\/p>\n<p id=\"fs-id1165039114554\">In this example, as in the preceding one, the system of interest is obvious. We see in later examples that choosing the system of interest is crucial\u2014and the choice is not always obvious.<\/p>\n<p id=\"fs-id1165039247454\">Newton\u2019s second law is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature.<\/p>\n<\/div>\n<div id=\"fs-id1165039048070\" class=\"check-understanding\">\n<div id=\"fs-id1165039069953\">\n<div id=\"fs-id1165039235667\">\n<p id=\"fs-id1165038988394\"><strong>Check Your Understanding<\/strong> A 550-kg sports car collides with a 2200-kg truck, and during the collision, the net force on each vehicle is the force exerted by the other. If the magnitude of the truck\u2019s acceleration is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-bb1d2861e543e666d8f15185d675ea64_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#49;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"63\" style=\"vertical-align: -4px;\" \/> what is the magnitude of the sports car\u2019s acceleration?<\/p>\n<\/div>\n<div id=\"fs-id1165039454293\">\n<p id=\"fs-id1165039273067\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-7446dec949758302d2b1db81f73aebeb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"59\" style=\"vertical-align: -4px;\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\" id=\"fs-id1165039187319\">\n<h3>Component Form of Newton\u2019s Second Law<\/h3>\n<p id=\"fs-id1165039340543\">We have developed Newton\u2019s second law and presented it as a vector equation in <a href=\"#fs-id1165039079819\" class=\"autogenerated-content\">(Figure)<\/a>. This vector equation can be written as three component equations:<\/p>\n<div id=\"fs-id1165039122533\" class=\"equation-callout\">\n<div id=\"fs-id1165039419225\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-02eaed4f04530147ceb4aed94f6d130e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#117;&#109;&#32;&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#120;&#125;&#61;&#109;&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#125;&#95;&#123;&#120;&#125;&#44;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#115;&#117;&#109;&#32;&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#121;&#125;&#61;&#109;&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#125;&#95;&#123;&#121;&#125;&#44;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#97;&#110;&#100;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#115;&#117;&#109;&#32;&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#122;&#125;&#61;&#109;&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#125;&#95;&#123;&#122;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"27\" width=\"379\" style=\"vertical-align: -6px;\" \/><\/div>\n<\/div>\n<p id=\"fs-id1165038966002\">The second law is a description of how a body responds mechanically to its environment. The influence of the environment is the net force <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-f7fada88279f8dfabebdace0a8470919_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"38\" style=\"vertical-align: -4px;\" \/> the body\u2019s response is the acceleration <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-79654a2961cc3cfd1420455790bb0d78_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"18\" style=\"vertical-align: -4px;\" \/> and the strength of the response is inversely proportional to the mass <em>m<\/em>. The larger the mass of an object, the smaller its response (its acceleration) to the influence of the environment (a given net force). Therefore, a body\u2019s mass is a measure of its inertia, as we explained in <a href=\"\/contents\/e3652cf1-835a-4a39-a951-80c909f22251\" class=\"target-chapter\">Newton\u2019s First Law<\/a>.<\/p>\n<div id=\"fs-id1165039045618\" class=\"textbox examples\">\n<p id=\"fs-id1165039225277\"><span>Force on a Soccer Ball<\/span><br \/>\nA 0.400-kg soccer ball is kicked across the field by a player; it undergoes acceleration given by <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-824ef3144cddc98851440368178f6b38_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#61;&#51;&#46;&#48;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#43;&#55;&#46;&#48;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"185\" style=\"vertical-align: -4px;\" \/> Find (a) the resultant force acting on the ball and (b) the magnitude and direction of the resultant force.<\/p>\n<p id=\"fs-id1165039123808\"><span>Strategy<\/span><br \/>\nThe vectors in <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-2641b7d53b94e986e5ee497e5e6670f6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: -1px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-28952002f9718cea3899bdac634978cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"9\" style=\"vertical-align: -4px;\" \/> format, which indicate force direction along the <em>x<\/em>-axis and the <em>y<\/em>-axis, respectively, are involved, so we apply Newton\u2019s second law in vector form.