{"id":375,"date":"2017-10-27T16:29:40","date_gmt":"2017-10-27T16:29:40","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/chapter\/gravitational-potential-energy\/"},"modified":"2017-11-08T03:24:26","modified_gmt":"2017-11-08T03:24:26","slug":"gravitational-potential-energy","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/chapter\/gravitational-potential-energy\/","title":{"raw":"Gravitational Potential Energy","rendered":"Gravitational Potential Energy"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3 itemprop=\"educationalUse\">Learning Objectives<\/h3>\n<ul>\n<li>Explain gravitational potential energy in terms of work done against gravity.<\/li>\n<li>Show that the gravitational potential energy of an object of mass [latex]m[\/latex] at height [latex]h[\/latex] on Earth is given by [latex]{\\text{PE}}_{\\text{g}}=\\text{mgh}[\/latex].<\/li>\n<li>Show how knowledge of the potential energy as a function of position can be used to simplify calculations and explain physical phenomena.<\/li>\n<\/ul>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id1833407\">\n<h1 data-type=\"title\">Work Done Against Gravity<\/h1>\n<p id=\"import-auto-id1389143\">Climbing stairs and lifting objects is work in both the scientific and everyday sense\u2014it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.<\/p>\n<p id=\"import-auto-id2003938\">Let us calculate the work done in lifting an object of mass [latex]m[\/latex] through a height [latex]h[\/latex], such as in <a href=\"#import-auto-id1697782\" class=\"autogenerated-content\">(Figure)<\/a>. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight [latex]\\text{mg}[\/latex]. The work done on the mass is then [latex]\\text{W = Fd = mgh}[\/latex]. We define this to be the <span data-type=\"term\">gravitational potential energy<\/span> [latex]\\left({\\text{PE}}_{\\text{g}}\\right)[\/latex] put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the [latex]{\\text{PE}}_{\\text{g}}[\/latex] gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word \u201csystem\u201d? Potential energy is a property of a system rather than of a single object\u2014due to its physical position. An object\u2019s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth\u2019s surface, but this point is arbitrary; what is important is the <em data-effect=\"italics\">difference<\/em> in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs. <\/p>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id1804087\">\n<h1 data-type=\"title\">Converting Between Potential Energy and Kinetic Energy<\/h1>\n<p id=\"import-auto-id2690190\">Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to [latex]\\text{mgh}[\/latex] on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of [latex]{\\text{PE}}_{\\text{g}}[\/latex] to [latex]\\text{KE}[\/latex] without explicitly considering the intermediate step of work. (See <a href=\"#fs-id2895434\" class=\"autogenerated-content\">(Figure)<\/a>.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1697782\">\n<div class=\"bc-figcaption figcaption\">(a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2005349\" data-alt=\"(a) The weight attached to the cuckoo clock is raised by a height h shown by a displacement vector d pointing upward. The weight is attached to a winding chain labeled with a force F vector pointing downward. Vector d is also shown in the same direction as force F. E in is equal to W and W is equal to m g h. (b) The weight attached to the cuckoo clock moves downward. E out is equal to m g h.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_01a.jpg\" data-media-type=\"image\/jpg\" alt=\"(a) The weight attached to the cuckoo clock is raised by a height h shown by a displacement vector d pointing upward. The weight is attached to a winding chain labeled with a force F vector pointing downward. Vector d is also shown in the same direction as force F. E in is equal to W and W is equal to m g h. (b) The weight attached to the cuckoo clock moves downward. E out is equal to m g h.\" width=\"350\"><\/span><\/p><\/div>\n<p id=\"import-auto-id1467683\">More precisely, we define the <em data-effect=\"italics\">change<\/em> in gravitational potential energy [latex]\\text{\u0394}{\\text{PE}}_{\\text{g}}[\/latex] to be<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{\u0394}{\\text{PE}}_{\\text{g}}=\\text{mgh},[\/latex]<\/div>\n<p id=\"import-auto-id1980996\">where, for simplicity, we denote the change in height by [latex]h[\/latex] rather than the usual [latex]\\text{\u0394}h[\/latex]. Note that [latex]h[\/latex] is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is <\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"fs-id1729924\">[latex]\\begin{array}{lll}\\text{mgh}&amp; =&amp; \\left(\\text{0.500 kg}\\right)\\left({\\text{9.80}\\phantom{\\rule{0.25em}{0ex}}\\text{m\/s}}^{2}\\right)\\left(\\text{1.00 m}\\right)\\\\ &amp; =&amp; \\text{4.90 kg}\\cdot {m}^{2}{\\text{\/s}}^{2}\\text{= 4.90 J.}\\end{array}[\/latex]<\/div>\n<p id=\"import-auto-id1369902\">Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, <em data-effect=\"italics\">without directly considering the force of gravity that does the work<\/em>.<\/p>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id1332926\">\n<h1 data-type=\"title\">Using Potential Energy to Simplify Calculations<\/h1>\n<p id=\"import-auto-id1838774\">The equation [latex]\\text{\u0394}{\\text{PE}}_{\\text{g}}=\\text{mgh}[\/latex] applies for any path that has a change in height of [latex]h[\/latex], not just when the mass is lifted straight up. (See <a href=\"#import-auto-id1556510\" class=\"autogenerated-content\">(Figure)<\/a>.) It is much easier to calculate [latex]\\text{mgh}[\/latex] (a simple multiplication) than it is to calculate the work done along a complicated path. The idea of gravitational potential energy has the double advantage that it is very broadly applicable and it makes calculations easier. From now on, we will consider that any change in vertical position [latex]h[\/latex] of a mass [latex]m[\/latex] is accompanied by a change in gravitational potential energy [latex]\\text{mgh}[\/latex], and we will avoid the equivalent but more difficult task of calculating work done by or against the gravitational force.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1556510\">\n<div class=\"bc-figcaption figcaption\">The change in gravitational potential energy [latex]\\left(\\text{\u0394}{\\text{PE}}_{\\text{g}}\\right)[\/latex] between points A and B is independent of the path. [latex]\\text{\u0394}{\\text{PE}}_{\\text{g}}=\\text{mgh}[\/latex] for any path between the two points. Gravity is one of a small class of forces where the work done by or against the force depends only on the starting and ending points, not on the path between them.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2693833\" data-alt=\"There is a four-story building. A person is carrying a television up the stairs of the building. The height of third story is h from the ground. A girl is standing outside the building and is lifting a similar television with the help of a pulley.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_02a.jpg\" data-media-type=\"image\/jpg\" alt=\"There is a four-story building. A person is carrying a television up the stairs of the building. The height of third story is h from the ground. A girl is standing outside the building and is lifting a similar television with the help of a pulley.\" width=\"175\"><\/span><\/p><\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id2491449\">\n<div data-type=\"title\" class=\"title\">The Force to Stop Falling<\/div>\n<p id=\"import-auto-id2407890\">A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the force on the knee joints.<\/p>\n<p id=\"import-auto-id1333083\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id2749977\">This person\u2019s energy is brought to zero in this situation by the work done on him by the floor as he stops. The initial [latex]{\\text{PE}}_{\\text{g}}[\/latex] is transformed into [latex]\\text{KE}[\/latex] as he falls. The work done by the floor reduces this kinetic energy to zero. <\/p>\n<p id=\"import-auto-id1445944\"><strong>Solution<\/strong><\/p>\n<p id=\"import-auto-id1583020\">The work done on the person by the floor as he stops is given by<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]W=\\text{Fd}\\phantom{\\rule{0.25em}{0ex}}\\text{cos}\\phantom{\\rule{0.25em}{0ex}}\\theta =-\\text{Fd},[\/latex]<\/div>\n<p id=\"import-auto-id2789210\">with a minus sign because the displacement while stopping and the force from floor are in opposite directions [latex]\\left(\\text{cos}\\phantom{\\rule{0.25em}{0ex}}\\theta =\\text{cos}\\phantom{\\rule{0.25em}{0ex}}\\text{180\u00ba}=-1\\right)[\/latex]. The floor removes energy from the system, so it does negative work. <\/p>\n<p id=\"eip-535\">The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height [latex]h[\/latex]:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"fs-id1449860\">[latex]\\text{KE}=-\\text{\u0394}{\\text{PE}}_{\\text{g}}=-\\text{mgh},[\/latex]<\/div>\n<p id=\"import-auto-id2496602\">The distance [latex]d[\/latex] that the person\u2019s knees bend is much smaller than the height [latex]h[\/latex] of the fall, so the additional change in gravitational potential energy during the knee bend is ignored. <\/p>\n<p id=\"eip-283\">The work [latex]W[\/latex] done by the floor on the person stops the person and brings the person\u2019s kinetic energy to zero:<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]W=-\\text{KE}=\\text{mgh}.