<\/p>\n<p id=\"fs-id1165039269338\"><span>Solution<\/span><\/p>\n<ol id=\"fs-id1165039107084\" type=\"a\">\n<li>We apply Newton\u2019s second law:\n<div><\/div>\n<div id=\"fs-id1165038962202\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-d206d0b40a64e05870075c360aa1c42d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#109;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#48;&#46;&#52;&#48;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#51;&#46;&#48;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#43;&#55;&#46;&#48;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#61;&#49;&#46;&#50;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#43;&#50;&#46;&#56;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"33\" width=\"511\" style=\"vertical-align: -12px;\" \/><\/div>\n<\/li>\n<li>Magnitude and direction are found using the components of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4f25a14466fdda757c31c8d609203ba4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"33\" style=\"vertical-align: -3px;\" \/>:\n<div><\/div>\n<div id=\"fs-id1165039293018\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-fd6454cad2f53226c14bbe31f62d963b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#92;&#115;&#113;&#114;&#116;&#123;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#49;&#46;&#50;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#43;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#50;&#46;&#56;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#125;&#61;&#51;&#46;&#48;&#53;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#97;&#110;&#100;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#104;&#101;&#116;&#97;&#32;&#61;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#97;&#110;&#125;&#125;&#94;&#123;&#45;&#49;&#125;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#46;&#56;&#48;&#125;&#123;&#49;&#46;&#50;&#48;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#61;&#54;&#54;&#46;&#56;&#92;&#116;&#101;&#120;&#116;&#123;&deg;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"33\" width=\"525\" style=\"vertical-align: -10px;\" \/><\/div>\n<\/li>\n<\/ol>\n<p id=\"fs-id1165039464375\"><span>Significance<\/span><br \/>\nWe must remember that Newton\u2019s second law is a vector equation. In (a), we are multiplying a vector by a scalar to determine the net force in vector form. While the vector form gives a compact representation of the force vector, it does not tell us how \u201cbig\u201d it is, or where it goes, in intuitive terms. In (b), we are determining the actual size (magnitude) of this force and the direction in which it travels.<\/p>\n<\/div>\n<div id=\"fs-id1165039443202\" class=\"textbox examples\">\n<p id=\"fs-id1165039475586\"><span>Mass of a Car<\/span><br \/>\nFind the mass of a car if a net force of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-366ae73edb35b902a0170d72630feda8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#45;&#54;&#48;&#48;&#46;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"86\" style=\"vertical-align: -4px;\" \/> produces an acceleration of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-116f4adce92a27810cb4e10b1fee0b66_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#45;&#48;&#46;&#50;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"93\" style=\"vertical-align: -4px;\" \/>.<\/p>\n<p id=\"fs-id1165039311437\"><span>Strategy<\/span><br \/>\nVector division is not defined, so <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4d9e9ffd8a8a072f050eb9769091f50f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;&#61;&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"100\" style=\"vertical-align: -4px;\" \/> cannot be performed. However, mass <em>m<\/em> is a scalar, so we can use the scalar form of Newton\u2019s second law, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-9159f841b088337abbfe96e66cb9a51b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;&#61;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#97;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"88\" style=\"vertical-align: -4px;\" \/>.<\/p>\n<p id=\"fs-id1165039295458\"><span>Solution<\/span><br \/>\nWe use <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-9159f841b088337abbfe96e66cb9a51b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;&#61;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#97;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"88\" style=\"vertical-align: -4px;\" \/> and substitute the magnitudes of the two vectors: <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-c2bb298f836254fe57aca4be60c6a8a2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#54;&#48;&#48;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"112\" style=\"vertical-align: -3px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-b0bbc47656eb027f958183248d5456c5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#61;&#48;&#46;&#50;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"102\" style=\"vertical-align: -4px;\" \/> Therefore,<\/p>\n<div id=\"fs-id1165039448064\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-508e31cbb19f8eb5acacd52db67f4027_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#125;&#123;&#97;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#54;&#48;&#48;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#125;&#123;&#48;&#46;&#50;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#61;&#51;&#48;&#48;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"28\" width=\"233\" style=\"vertical-align: -12px;\" \/><\/div>\n<p id=\"fs-id1165039345542\"><span>Significance<\/span><br \/>\nForce and acceleration were given in the <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-2641b7d53b94e986e5ee497e5e6670f6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: -1px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-28952002f9718cea3899bdac634978cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"9\" style=\"vertical-align: -4px;\" \/> format, but the answer, mass <em>m<\/em>, is a scalar and thus is not given in <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-2641b7d53b94e986e5ee497e5e6670f6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: -1px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-28952002f9718cea3899bdac634978cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"9\" style=\"vertical-align: -4px;\" \/> form.<\/p>\n<\/div>\n<div id=\"fs-id1165039464584\" class=\"textbox examples\">\n<p id=\"fs-id1165039464586\"><span>Several Forces on a Particle<\/span><br \/>\nA particle of mass <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-e24172b5e51995676a3238aa02973b04_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;&#61;&#52;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"83\" style=\"vertical-align: -3px;\" \/> is acted upon by four forces of magnitudes. <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-40381c8c5cc5ffec08b356974b392dbc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#49;&#125;&#61;&#49;&#48;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#44;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#70;&#125;&#95;&#123;&#50;&#125;&#61;&#52;&#48;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#44;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#70;&#125;&#95;&#123;&#51;&#125;&#61;&#53;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#44;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#97;&#110;&#100;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#70;&#125;&#95;&#123;&#52;&#125;&#61;&#50;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"413\" style=\"vertical-align: -4px;\" \/>, with the directions as shown in the free-body diagram in <a href=\"#CNX_UPhysics_05_03_FourForces\" class=\"autogenerated-content\">(Figure)<\/a>. What is the acceleration of the particle?<\/p>\n<div class=\"bc-figure figure\" id=\"CNX_UPhysics_05_03_FourForces\">\n<div class=\"bc-figcaption figcaption\">Four forces in the <em>xy<\/em>-plane are applied to a 4.0-kg particle.<\/div>\n<p><span id=\"fs-id1165039374163\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_FourForces.jpg\" alt=\"A particle is shown in the xy plane. Force F1 is at an angle of 30 degrees with the positive x axis, force F2 is in the downward direction, force F3 points left and force F4 points upwards.\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-id1165039296260\"><span>Strategy<\/span><br \/>\nBecause this is a two-dimensional problem, we must use a free-body diagram. First, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-89449e86976ad96bece37257ae45b82a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#49;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"20\" style=\"vertical-align: -4px;\" \/> must be resolved into <em>x<\/em>&#8211; and <em>y<\/em>-components. We can then apply the second law in each direction.<\/p>\n<p id=\"fs-id1165039464479\"><span>Solution<\/span><br \/>\nWe draw a free-body diagram as shown in <a href=\"#CNX_UPhysics_05_03_FourForces\" class=\"autogenerated-content\">(Figure)<\/a>. Now we apply Newton\u2019s second law. We consider all vectors resolved into <em>x<\/em>&#8211; and <em>y<\/em>-components:<\/p>\n<div id=\"fs-id1165039340892\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-d7725a80bbeccfd7d37a205ab0f1a750_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#125;&#92;&#115;&#117;&#109;&#32;&#123;&#70;&#125;&#95;&#123;&#120;&#125;&#61;&#109;&#123;&#97;&#125;&#95;&#123;&#120;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#115;&#117;&#109;&#32;&#123;&#70;&#125;&#95;&#123;&#121;&#125;&#61;&#109;&#123;&#97;&#125;&#95;&#123;&#121;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#123;&#70;&#125;&#95;&#123;&#49;&#120;&#125;&#45;&#123;&#70;&#125;&#95;&#123;&#51;&#120;&#125;&#61;&#109;&#123;&#97;&#125;&#95;&#123;&#120;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#123;&#70;&#125;&#95;&#123;&#49;&#121;&#125;&#43;&#123;&#70;&#125;&#95;&#123;&#52;&#121;&#125;&#45;&#123;&#70;&#125;&#95;&#123;&#50;&#121;&#125;&#61;&#109;&#123;&#97;&#125;&#95;&#123;&#121;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#123;&#70;&#125;&#95;&#123;&#49;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#111;&#115;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#51;&#48;&#92;&#116;&#101;&#120;&#116;&#123;&deg;&#125;&#45;&#123;&#70;&#125;&#95;&#123;&#51;&#120;&#125;&#61;&#109;&#123;&#97;&#125;&#95;&#123;&#120;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#123;&#70;&#125;&#95;&#123;&#49;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#105;&#110;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#51;&#48;&#92;&#116;&#101;&#120;&#116;&#123;&deg;&#125;&#43;&#123;&#70;&#125;&#95;&#123;&#52;&#121;&#125;&#45;&#123;&#70;&#125;&#95;&#123;&#50;&#121;&#125;&#61;&#109;&#123;&#97;&#125;&#95;&#123;&#121;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#92;&#108;&#101;&#102;&#116;&#40;&#49;&#48;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#111;&#115;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#51;&#48;&#92;&#116;&#101;&#120;&#116;&#123;&deg;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#45;&#53;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#123;&#97;&#125;&#95;&#123;&#120;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#108;&#101;&#102;&#116;&#40;&#49;&#48;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#105;&#110;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#51;&#48;&#92;&#116;&#101;&#120;&#116;&#123;&deg;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#43;&#50;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#45;&#52;&#48;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#123;&#97;&#125;&#95;&#123;&#121;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#123;&#97;&#125;&#95;&#123;&#120;&#125;&#61;&#48;&#46;&#57;&#50;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#46;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#123;&#97;&#125;&#95;&#123;&#121;&#125;&#61;&#45;&#56;&#46;&#51;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#46;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"690\" style=\"vertical-align: -50px;\" \/><\/div>\n<p id=\"fs-id1165039196016\">Thus, the net acceleration is<\/p>\n<div id=\"fs-id1165039443012\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4c52ae341f5c1694473bf4103c92be96_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#48;&#46;&#57;&#50;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#45;&#56;&#46;&#51;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"33\" width=\"198\" style=\"vertical-align: -12px;\" \/><\/div>\n<p id=\"fs-id1165039458298\">which is a vector of magnitude <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-79586ed9cc36ddb07fffe8db6218d549_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#56;&#46;&#52;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"64\" style=\"vertical-align: -4px;\" \/> directed at <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-454be5a84b3a2e01b64423506250ce05_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#55;&#54;&#92;&#116;&#101;&#120;&#116;&#123;&deg;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"27\" style=\"vertical-align: 0px;\" \/> to the positive <em>x<\/em>-axis.