[\/latex]<\/div>\n<p id=\"import-auto-id1997664\">Combining this equation with the expression for [latex]W[\/latex] gives <\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"fs-id2124284\">[latex]-\\text{Fd}=\\text{mgh}.[\/latex]<\/div>\n<p id=\"import-auto-id1400348\">Recalling that [latex]h[\/latex] is negative because the person fell <em data-effect=\"italics\">down<\/em>, the force on the knee joints is given by <\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-118\">[latex]F=-\\frac{\\text{mgh}}{d}=-\\frac{\\left(\\text{60.0 kg}\\right)\\left(\\text{9.80 m}{\\text{\/s}}^{2}\\right)\\left(-3\\text{.}\\text{00 m}\\right)}{5\\text{.}\\text{00}\u00d7{\\text{10}}^{-3}\\phantom{\\rule{0.25em}{0ex}}\\text{m}}=3\\text{.}\\text{53}\u00d7{\\text{10}}^{5}\\phantom{\\rule{0.25em}{0ex}}N.[\/latex]<\/div>\n<p id=\"import-auto-id1143579\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id1515219\">Such a large force (500 times more than the person\u2019s weight) over the short impact time is enough to break bones. A much better way to cushion the shock is by bending the legs or rolling on the ground, increasing the time over which the force acts. A bending motion of 0.5 m this way yields a force 100 times smaller than in the example. A kangaroo's hopping shows this method in action. The kangaroo is the only large animal to use hopping for locomotion, but the shock in hopping is cushioned by the bending of its hind legs in each jump.(See <a href=\"#fs-id1503032\" class=\"autogenerated-content\">(Figure)<\/a>.)<\/p>\n<\/div>\n<div class=\"bc-figure figure\" id=\"fs-id1503032\">\n<div class=\"bc-figcaption figcaption\">The work done by the ground upon the kangaroo reduces its kinetic energy to zero as it lands. However, by applying the force of the ground on the hind legs over a longer distance, the impact on the bones is reduced. (credit: Chris Samuel, Flickr)<\/div>\n<p><span data-type=\"media\" id=\"fs-id1697100\" data-alt=\"A hopping kangaroo is shown landing on the ground in one photograph and in the air just after taking another jump in the second photograph.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_03.jpg\" data-media-type=\"image\/jpg\" alt=\"A hopping kangaroo is shown landing on the ground in one photograph and in the air just after taking another jump in the second photograph.\" height=\"300\"><\/span><\/p><\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id2895434\">\n<div data-type=\"title\" class=\"title\">Finding the Speed of a Roller Coaster from its Height<\/div>\n<p id=\"import-auto-id1547593\">(a) What is the final speed of the roller coaster shown in <a href=\"#import-auto-id1349448\" class=\"autogenerated-content\">(Figure)<\/a> if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m\/s?<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1349448\">\n<div class=\"bc-figcaption figcaption\">The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the roller-coaster-Earth system\u2019s gravitational potential energy is converted to kinetic energy. If work done by friction is negligible, all [latex]\\text{\u0394}{\\text{PE}}_{\\text{g}}[\/latex] is converted to [latex]\\text{KE}[\/latex].<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1892950\" data-alt=\"A roller coaster track is shown with a car about to go downhill. The initial height of the roller coaster car on the track is twenty-five meters from the lowest part of the track and its speed v sub zero is equal to zero. The roller coaster\u2019s height from the level part of the track is twenty meters. The finish point of the car is on the level part of the track and the speed at that point is unknown.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_04a.jpg\" data-media-type=\"image\/jpg\" alt=\"A roller coaster track is shown with a car about to go downhill. The initial height of the roller coaster car on the track is twenty-five meters from the lowest part of the track and its speed v sub zero is equal to zero. The roller coaster\u2019s height from the level part of the track is twenty meters. The finish point of the car is on the level part of the track and the speed at that point is unknown.\" height=\"200\"><\/span><\/p><\/div>\n<p id=\"import-auto-id1848019\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1521922\">The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal force, which is perpendicular to the direction of motion and does no work. The net work on the roller coaster is then done by gravity alone. The <em data-effect=\"italics\">loss<\/em> of gravitational potential energy from moving <em data-effect=\"italics\">downward<\/em> through a distance [latex]h[\/latex] equals the <em data-effect=\"italics\">gain<\/em> in kinetic energy. This can be written in equation form as [latex]-\\text{\u0394}{\\text{PE}}_{\\text{g}}=\\text{\u0394}\\text{KE}[\/latex]. Using the equations for [latex]{\\text{PE}}_{\\text{g}}[\/latex] and [latex]\\text{KE}[\/latex], we can solve for the final speed [latex]v[\/latex], which is the desired quantity.<\/p>\n<p id=\"import-auto-id2009732\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1472608\">Here the initial kinetic energy is zero, so that [latex]\\text{\u0394KE}=\\frac{1}{2}{\\text{mv}}^{2}[\/latex]. The equation for change in potential energy states that [latex]{\\text{\u0394PE}}_{\\text{g}}=\\text{mgh}[\/latex]. Since [latex]h[\/latex] is negative in this case, we will rewrite this as [latex]{\\text{\u0394PE}}_{\\text{g}}=-\\text{mg}\\mid h\\mid [\/latex] to show the minus sign clearly. Thus,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"fs-id1349094\">[latex]-\\text{\u0394}{\\text{PE}}_{\\text{g}}=\\text{\u0394}\\text{KE}[\/latex]<\/div>\n<p id=\"import-auto-id1848851\">becomes<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-579\">[latex]\\text{mg}\\mid h\\mid =\\frac{1}{2}{\\text{mv}}^{2}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id2085682\">Solving for [latex]v[\/latex], we find that mass cancels and that<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]v=\\sqrt{2g\\mid h\\mid }.[\/latex]<\/div>\n<p id=\"import-auto-id1337974\">Substituting known values,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-230\">[latex]\\begin{array}{lll}v&amp; =&amp; \\sqrt{2\\left(9\\text{.}\\text{80 m}{\\text{\/s}}^{2}\\right)\\left(\\text{20.0 m}\\right)}\\\\ &amp; =&amp; \\text{19}\\text{.8 m\/s.}\\end{array}[\/latex]<\/div>\n<p id=\"import-auto-id949808\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id2821869\">Again [latex]-{\\text{\u0394PE}}_{\\text{g}}=\\text{\u0394KE}[\/latex]. In this case there is initial kinetic energy, so [latex]\\text{\u0394KE}=\\frac{1}{2}m{v}^{2}-\\frac{1}{2}m{{v}_{0}}^{2}[\/latex]. Thus,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-90\">[latex]\\text{mg}\\mid h\\mid =\\frac{1}{2}{\\text{mv}}^{2}-\\frac{1}{2}m{{v}_{0}}^{2}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id1985822\">Rearranging gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-484\">[latex]\\frac{1}{2}{\\text{mv}}^{2}=\\text{mg}\\mid h\\mid +\\frac{1}{2}m{{v}_{0}}^{2}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id1554938\">This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass again cancels, and<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-436\">[latex]v=\\sqrt{2g\\mid h\\mid +{{v}_{0}}^{2}}.[\/latex]<\/div>\n<p id=\"import-auto-id2124254\">This equation is very similar to the kinematics equation [latex]v=\\sqrt{{{v}_{0}}^{2}+2\\text{ad}}[\/latex], but it is more general\u2014the kinematics equation is valid only for constant acceleration, whereas our equation above is valid for any path regardless of whether the object moves with a constant acceleration. Now, substituting known values gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-388\">[latex]\\begin{array}{lll}v&amp; =&amp; \\sqrt{2\\left(9\\text{.}\\text{80}\\phantom{\\rule{0.25em}{0ex}}{\\text{m\/s}}^{2}\\right)\\left(\\text{20}\\text{.0 m}\\right)+\\left(5\\text{.00 m\/s}{\\right)}^{2}}\\\\ &amp; =&amp; \\text{20.4 m\/s.}\\end{array}[\/latex]<\/div>\n<p id=\"import-auto-id1747146\"><strong>Discussion and Implications<\/strong><\/p>\n<p id=\"import-auto-id1040332\">First, note that mass cancels. This is quite consistent with observations made in <a href=\"\/contents\/9e486db8-120e-4507-b265-e5248007faba\">Falling Objects<\/a> that all objects fall at the same rate if friction is negligible. Second, only the speed of the roller coaster is considered; there is no information about its direction at any point. This reveals another general truth. When friction is negligible, the speed of a falling body depends only on its initial speed and height, and not on its mass or the path taken. For example, the roller coaster will have the same final speed whether it falls 20.0 m straight down or takes a more complicated path like the one in the figure. Third, and perhaps unexpectedly, the final speed in part (b) is greater than in part (a), but by far less than 5.00 m\/s. Finally, note that speed can be found at <em data-effect=\"italics\">any<\/em> height along the way by simply using the appropriate value of [latex]h[\/latex] at the point of interest.<\/p>\n<\/div>\n<p id=\"import-auto-id1645227\">We have seen that work done by or against the gravitational force depends only on the starting and ending points, and not on the path between, allowing us to define the simplifying concept of gravitational potential energy. We can do the same thing for a few other forces, and we will see that this leads to a formal definition of the law of conservation of energy.<\/p>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id1625417\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Making Connections: Take-Home Investigation\u2014Converting Potential to Kinetic Energy<\/div>\n<p id=\"import-auto-id1842650\">One can study the conversion of gravitational potential energy into kinetic energy in this experiment. On a smooth, level surface, use a ruler of the kind that has a groove running along its length and a book to make an incline (see <a href=\"#import-auto-id1965985\" class=\"autogenerated-content\">(Figure)<\/a>). Place a marble at the 10-cm position on the ruler and let it roll down the ruler. When it hits the level surface, measure the time it takes to roll one meter. Now place the marble at the 20-cm and the 30-cm positions and again measure the times it takes to roll 1 m on the level surface. Find the velocity of the marble on the level surface for all three positions. Plot velocity squared versus the distance traveled by the marble. What is the shape of each plot? If the shape is a straight line, the plot shows that the marble\u2019s kinetic energy at the bottom is proportional to its potential energy at the release point. <\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1965985\">\n<div class=\"bc-figcaption figcaption\">A marble rolls down a ruler, and its speed on the level surface is measured. <\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1698574\" data-alt=\"A book is lying on the table and one end of a ruler rests on the edge of this book while the other end rests on the table, making it an incline. A marble is shown rolling down the ruler.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_05a.jpg\" data-media-type=\"image\/jpg\" alt=\"A book is lying on the table and one end of a ruler rests on the edge of this book while the other end rests on the table, making it an incline. A marble is shown rolling down the ruler.\" height=\"150\"><\/span><\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"section-summary\" data-depth=\"1\" id=\"fs-id1683344\">\n<h1 data-type=\"title\">Section Summary<\/h1>\n<ul id=\"fs-id2406525\">\n<li id=\"import-auto-id2742012\">Work done against gravity in lifting an object becomes potential energy of the object-Earth system.<\/li>\n<li id=\"import-auto-id2449367\">The change in gravitational potential energy, [latex]\\Delta {\\text{PE}}_{\\text{g}}[\/latex], is [latex]{\\text{\u0394PE}}_{g}=\\text{mgh}[\/latex], with [latex]h[\/latex] being the increase in height and [latex]g[\/latex] the acceleration due to gravity. <\/li>\n<li id=\"import-auto-id2746176\">The gravitational potential energy of an object near Earth\u2019s surface is due to its position in the mass-Earth system. Only differences in gravitational potential energy, [latex]{\\text{\u0394PE}}_{g}[\/latex], have physical significance.<\/li>\n<li id=\"import-auto-id1527618\">As an object descends without friction, its gravitational potential energy changes into kinetic energy corresponding to increasing speed, so that [latex]\\text{\u0394KE}\\text{= \u2212}{\\text{\u0394PE}}_{\\text{g}}[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div class=\"conceptual-questions\" data-depth=\"1\" id=\"fs-id2011989\" data-element-type=\"conceptual-questions\">\n<h1 data-type=\"title\">Conceptual Questions<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1415865\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1753129\">\n<p id=\"import-auto-id2111705\">In <a href=\"#fs-id2895434\" class=\"autogenerated-content\">(Figure)<\/a>, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m\/s downhill. Suppose the roller coaster had had an initial speed of 5 m\/s <em data-effect=\"italics\">uphill<\/em> instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m below the start. We would find in that case that it had the same final speed. Explain in terms of conservation of energy. <\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1759147\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2183193\">\n<p id=\"import-auto-id1844102\">Does the work you do on a book when you lift it onto a shelf depend on the path taken? On the time taken? On the height of the shelf? On the mass of the book? <\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"problems-exercises\" data-depth=\"1\" data-element-type=\"problems-exercises\">\n<h1 data-type=\"title\">Problems &amp; Exercises<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1942940\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1426768\">\n<p id=\"import-auto-id2896309\">A hydroelectric power facility (see <a href=\"#import-auto-id1889215\" class=\"autogenerated-content\">(Figure)<\/a>) converts the gravitational potential energy of water behind a dam to electric energy. (a) What is the gravitational potential energy relative to the generators of a lake of volume [latex]\\text{50}\\text{.}0 k{\\text{m}}^{3}[\/latex] ([latex]\\text{mass}=5\\text{.}\\text{00}\u00d7{\\text{10}}^{\\text{13}}\\phantom{\\rule{0.25em}{0ex}}\\text{kg}\\right)[\/latex], given that the lake has an average height of 40.0 m above the generators? (b) Compare this with the energy stored in a 9-megaton fusion bomb.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1889215\">\n<div class=\"bc-figcaption figcaption\">Hydroelectric facility (credit: Denis Belevich, Wikimedia Commons)<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1707793\" data-alt=\"A dam with water flowing down its gates.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_06a.jpg\" data-media-type=\"image\/jpg\" alt=\"A dam with water flowing down its gates.\" height=\"250\"><\/span><\/p><\/div>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1391409\">\n<p id=\"import-auto-id2555983\">(a) [latex]1\\text{.}\\text{96}\u00d7{\\text{10}}^{\\text{16}}\\phantom{\\rule{0.25em}{0ex}}\\text{J}[\/latex]<\/p>\n<p id=\"import-auto-id1745609\">(b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1332867\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1889598\">\n<p id=\"import-auto-id1349094\">(a) How much gravitational potential energy (relative to the ground on which it is built) is stored in the Great Pyramid of Cheops, given that its mass is about [latex]7\\mathrm{&nbsp;\u00d7&nbsp;}{\\text{10}}^{9}\\text{&nbsp;kg}[\/latex] and its center of mass is 36.5 m above the surrounding ground? (b) How does this energy compare with the daily food intake of a person?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1796084\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1980316\">\n<p id=\"import-auto-id1895561\">Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a) How much work did the bird do on the snake? (b) How much work did it do to raise its own center of mass to the branch?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2124246\">\n<p id=\"import-auto-id1472089\">(a) 1.8 J<\/p>\n<p id=\"import-auto-id1170503\">(b) 8.6 J<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2189211\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1836678\">\n<p id=\"import-auto-id1965941\">In <a href=\"#fs-id2895434\" class=\"autogenerated-content\">(Figure)<\/a>, we found that the speed of a roller coaster that had descended 20.0 m was only slightly greater when it had an initial speed of 5.00 m\/s than when it started from rest. This implies that [latex]\\text{\u0394}{\\text{PE&nbsp;&gt;&gt;&nbsp;KE}}_{\\text{i}}[\/latex]. Confirm this statement by taking the ratio of [latex]\\text{\u0394}\\text{PE}[\/latex] to [latex]{\\text{KE}}_{\\text{i}}[\/latex]. (Note that mass cancels.)<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2097867\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2047156\">\n<p id=\"import-auto-id2092581\">A 100-g toy car is propelled by a compressed spring that starts it moving. The car follows the curved track in <a href=\"#import-auto-id1299469\" class=\"autogenerated-content\">(Figure)<\/a>. Show that the final speed of the toy car is 0.687 m\/s if its initial speed is 2.00 m\/s and it coasts up the frictionless slope, gaining 0.180 m in altitude.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1299469\">\n<div class=\"bc-figcaption figcaption\">A toy car moves up a sloped track. (credit: Leszek Leszczynski, Flickr)<\/div>\n<p><span data-type=\"media\" data-alt=\"A toy car is moving up a curved track.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_07a.jpg\" data-media-type=\"image\/png\" alt=\"A toy car is moving up a curved track.