<\/p>\n<p id=\"fs-id1165039082704\"><span>Significance<\/span><br \/>\nNumerous examples in everyday life can be found that involve three or more forces acting on a single object, such as cables running from the Golden Gate Bridge or a football player being tackled by three defenders. We can see that the solution of this example is just an extension of what we have already done.<\/p>\n<\/div>\n<div id=\"fs-id1165039084354\" class=\"check-understanding\">\n<div id=\"fs-id1165039026512\">\n<div id=\"fs-id1165039026514\">\n<p id=\"fs-id1165039464613\"><strong>Check Your Understanding<\/strong> A car has forces acting on it, as shown below. The mass of the car is 1000.0 kg. The road is slick, so friction can be ignored. (a) What is the net force on the car? (b) What is the acceleration of the car?<\/p>\n<p><span id=\"fs-id1165039235873\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_RedCar3_img.jpg\" alt=\"The top view of a car is shown. Two force vectors originate from the car and point upwards and outwards. A force of 450 newtons makes an angle of 30 degrees with the straight line motion of the car, towards the right. Another force of 360 newtons makes an angle of 10 degrees with the straight line motion of the car, towards the left.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1165039463406\">\n<p id=\"fs-id1165039483330\">a. <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-7ced7f7d6545de33db95c6db7c00b681_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#49;&#53;&#57;&#46;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#43;&#55;&#55;&#48;&#46;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"140\" style=\"vertical-align: -4px;\" \/>; b. <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-d4039ed9a63a9873ad5a6a03457c1d30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#48;&#46;&#49;&#53;&#57;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#43;&#48;&#46;&#55;&#55;&#48;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"159\" style=\"vertical-align: -4px;\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\" id=\"fs-id1165039485321\">\n<h3>Newton\u2019s Second Law and Momentum<\/h3>\n<p id=\"fs-id1165039420955\">Newton actually stated his second law in terms of momentum: \u201cThe instantaneous rate at which a body\u2019s momentum changes is equal to the net force acting on the body.\u201d (\u201cInstantaneous rate\u201d implies that the derivative is involved.) This can be given by the vector equation<\/p>\n<div id=\"fs-id1165039420958\" class=\"equation-callout\">\n<div id=\"fs-id1165039247034\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-a91742600ede2380e572600b0b5b5045_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#100;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#112;&#125;&#125;&#123;&#100;&#116;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"28\" width=\"84\" style=\"vertical-align: -6px;\" \/><\/div>\n<\/div>\n<p id=\"fs-id1165039440108\">This means that Newton\u2019s second law addresses the central question of motion: What causes a change in motion of an object? Momentum was described by Newton as \u201cquantity of motion,\u201d a way of combining both the velocity of an object and its mass. We devote <a href=\"\/contents\/647dc0b6-01e5-4e1c-893e-4ada7841ba77\" class=\"target-chapter\">Linear Momentum and Collisions<\/a> to the study of <span class=\"no-emphasis\">momentum<\/span>.<\/p>\n<p id=\"fs-id1165039484763\">For now, it is sufficient to define <em>momentum<\/em> <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-75fc482dd6ca1b9336d7baf560e8063a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#112;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"13\" style=\"vertical-align: -2px;\" \/> as the product of the mass of the object <em>m<\/em> and its velocity <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-08c90ae1637f7806fc89c3ad30294a55_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#118;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"13\" style=\"vertical-align: 1px;\" \/>:<\/p>\n<div id=\"fs-id1165039081523\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4de60402cf7e5b8e9761b7830b368477_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#112;&#125;&#61;&#109;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#118;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"76\" style=\"vertical-align: -2px;\" \/><\/div>\n<p id=\"fs-id1165039341205\">Since velocity is a vector, so is momentum.<\/p>\n<p id=\"fs-id1165039341208\">It is easy to visualize momentum. A train moving at 10 m\/s has more momentum than one that moves at 2 m\/s. In everyday life, we speak of one sports team as \u201chaving momentum\u201d when they score points against the opposing team.