\" height=\"200\"><\/span><\/p><\/div>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2122741\">\n<div data-type=\"equation\" class=\"equation\" id=\"eip-id2183293\">[latex]{v}_{f}=\\sqrt{2\\text{gh}+{{v}_{0}}^{2}}=\\sqrt{2\\left(\\text{9.80 m}{\\text{\/s}}^{2}\\right)\\left(-0\\text{.180 m}\\right)+\\left(2\\text{.00 m\/s}{\\right)}^{2}}=0\\text{.687 m\/s}[\/latex]<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2175922\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id732793\">\n<p id=\"import-auto-id2393516\">In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills.) To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a [latex]\\text{30\u00ba}[\/latex] slope neglecting friction: (a) Starting from rest. (b) Starting with an initial speed of 2.50 m\/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"glossary\" class=\"textbox shaded\">\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\n<dl class=\"definition\" id=\"fs-id1672281\">\n<dt>gravitational potential energy<\/dt>\n<dd id=\"fs-id1239787\">the energy an object has due to its position in a gravitational field<\/dd>\n<\/dl>\n<\/div>\n\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3 itemprop=\"educationalUse\">Learning Objectives<\/h3>\n<ul>\n<li>Explain gravitational potential energy in terms of work done against gravity.<\/li>\n<li>Show that the gravitational potential energy of an object of mass <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-6b41df788161942c6f98604d37de8098_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"15\" style=\"vertical-align: 0px;\" \/> at height <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> on Earth is given by <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-fb165741d6b9363d69d79d49a036be90_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"89\" style=\"vertical-align: -6px;\" \/>.<\/li>\n<li>Show how knowledge of the potential energy as a function of position can be used to simplify calculations and explain physical phenomena.<\/li>\n<\/ul>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id1833407\">\n<h1 data-type=\"title\">Work Done Against Gravity<\/h1>\n<p id=\"import-auto-id1389143\">Climbing stairs and lifting objects is work in both the scientific and everyday sense\u2014it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.<\/p>\n<p id=\"import-auto-id2003938\">Let us calculate the work done in lifting an object of mass <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-6b41df788161942c6f98604d37de8098_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"15\" style=\"vertical-align: 0px;\" \/> through a height <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/>, such as in <a href=\"#import-auto-id1697782\" class=\"autogenerated-content\">(Figure)<\/a>. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-649ee3316590a2ad3b7d61c9dfffad9b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"24\" style=\"vertical-align: -3px;\" \/>. The work done on the mass is then <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-6487969fe01960635b7eefff3ae883db_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#87;&#32;&#61;&#32;&#70;&#100;&#32;&#61;&#32;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"125\" style=\"vertical-align: -3px;\" \/>. We define this to be the <span data-type=\"term\">gravitational potential energy<\/span> <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-2ca4c67d6956da3d13a734eb7b914c26_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"44\" style=\"vertical-align: -6px;\" \/> put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-e73c9ff5cb76b1e812833503100f8466_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"31\" style=\"vertical-align: -6px;\" \/> gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word \u201csystem\u201d? Potential energy is a property of a system rather than of a single object\u2014due to its physical position. An object\u2019s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth\u2019s surface, but this point is arbitrary; what is important is the <em data-effect=\"italics\">difference<\/em> in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs. <\/p>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id1804087\">\n<h1 data-type=\"title\">Converting Between Potential Energy and Kinetic Energy<\/h1>\n<p id=\"import-auto-id2690190\">Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c6103d60cd471f934368e423f05e3741_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"34\" style=\"vertical-align: -3px;\" \/> on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-e73c9ff5cb76b1e812833503100f8466_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"31\" style=\"vertical-align: -6px;\" \/> to <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-909db8dd14ec527e98eef010c3baba6e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"26\" style=\"vertical-align: -1px;\" \/> without explicitly considering the intermediate step of work. (See <a href=\"#fs-id2895434\" class=\"autogenerated-content\">(Figure)<\/a>.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1697782\">\n<div class=\"bc-figcaption figcaption\">(a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2005349\" data-alt=\"(a) The weight attached to the cuckoo clock is raised by a height h shown by a displacement vector d pointing upward. The weight is attached to a winding chain labeled with a force F vector pointing downward. Vector d is also shown in the same direction as force F. E in is equal to W and W is equal to m g h. (b) The weight attached to the cuckoo clock moves downward. E out is equal to m g h.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_01a.jpg\" data-media-type=\"image\/jpg\" alt=\"(a) The weight attached to the cuckoo clock is raised by a height h shown by a displacement vector d pointing upward. The weight is attached to a winding chain labeled with a force F vector pointing downward. Vector d is also shown in the same direction as force F. E in is equal to W and W is equal to m g h. (b) The weight attached to the cuckoo clock moves downward. E out is equal to m g h.\" width=\"350\" \/><\/span><\/p>\n<\/div>\n<p id=\"import-auto-id1467683\">More precisely, we define the <em data-effect=\"italics\">change<\/em> in gravitational potential energy <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4f3de681b27bd9f1619f21ecfa104cc6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"31\" style=\"vertical-align: -6px;\" \/> to be<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-d7337733ba5fc5773cd076b344e9d6ab_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"93\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1980996\">where, for simplicity, we denote the change in height by <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> rather than the usual <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-90a1ffbf266233d8a3c405f69ebc1c88_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/>. Note that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is <\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"fs-id1729924\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-6212ac0a7847dd102e8effe8ecd076a8_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;&#108;&#108;&#108;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#38;&#32;&#61;&#38;&#32;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#46;&#53;&#48;&#48;&#32;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#57;&#46;&#56;&#48;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#46;&#48;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#92;&#32;&#38;&#32;&#61;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#52;&#46;&#57;&#48;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#109;&#125;&#94;&#123;&#50;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#61;&#32;&#52;&#46;&#57;&#48;&#32;&#74;&#46;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"52\" width=\"324\" style=\"vertical-align: -20px;\" \/><\/div>\n<p id=\"import-auto-id1369902\">Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, <em data-effect=\"italics\">without directly considering the force of gravity that does the work<\/em>.<\/p>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id1332926\">\n<h1 data-type=\"title\">Using Potential Energy to Simplify Calculations<\/h1>\n<p id=\"import-auto-id1838774\">The equation <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-70ddf2cb950b0c66875a7443cf5ad36c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"89\" style=\"vertical-align: -6px;\" \/> applies for any path that has a change in height of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/>, not just when the mass is lifted straight up. (See <a href=\"#import-auto-id1556510\" class=\"autogenerated-content\">(Figure)<\/a>.) It is much easier to calculate <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c6103d60cd471f934368e423f05e3741_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"34\" style=\"vertical-align: -3px;\" \/> (a simple multiplication) than it is to calculate the work done along a complicated path. The idea of gravitational potential energy has the double advantage that it is very broadly applicable and it makes calculations easier. From now on, we will consider that any change in vertical position <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> of a mass <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-6b41df788161942c6f98604d37de8098_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"15\" style=\"vertical-align: 0px;\" \/> is accompanied by a change in gravitational potential energy <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c6103d60cd471f934368e423f05e3741_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"34\" style=\"vertical-align: -3px;\" \/>, and we will avoid the equivalent but more difficult task of calculating work done by or against the gravitational force.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1556510\">\n<div class=\"bc-figcaption figcaption\">The change in gravitational potential energy <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-33097e6f8e47a4e48bea3436e6791f7d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"44\" style=\"vertical-align: -6px;\" \/> between points A and B is independent of the path. <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-70ddf2cb950b0c66875a7443cf5ad36c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"89\" style=\"vertical-align: -6px;\" \/> for any path between the two points. Gravity is one of a small class of forces where the work done by or against the force depends only on the starting and ending points, not on the path between them.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2693833\" data-alt=\"There is a four-story building. A person is carrying a television up the stairs of the building. The height of third story is h from the ground. A girl is standing outside the building and is lifting a similar television with the help of a pulley.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_02a.jpg\" data-media-type=\"image\/jpg\" alt=\"There is a four-story building. A person is carrying a television up the stairs of the building. The height of third story is h from the ground. A girl is standing outside the building and is lifting a similar television with the help of a pulley.\" width=\"175\" \/><\/span><\/p>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id2491449\">\n<div data-type=\"title\" class=\"title\">The Force to Stop Falling<\/div>\n<p id=\"import-auto-id2407890\">A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the force on the knee joints.<\/p>\n<p id=\"import-auto-id1333083\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id2749977\">This person\u2019s energy is brought to zero in this situation by the work done on him by the floor as he stops. The initial <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-e73c9ff5cb76b1e812833503100f8466_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"31\" style=\"vertical-align: -6px;\" \/> is transformed into <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-909db8dd14ec527e98eef010c3baba6e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"26\" style=\"vertical-align: -1px;\" \/> as he falls. The work done by the floor reduces this kinetic energy to zero. <\/p>\n<p id=\"import-auto-id1445944\"><strong>Solution<\/strong><\/p>\n<p id=\"import-auto-id1583020\">The work done on the person by the floor as he stops is given by<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-0d0e77f48d65f668cbdda69e5ba5ab05_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#87;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#100;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#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;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#104;&#101;&#116;&#97;&#32;&#61;&#45;&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#100;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"169\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"import-auto-id2789210\">with a minus sign because the displacement while stopping and the force from floor are in opposite directions <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-7439cdf37ac55c837df22916713b6277_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#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;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#104;&#101;&#116;&#97;&#32;&#61;&#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;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#56;&#48;&ordm;&#125;&#61;&#45;&#49;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"174\" style=\"vertical-align: -4px;\" \/>. The floor removes energy from the system, so it does negative work. <\/p>\n<p id=\"eip-535\">The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/>:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"fs-id1449860\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-7336c381046dbde0d09c6aff34ad9460_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#61;&#45;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#45;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"170\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id2496602\">The distance <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4e8716946f6a868f015e0d62f28bc540_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#100;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> that the person\u2019s knees bend is much smaller than the height <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> of the fall, so the additional change in gravitational potential energy during the knee bend is ignored. <\/p>\n<p id=\"eip-283\">The work <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4caed22919a1780df1b6310b338b904e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#87;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"19\" style=\"vertical-align: 0px;\" \/> done by the floor on the person stops the person and brings the person\u2019s kinetic energy to zero:<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-9c5442ee25aac691e6b0dc6b6cc2bd44_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#87;&#61;&#45;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"144\" style=\"vertical-align: -3px;\" \/><\/div>\n<p id=\"import-auto-id1997664\">Combining this equation with the expression for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4caed22919a1780df1b6310b338b904e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#87;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"19\" style=\"vertical-align: 0px;\" \/> gives <\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"fs-id2124284\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-d2d6864fa3dee874eaf18d6882385087_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#45;&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#100;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"95\" style=\"vertical-align: -3px;\" \/><\/div>\n<p id=\"import-auto-id1400348\">Recalling that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> is negative because the person fell <em data-effect=\"italics\">down<\/em>, the force on the knee joints is given by <\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-118\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-b1c4c45dfa1cc8b9cb4dbc5dfd277a5d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#70;&#61;&#45;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#125;&#123;&#100;&#125;&#61;&#45;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#54;&#48;&#46;&#48;&#32;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#57;&#46;&#56;&#48;&#32;&#109;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#45;&#51;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#123;&#53;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#48;&#125;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#45;&#51;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#61;&#51;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#51;&#125;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#53;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#78;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"32\" width=\"416\" style=\"vertical-align: -8px;\" \/><\/div>\n<p id=\"import-auto-id1143579\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id1515219\">Such a large force (500 times more than the person\u2019s weight) over the short impact time is enough to break bones. A much better way to cushion the shock is by bending the legs or rolling on the ground, increasing the time over which the force acts. A bending motion of 0.5 m this way yields a force 100 times smaller than in the example. A kangaroo&#8217;s hopping shows this method in action. The kangaroo is the only large animal to use hopping for locomotion, but the shock in hopping is cushioned by the bending of its hind legs in each jump.(See <a href=\"#fs-id1503032\" class=\"autogenerated-content\">(Figure)<\/a>.)<\/p>\n<\/div>\n<div class=\"bc-figure figure\" id=\"fs-id1503032\">\n<div class=\"bc-figcaption figcaption\">The work done by the ground upon the kangaroo reduces its kinetic energy to zero as it lands. However, by applying the force of the ground on the hind legs over a longer distance, the impact on the bones is reduced. (credit: Chris Samuel, Flickr)<\/div>\n<p><span data-type=\"media\" id=\"fs-id1697100\" data-alt=\"A hopping kangaroo is shown landing on the ground in one photograph and in the air just after taking another jump in the second photograph.