<\/p>\n<p id=\"fs-id1165039485414\">If we substitute <a href=\"#fs-id1165039081523\" class=\"autogenerated-content\">(Figure)<\/a> into <a href=\"#fs-id1165039247034\" class=\"autogenerated-content\">(Figure)<\/a>, we obtain<\/p>\n<div id=\"fs-id1165039412912\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-30d9937c84d7feda1c55d29278b5e7b3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#100;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#112;&#125;&#125;&#123;&#100;&#116;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#100;&#92;&#108;&#101;&#102;&#116;&#40;&#109;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#118;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#123;&#100;&#116;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"35\" width=\"159\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"fs-id1165039461971\">When <em>m<\/em> is constant, we have<\/p>\n<div id=\"fs-id1165039340576\" class=\"unnumbered\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-3827bb7f7431513e8ae27ac278e0d60f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#109;&#92;&#102;&#114;&#97;&#99;&#123;&#100;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#118;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#123;&#100;&#116;&#125;&#61;&#109;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"35\" width=\"179\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"fs-id1165039495422\">Thus, we see that the momentum form of Newton\u2019s second law reduces to the form given earlier in this section.<\/p>\n<div id=\"fs-id1165039495428\" class=\"media-2\">\n<p id=\"fs-id1165039209657\">Explore the <a href=\"https:\/\/openstaxcollege.org\/l\/21forcesatwork\">forces at work<\/a> when <a href=\"https:\/\/openstaxcollege.org\/l\/21pullacart\">pulling a cart<\/a> or pushing a refrigerator, crate, or person. Create an <a href=\"https:\/\/openstaxcollege.org\/l\/21forcemotion\">applied force<\/a> and see how it makes objects move. Put <a href=\"https:\/\/openstaxcollege.org\/l\/21ramp\">an object on a ramp<\/a> and see how it affects its motion.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\" id=\"fs-id1165039440646\">\n<h3>Summary<\/h3>\n<ul id=\"fs-id1165039440654\">\n<li>An external force acts on a system from outside the system, as opposed to internal forces, which act between components within the system.<\/li>\n<li>Newton\u2019s second law of motion says that the net external force on an object with a certain mass is directly proportional to and in the same direction as the acceleration of the object.<\/li>\n<li>Newton\u2019s second law can also describe net force as the instantaneous rate of change of momentum. Thus, a net external force causes nonzero acceleration.<\/li>\n<\/ul>\n<\/div>\n<div class=\"review-conceptual-questions\" id=\"fs-id1165039484961\">\n<h3>Conceptual Questions<\/h3>\n<div id=\"fs-id1165039484968\">\n<div id=\"fs-id1165039483720\">\n<p id=\"fs-id1165039483722\">Why can we neglect forces such as those holding a body together when we apply Newton\u2019s second law?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039479736\">\n<div id=\"fs-id1165039479738\">\n<p id=\"fs-id1165039479740\">A rock is thrown straight up. At the top of the trajectory, the velocity is momentarily zero. Does this imply that the force acting on the object is zero? Explain your answer.<\/p>\n<\/div>\n<div id=\"fs-id1165039351636\">\n<p id=\"fs-id1165039351638\">No. If the force were zero at this point, then there would be nothing to change the object\u2019s momentary zero velocity. Since we do not observe the object hanging motionless in the air, the force could not be zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"review-problems\" id=\"fs-id1165039475560\">\n<h3>Problems<\/h3>\n<div id=\"fs-id1165039477558\">\n<div id=\"fs-id1165039477560\">\n<p id=\"fs-id1165039477562\">Andrea, a 63.0-kg sprinter, starts a race with an acceleration of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-ff2e00380a692bb13eb1d92d32bab7c9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#46;&#50;&#48;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"82\" style=\"vertical-align: -4px;\" \/>. What is the net external force on her?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039507155\">\n<div id=\"fs-id1165039485048\">\n<p id=\"fs-id1165039485051\">If the sprinter from the previous problem accelerates at that rate for 20.00 m and then maintains that velocity for the remainder of a 100.00-m dash, what will her time be for the race?<\/p>\n<\/div>\n<div id=\"fs-id1165039485057\">\n<p id=\"fs-id1165039485059\">Running from rest, the sprinter attains a velocity of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-4f1c3fbeadff0dae4fa6695bd1efb66c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;&#61;&#49;&#50;&#46;&#57;&#54;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"108\" style=\"vertical-align: -4px;\" \/>, at end of acceleration. We find the time for acceleration using <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-a5216116a17e03ef88482a2f2a161e3c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#120;&#61;&#50;&#48;&#46;&#48;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#61;&#48;&#43;&#48;&#46;&#53;&#97;&#123;&#116;&#125;&#95;&#123;&#49;&#125;&#123;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"200\" style=\"vertical-align: -4px;\" \/>, or <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-69757bd5a983174f689cff73f37fc84c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#116;&#125;&#95;&#123;&#49;&#125;&#61;&#51;&#46;&#48;&#56;&#54;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"93\" style=\"vertical-align: -4px;\" \/> For maintained velocity, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-6ac480e2926e24d1491171b0971bd644_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#120;&#125;&#95;&#123;&#50;&#125;&#61;&#118;&#123;&#116;&#125;&#95;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"64\" style=\"vertical-align: -3px;\" \/>, or <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-555c6194d06bd42f0c1969c81113e7dd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#116;&#125;&#95;&#123;&#50;&#125;&#61;&#123;&#120;&#125;&#95;&#123;&#50;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#118;&#61;&#56;&#48;&#46;&#48;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#49;&#50;&#46;&#57;&#54;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#61;&#54;&#46;&#49;&#55;&#51;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"314\" style=\"vertical-align: -4px;\" \/>. <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-84b7fd1342d70c49a9bc2d5ab1c82853_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#84;&#111;&#116;&#97;&#108;&#32;&#116;&#105;&#109;&#101;&#125;&#61;&#57;&#46;&#50;&#53;&#57;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"156\" style=\"vertical-align: -1px;\" \/>.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039340333\">\n<div id=\"fs-id1165039340335\">\n<p id=\"fs-id1165039486251\">A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of his cart\u2019s acceleration.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039487882\">\n<div id=\"fs-id1165039487884\">\n<p id=\"fs-id1165039487886\">Astronauts in orbit are apparently weightless. This means that a clever method of measuring the mass of astronauts is needed to monitor their mass gains or losses, and adjust their diet. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted, and an astronaut\u2019s acceleration is measured to be <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-ce834d3c9051aba3f2369880236f14cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#48;&#46;&#56;&#57;&#51;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"82\" style=\"vertical-align: -4px;\" \/>. (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which she orbits experiences an equal and opposite force. Use this knowledge to find an equation for the acceleration of the system (astronaut and spaceship) that would be measured by a nearby observer. (c) Discuss how this would affect the measurement of the astronaut\u2019s acceleration. Propose a method by which recoil of the vehicle is avoided.<\/p>\n<\/div>\n<div id=\"fs-id1165039234638\">\n<p id=\"fs-id1165039495414\">a. <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-8fe6431b7c1b89b25b7a537b88a875b6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;&#61;&#53;&#54;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"92\" style=\"vertical-align: -3px;\" \/>; b. <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-0da18efd8fa240065bc3de8c784355e1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#97;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#101;&#97;&#115;&#125;&#125;&#61;&#123;&#97;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#97;&#115;&#116;&#114;&#111;&#125;&#125;&#43;&#123;&#97;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#104;&#105;&#112;&#125;&#125;&#44;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#119;&#104;&#101;&#114;&#101;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#97;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#104;&#105;&#112;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#109;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#97;&#115;&#116;&#114;&#111;&#125;&#125;&#123;&#97;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#97;&#115;&#116;&#114;&#111;&#125;&#125;&#125;&#123;&#123;&#109;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#104;&#105;&#112;&#125;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"355\" style=\"vertical-align: -11px;\" \/>; c. If the force could be exerted on the astronaut by another source (other than the spaceship), then the spaceship would not experience a recoil.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039125323\">\n<div id=\"fs-id1165039125326\">\n<p id=\"fs-id1165039125328\">In <a href=\"#CNX_UPhysics_05_03_Mower\" class=\"autogenerated-content\">(Figure)<\/a>, the net external force on the 24-kg mower is given as 51 N. If the force of friction opposing the motion is 24 N, what force <em>F<\/em> (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m\/s when the force <em>F<\/em> is removed. How far will the mower go before stopping?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039485302\">\n<div id=\"fs-id1165039485304\">\n<p id=\"fs-id1165039485307\">The rocket sled shown below decelerates at a rate of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-54a26f34bac1a565ba7ff50cb15bfcc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#49;&#57;&#54;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"67\" style=\"vertical-align: -4px;\" \/>. What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-85d42f569c290e596630eb433aa0d8cf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#46;&#49;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#51;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"63\" style=\"vertical-align: -1px;\" \/> kg.<\/p>\n<p><span id=\"fs-id1165039484359\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_RocketSld1.jpg\" alt=\"Figure shows a rocket sled pointing right. Frictional force f points left. Upward force N and downward force w are equal in magnitude.