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_03.jpg\" data-media-type=\"image\/jpg\" alt=\"A hopping kangaroo is shown landing on the ground in one photograph and in the air just after taking another jump in the second photograph.\" height=\"300\" \/><\/span><\/p>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id2895434\">\n<div data-type=\"title\" class=\"title\">Finding the Speed of a Roller Coaster from its Height<\/div>\n<p id=\"import-auto-id1547593\">(a) What is the final speed of the roller coaster shown in <a href=\"#import-auto-id1349448\" class=\"autogenerated-content\">(Figure)<\/a> if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m\/s?<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1349448\">\n<div class=\"bc-figcaption figcaption\">The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the roller-coaster-Earth system\u2019s gravitational potential energy is converted to kinetic energy. If work done by friction is negligible, all <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4f3de681b27bd9f1619f21ecfa104cc6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"31\" style=\"vertical-align: -6px;\" \/> is converted to <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-909db8dd14ec527e98eef010c3baba6e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"26\" style=\"vertical-align: -1px;\" \/>.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1892950\" data-alt=\"A roller coaster track is shown with a car about to go downhill. The initial height of the roller coaster car on the track is twenty-five meters from the lowest part of the track and its speed v sub zero is equal to zero. The roller coaster\u2019s height from the level part of the track is twenty meters. The finish point of the car is on the level part of the track and the speed at that point is unknown.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_04a.jpg\" data-media-type=\"image\/jpg\" alt=\"A roller coaster track is shown with a car about to go downhill. The initial height of the roller coaster car on the track is twenty-five meters from the lowest part of the track and its speed v sub zero is equal to zero. The roller coaster\u2019s height from the level part of the track is twenty meters. The finish point of the car is on the level part of the track and the speed at that point is unknown.\" height=\"200\" \/><\/span><\/p>\n<\/div>\n<p id=\"import-auto-id1848019\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1521922\">The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal force, which is perpendicular to the direction of motion and does no work. The net work on the roller coaster is then done by gravity alone. The <em data-effect=\"italics\">loss<\/em> of gravitational potential energy from moving <em data-effect=\"italics\">downward<\/em> through a distance <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> equals the <em data-effect=\"italics\">gain<\/em> in kinetic energy. This can be written in equation form as <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-5c837f4b60264ee99320564e6c398566_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#45;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"94\" style=\"vertical-align: -6px;\" \/>. Using the equations for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-e73c9ff5cb76b1e812833503100f8466_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"31\" style=\"vertical-align: -6px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-909db8dd14ec527e98eef010c3baba6e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"26\" style=\"vertical-align: -1px;\" \/>, we can solve for the final speed <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ef71511c70f0e4b25cc6bd69f3bc20c2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\" \/>, which is the desired quantity.<\/p>\n<p id=\"import-auto-id2009732\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1472608\">Here the initial kinetic energy is zero, so that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-e6791118de84c5cb7205dd1abb25e9cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#75;&#69;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"91\" style=\"vertical-align: -6px;\" \/>. The equation for change in potential energy states that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-01d55a194885125eb6fb8ea8e1627751_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"89\" style=\"vertical-align: -6px;\" \/>. Since <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> is negative in this case, we will rewrite this as <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-a72ba42d00c9d8a2e70863bbdb1be0a0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#45;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#92;&#109;&#105;&#100;&#32;&#104;&#92;&#109;&#105;&#100;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"126\" style=\"vertical-align: -6px;\" \/> to show the minus sign clearly. Thus,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"fs-id1349094\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-5c837f4b60264ee99320564e6c398566_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#45;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"94\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1848851\">becomes<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-579\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-53792d9f3be7a223184ad2294f02ce4c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#92;&#109;&#105;&#100;&#32;&#104;&#92;&#109;&#105;&#100;&#32;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"123\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id2085682\">Solving for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ef71511c70f0e4b25cc6bd69f3bc20c2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\" \/>, we find that mass cancels and that<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-01db31ec9441ca24a61974cd52d13244_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;&#61;&#92;&#115;&#113;&#114;&#116;&#123;&#50;&#103;&#92;&#109;&#105;&#100;&#32;&#104;&#92;&#109;&#105;&#100;&#32;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"107\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1337974\">Substituting known values,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-230\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-28bc20523a2c3958b8e92c555658feb5_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;&#108;&#108;&#108;&#125;&#118;&#38;&#32;&#61;&#38;&#32;&#92;&#115;&#113;&#114;&#116;&#123;&#50;&#92;&#108;&#101;&#102;&#116;&#40;&#57;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#56;&#48;&#32;&#109;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#48;&#46;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#92;&#92;&#32;&#38;&#32;&#61;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#57;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#56;&#32;&#109;&#47;&#115;&#46;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"63\" width=\"251\" style=\"vertical-align: -26px;\" \/><\/div>\n<p id=\"import-auto-id949808\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id2821869\">Again <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-80283e971828e3f5c08be34225ad1ce0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#45;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"94\" style=\"vertical-align: -6px;\" \/>. In this case there is initial kinetic energy, so <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-47eef80035f3b618bc7d8178a1e62fb2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#75;&#69;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#109;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#45;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#109;&#123;&#123;&#118;&#125;&#95;&#123;&#48;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"164\" style=\"vertical-align: -6px;\" \/>. Thus,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-90\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-420f375c943c49f5d2ad47ac635d290e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#92;&#109;&#105;&#100;&#32;&#104;&#92;&#109;&#105;&#100;&#32;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#45;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#109;&#123;&#123;&#118;&#125;&#95;&#123;&#48;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"195\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1985822\">Rearranging gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-484\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c13aaa0111dd0c57223a8c2e0fc6dc50_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#92;&#109;&#105;&#100;&#32;&#104;&#92;&#109;&#105;&#100;&#32;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#109;&#123;&#123;&#118;&#125;&#95;&#123;&#48;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"195\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1554938\">This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass again cancels, and<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-436\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-6fef31ca69a9e833b66f71279b73a86e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;&#61;&#92;&#115;&#113;&#114;&#116;&#123;&#50;&#103;&#92;&#109;&#105;&#100;&#32;&#104;&#92;&#109;&#105;&#100;&#32;&#43;&#123;&#123;&#118;&#125;&#95;&#123;&#48;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"150\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id2124254\">This equation is very similar to the kinematics equation <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-7f1b1173198947400b078b2e83cbbe30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;&#61;&#92;&#115;&#113;&#114;&#116;&#123;&#123;&#123;&#118;&#125;&#95;&#123;&#48;&#125;&#125;&#94;&#123;&#50;&#125;&#43;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&#97;&#100;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"122\" style=\"vertical-align: -3px;\" \/>, but it is more general\u2014the kinematics equation is valid only for constant acceleration, whereas our equation above is valid for any path regardless of whether the object moves with a constant acceleration. Now, substituting known values gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-388\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-13e798c2898d4203b34492321de08012_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;&#108;&#108;&#108;&#125;&#118;&#38;&#32;&#61;&#38;&#32;&#92;&#115;&#113;&#114;&#116;&#123;&#50;&#92;&#108;&#101;&#102;&#116;&#40;&#57;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#56;&#48;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#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;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#48;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#43;&#92;&#108;&#101;&#102;&#116;&#40;&#53;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#48;&#48;&#32;&#109;&#47;&#115;&#125;&#123;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#125;&#92;&#92;&#32;&#38;&#32;&#61;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#48;&#46;&#52;&#32;&#109;&#47;&#115;&#46;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"66\" width=\"360\" style=\"vertical-align: -27px;\" \/><\/div>\n<p id=\"import-auto-id1747146\"><strong>Discussion and Implications<\/strong><\/p>\n<p id=\"import-auto-id1040332\">First, note that mass cancels. This is quite consistent with observations made in <a href=\"\/contents\/9e486db8-120e-4507-b265-e5248007faba\">Falling Objects<\/a> that all objects fall at the same rate if friction is negligible. Second, only the speed of the roller coaster is considered; there is no information about its direction at any point. This reveals another general truth. When friction is negligible, the speed of a falling body depends only on its initial speed and height, and not on its mass or the path taken. For example, the roller coaster will have the same final speed whether it falls 20.0 m straight down or takes a more complicated path like the one in the figure. Third, and perhaps unexpectedly, the final speed in part (b) is greater than in part (a), but by far less than 5.00 m\/s. Finally, note that speed can be found at <em data-effect=\"italics\">any<\/em> height along the way by simply using the appropriate value of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> at the point of interest.<\/p>\n<\/div>\n<p id=\"import-auto-id1645227\">We have seen that work done by or against the gravitational force depends only on the starting and ending points, and not on the path between, allowing us to define the simplifying concept of gravitational potential energy. We can do the same thing for a few other forces, and we will see that this leads to a formal definition of the law of conservation of energy.<\/p>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id1625417\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Making Connections: Take-Home Investigation\u2014Converting Potential to Kinetic Energy<\/div>\n<p id=\"import-auto-id1842650\">One can study the conversion of gravitational potential energy into kinetic energy in this experiment. On a smooth, level surface, use a ruler of the kind that has a groove running along its length and a book to make an incline (see <a href=\"#import-auto-id1965985\" class=\"autogenerated-content\">(Figure)<\/a>). Place a marble at the 10-cm position on the ruler and let it roll down the ruler. When it hits the level surface, measure the time it takes to roll one meter. Now place the marble at the 20-cm and the 30-cm positions and again measure the times it takes to roll 1 m on the level surface. Find the velocity of the marble on the level surface for all three positions. Plot velocity squared versus the distance traveled by the marble. What is the shape of each plot? If the shape is a straight line, the plot shows that the marble\u2019s kinetic energy at the bottom is proportional to its potential energy at the release point. <\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1965985\">\n<div class=\"bc-figcaption figcaption\">A marble rolls down a ruler, and its speed on the level surface is measured. <\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1698574\" data-alt=\"A book is lying on the table and one end of a ruler rests on the edge of this book while the other end rests on the table, making it an incline. A marble is shown rolling down the ruler.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_05a.jpg\" data-media-type=\"image\/jpg\" alt=\"A book is lying on the table and one end of a ruler rests on the edge of this book while the other end rests on the table, making it an incline. A marble is shown rolling down the ruler.\" height=\"150\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"section-summary\" data-depth=\"1\" id=\"fs-id1683344\">\n<h1 data-type=\"title\">Section Summary<\/h1>\n<ul id=\"fs-id2406525\">\n<li id=\"import-auto-id2742012\">Work done against gravity in lifting an object becomes potential energy of the object-Earth system.<\/li>\n<li id=\"import-auto-id2449367\">The change in gravitational potential energy, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-0ef54fab901a3dac8e1946a8f2822bab_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#68;&#101;&#108;&#116;&#97;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"46\" style=\"vertical-align: -6px;\" \/>, is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f52ed9d94bea01a413fcd4e5d4a327e9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#80;&#69;&#125;&#125;&#95;&#123;&#103;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"90\" style=\"vertical-align: -6px;\" \/>, with <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> being the increase in height and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-d208fd391fa57c168dc0f151de829fee_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#103;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\" \/> the acceleration due to gravity. <\/li>\n<li id=\"import-auto-id2746176\">The gravitational potential energy of an object near Earth\u2019s surface is due to its position in the mass-Earth system. Only differences in gravitational potential energy, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-2a9a72c3240c592650a0d43d8bcdd218_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#80;&#69;&#125;&#125;&#95;&#123;&#103;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"31\" style=\"vertical-align: -6px;\" \/>, have physical significance.<\/li>\n<li id=\"import-auto-id1527618\">As an object descends without friction, its gravitational potential energy changes into kinetic energy corresponding to increasing speed, so that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-b3122a0a2edb57c27671dc9bd8214482_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#75;&#69;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#61;&#32;&#8722;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -6px;\" \/>.<\/li>\n<\/ul>\n<\/div>\n<div class=\"conceptual-questions\" data-depth=\"1\" id=\"fs-id2011989\" data-element-type=\"conceptual-questions\">\n<h1 data-type=\"title\">Conceptual Questions<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1415865\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1753129\">\n<p id=\"import-auto-id2111705\">In <a href=\"#fs-id2895434\" class=\"autogenerated-content\">(Figure)<\/a>, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m\/s downhill. Suppose the roller coaster had had an initial speed of 5 m\/s <em data-effect=\"italics\">uphill<\/em> instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m below the start. We would find in that case that it had the same final speed. Explain in terms of conservation of energy. <\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1759147\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2183193\">\n<p id=\"import-auto-id1844102\">Does the work you do on a book when you lift it onto a shelf depend on the path taken? On the time taken? On the height of the shelf? On the mass of the book? <\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"problems-exercises\" data-depth=\"1\" data-element-type=\"problems-exercises\">\n<h1 data-type=\"title\">Problems &amp; Exercises<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1942940\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1426768\">\n<p id=\"import-auto-id2896309\">A hydroelectric power facility (see <a href=\"#import-auto-id1889215\" class=\"autogenerated-content\">(Figure)<\/a>) converts the gravitational potential energy of water behind a dam to electric energy. (a) What is the gravitational potential energy relative to the generators of a lake of volume <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-9bb7a45361f8e3c68baedb410e09654a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#48;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#48;&#32;&#107;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#51;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"63\" style=\"vertical-align: 0px;\" \/> (<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ea9431b316d7d658908b643840e3df13_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#97;&#115;&#115;&#125;&#61;&#53;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#48;&#125;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#51;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#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;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"148\" style=\"vertical-align: -3px;\" \/>, given that the lake has an average height of 40.0 m above the generators? (b) Compare this with the energy stored in a 9-megaton fusion bomb.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1889215\">\n<div class=\"bc-figcaption figcaption\">Hydroelectric facility (credit: Denis Belevich, Wikimedia Commons)<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1707793\" data-alt=\"A dam with water flowing down its gates.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_06a.jpg\" data-media-type=\"image\/jpg\" alt=\"A dam with water flowing down its gates.\" height=\"250\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1391409\">\n<p id=\"import-auto-id2555983\">(a) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ccc49a45bcad4d8bc4f362c35bf51fca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#49;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#57;&#54;&#125;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#54;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#74;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"76\" style=\"vertical-align: -1px;\" \/><\/p>\n<p id=\"import-auto-id1745609\">(b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1332867\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1889598\">\n<p id=\"import-auto-id1349094\">(a) How much gravitational potential energy (relative to the ground on which it is built) is stored in the Great Pyramid of Cheops, given that its mass is about <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-9017063f7df063dbd90e47b91ddaa376_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#55;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#32;&times;&#32;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#57;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#32;&#107;&#103;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"58\" style=\"vertical-align: -3px;\" \/> and its center of mass is 36.5 m above the surrounding ground? (b) How does this energy compare with the daily food intake of a person?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1796084\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1980316\">\n<p id=\"import-auto-id1895561\">Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a) How much work did the bird do on the snake? (b) How much work did it do to raise its own center of mass to the branch?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2124246\">\n<p id=\"import-auto-id1472089\">(a) 1.8 J<\/p>\n<p id=\"import-auto-id1170503\">(b) 8.6 J<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2189211\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1836678\">\n<p id=\"import-auto-id1965941\">In <a href=\"#fs-id2895434\" class=\"autogenerated-content\">(Figure)<\/a>, we found that the speed of a roller coaster that had descended 20.0 m was only slightly greater when it had an initial speed of 5.00 m\/s than when it started from rest. This implies that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-dd30fcdcae51635be420d6827f6b1c3f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#32;&#62;&#62;&#32;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#105;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"83\" style=\"vertical-align: -5px;\" \/>. Confirm this statement by taking the ratio of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-08de69141512ad5611090977d33d93c5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"24\" style=\"vertical-align: -1px;\" \/> to <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3414ecd35b8bb4ec5bfe9cba398b6944_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#105;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"30\" style=\"vertical-align: -4px;\" \/>. (Note that mass cancels.)<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2097867\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2047156\">\n<p id=\"import-auto-id2092581\">A 100-g toy car is propelled by a compressed spring that starts it moving. The car follows the curved track in <a href=\"#import-auto-id1299469\" class=\"autogenerated-content\">(Figure)<\/a>. Show that the final speed of the toy car is 0.687 m\/s if its initial speed is 2.00 m\/s and it coasts up the frictionless slope, gaining 0.180 m in altitude.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1299469\">\n<div class=\"bc-figcaption figcaption\">A toy car moves up a sloped track. (credit: Leszek Leszczynski, Flickr)<\/div>\n<p><span data-type=\"media\" data-alt=\"A toy car is moving up a curved track.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_08_03_07a.jpg\" data-media-type=\"image\/png\" alt=\"A toy car is moving up a curved track.\" height=\"200\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2122741\">\n<div data-type=\"equation\" class=\"equation\" id=\"eip-id2183293\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-db93b8f720f7f358a13506805bbf1027_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#118;&#125;&#95;&#123;&#102;&#125;&#61;&#92;&#115;&#113;&#114;&#116;&#123;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#104;&#125;&#43;&#123;&#123;&#118;&#125;&#95;&#123;&#48;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#61;&#92;&#115;&#113;&#114;&#116;&#123;&#50;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#57;&#46;&#56;&#48;&#32;&#109;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#45;&#48;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#49;&#56;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#43;&#92;&#108;&#101;&#102;&#116;&#40;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#48;&#48;&#32;&#109;&#47;&#115;&#125;&#123;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#125;&#61;&#48;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#54;&#56;&#55;&#32;&#109;&#47;&#115;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"66\" width=\"582\" style=\"vertical-align: -4px;\" \/><\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2175922\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id732793\">\n<p id=\"import-auto-id2393516\">In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills.) To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-7e404b535ec15d8300b0b1148704da97_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#51;&#48;&ordm;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"18\" style=\"vertical-align: 0px;\" \/> slope neglecting friction: (a) Starting from rest. (b) Starting with an initial speed of 2.50 m\/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"glossary\" class=\"textbox shaded\">\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\n<dl class=\"definition\" id=\"fs-id1672281\">\n<dt>gravitational potential energy<\/dt>\n<dd id=\"fs-id1239787\">the energy an object has due to its position in a gravitational field<\/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-375","chapter","type-chapter","status-publish","hentry","license-all-rights-reserved"],"part":353,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/375","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/users\/211"}],"version-history":[{"count":1,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/375\/revisions"}],"predecessor-version":[{"id":376,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/375\/revisions\/376"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/parts\/353"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/375\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/media?parent=375"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapter-type?post=375"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/contributor?post=375"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/license?post=375"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}