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1165039498612\">\n<p id=\"fs-id1165039498614\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-f0a4afea9ceb61458194db59b6da25c4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#52;&#46;&#49;&#50;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#53;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"136\" style=\"vertical-align: -3px;\" \/><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039477539\">\n<div id=\"fs-id1165039477541\">\n<p id=\"fs-id1165039485070\">If the rocket sled shown in the previous problem starts with only one rocket burning, what is the magnitude of this acceleration? Assume that the mass of the system is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-85d42f569c290e596630eb433aa0d8cf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#46;&#49;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#51;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"63\" style=\"vertical-align: -1px;\" \/> kg, the thrust <em>T<\/em> is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-01f19b6f7b92ff31943619269046fcc6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#46;&#52;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#52;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"85\" style=\"vertical-align: -4px;\" \/> and the force of friction opposing the motion is 650.0 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039470570\">\n<div id=\"fs-id1165039470572\">\n<p id=\"fs-id1165039470574\">What is the deceleration of the rocket sled if it comes to rest in 1.10 s from a speed of 1000.0 km\/h? (Such deceleration caused one test subject to black out and have temporary blindness.)<\/p>\n<\/div>\n<div id=\"fs-id1165039470658\">\n<p id=\"fs-id1165039470661\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-a88857a72d9d14bb348112a4f7e78409_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#61;&#123;&#50;&#53;&#51;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"101\" style=\"vertical-align: -4px;\" \/><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039485447\">\n<div id=\"fs-id1165039485449\">\n<p id=\"fs-id1165039485452\">Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second exerts a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (See the free-body diagram.) (b) Calculate the acceleration. (c) What would the acceleration be if friction were 15.0 N?<\/p>\n<p><span id=\"fs-id1165039485459\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_FBDwagon_img.jpg\" alt=\"Figure shows a free body diagram. Force Fr points right, force N points upwards, forces Fl and f point left and force w points downwards.\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039352949\">\n<div id=\"fs-id1165039352951\">\n<p id=\"fs-id1165039352954\">A powerful motorcycle can produce an acceleration of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-674de2328b28da168450b1b657a55cd8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#51;&#46;&#53;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"73\" style=\"vertical-align: -4px;\" \/> while traveling at 90.0 km\/h. At that speed, the forces resisting motion, including friction and air resistance, total 400.0 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force that motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?<\/p>\n<\/div>\n<div id=\"fs-id1165039486236\">\n<p id=\"fs-id1165039486238\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-2d8c1f769fea614aee6a1fa131f6e752_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#70;&#45;&#102;&#61;&#109;&#97;&#8658;&#70;&#61;&#49;&#46;&#50;&#54;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#51;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"268\" style=\"vertical-align: -4px;\" \/><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039324241\">\n<div id=\"fs-id1165039324243\">\n<p id=\"fs-id1165039447938\">A car with a mass of 1000.0 kg accelerates from 0 to 90.0 km\/h in 10.0 s. (a) What is its acceleration? (b) What is the net force on the car?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039466306\">\n<div id=\"fs-id1165039466308\">\n<p id=\"fs-id1165039466310\">The driver in the previous problem applies the brakes when the car is moving at 90.0 km\/h, and the car comes to rest after traveling 40.0 m. What is the net force on the car during its deceleration?<\/p>\n<\/div>\n<div id=\"fs-id1165039466316\">\n<p id=\"fs-id1165039466318\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-a331fd2b10d2ba6db89b5bda0bca09b7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#125;&#92;&#92;&#32;&#92;&#92;&#32;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#61;&#123;&#118;&#125;&#95;&#123;&#48;&#125;&#94;&#123;&#50;&#125;&#43;&#50;&#97;&#120;&#8658;&#97;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#8722;&#125;&#55;&#46;&#56;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#45;&#55;&#46;&#56;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#51;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"213\" style=\"vertical-align: -36px;\" \/><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039462140\">\n<div id=\"fs-id1165039462142\">\n<p id=\"fs-id1165039462144\">An 80.0-kg passenger in an SUV traveling at <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-a67fa4725e4584010e9b6f54a2164d30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#49;&#46;&#48;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&times;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#49;&#48;&#125;&#94;&#123;&#51;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"62\" style=\"vertical-align: -1px;\" \/> km\/h is wearing a seat belt. The driver slams on the brakes and the SUV stops in 45.0 m. Find the force of the seat belt on the passenger.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039461707\">\n<div id=\"fs-id1165039461709\">\n<p id=\"fs-id1165039461711\">A particle of mass 2.0 kg is acted on by a single force <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-2b7c8b0a7c844fa6e9765de0a08d2206_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#49;&#125;&#61;&#49;&#56;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"98\" style=\"vertical-align: -4px;\" \/> (a) What is the particle\u2019s acceleration? (b) If the particle starts at rest, how far does it travel in the first 5.0 s?<\/p>\n<\/div>\n<div id=\"fs-id1165039483687\">\n<p id=\"fs-id1165039483689\">a. <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-13db072a7a3a17411a6210c3884792a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#125;&#61;&#109;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#8658;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#97;&#125;&#61;&#57;&#46;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"212\" style=\"vertical-align: -4px;\" \/>; b. The acceleration has magnitude <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-3692279d662c4cfc6e9e657e58af3a54_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#57;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"64\" style=\"vertical-align: -4px;\" \/>, so <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-34511726bebec346b3e386ee54402b77_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#120;&#61;&#49;&#49;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"79\" style=\"vertical-align: -1px;\" \/>.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039477078\">\n<div id=\"fs-id1165039477080\">\n<p id=\"fs-id1165039477082\">Suppose that the particle of the previous problem also experiences forces <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-be7141483fdd6b22627a78f48d054d93_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#50;&#125;&#61;&#45;&#49;&#53;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"107\" style=\"vertical-align: -3px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-2cc1a8582cd25b18ffcdf806cf3de179_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#92;&#116;&#111;&#32;&#125;&#123;&#70;&#125;&#125;&#95;&#123;&#51;&#125;&#61;&#54;&#46;&#48;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"105\" style=\"vertical-align: -4px;\" \/> What is its acceleration in this case?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039478494\">\n<div id=\"fs-id1165039478496\">\n<p id=\"fs-id1165039478498\">Find the acceleration of the body of mass 5.0 kg shown below.<\/p>\n<p><span id=\"fs-id1165039478502\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_BodyMass1_img.jpg\" alt=\"Figure shows a circle labeled m in the xy plane. Three arrows originate from it. One points right and is labeled 10 i newtons. Another points left and is labeled -2 i newtons. The third points downwards and is labeled \u2013 4 j newtons.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1165039478514\">\n<p id=\"fs-id1165039478516\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-6ab0df456eaddd4c3382d585e97b642b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#49;&#46;&#54;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#105;&#125;&#45;&#48;&#46;&#56;&#92;&#115;&#116;&#97;&#99;&#107;&#114;&#101;&#108;&#123;&#94;&#125;&#123;&#106;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"129\" style=\"vertical-align: -4px;\" \/><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039484519\">\n<div id=\"fs-id1165039484521\">\n<p id=\"fs-id1165039484523\">In the following figure, the horizontal surface on which this block slides is frictionless. If the two forces acting on it each have magnitude <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-e40214550d731b15cd9a3a44155523a1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#70;&#61;&#51;&#48;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"86\" style=\"vertical-align: 0px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-content\/ql-cache\/quicklatex.com-454807dd2e4197c4991ffa7b69db8148_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#77;&#61;&#49;&#48;&#46;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#107;&#103;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"96\" style=\"vertical-align: -3px;\" \/>, what is the magnitude of the resulting acceleration of the block?<\/p>\n<p><span id=\"fs-id1165039484557\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/universityphysicssandbox\/wp-content\/uploads\/sites\/289\/2017\/11\/CNX_UPhysics_05_03_Prob5-12_img.jpg\" alt=\"Figure shows a box labeled M resting on a surface. An arrow forming an angle of minus 30 degrees with the horizontal is labeled F and points towards the box. Another arrow labeled F points right.\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165039477315\">\n<dt>Newton\u2019s second law of motion<\/dt>\n<dd id=\"fs-id1165039477320\">acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system and is inversely proportional to its mass<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":211,"menu_order":1,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"all-rights-reserved"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-241","chapter","type-chapter","status-publish","hentry","license-all-rights-reserved"],"part":213,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/pressbooks\/v2\/chapters\/241","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/wp\/v2\/users\/211"}],"version-history":[{"count":1,"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/pressbooks\/v2\/chapters\/241\/revisions"}],"predecessor-version":[{"id":242,"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/pressbooks\/v2\/chapters\/241\/revisions\/242"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/pressbooks\/v2\/parts\/213"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/pressbooks\/v2\/chapters\/241\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/wp\/v2\/media?parent=241"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/pressbooks\/v2\/chapter-type?post=241"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/wp\/v2\/contributor?post=241"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/universityphysicssandboxbook1\/wp-json\/wp\/v2\/license?post=241"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}