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.

Strategy

This person’s 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.

Solution

The work done on the person by the floor as he stops is given by

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{180º}=-1\right)[/latex]. The floor removes energy from the system, so it does negative work.

The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height [latex]h[/latex]:

The distance [latex]d[/latex] that the person’s 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.

The work [latex]W[/latex] done by the floor on the person stops the person and brings the person’s kinetic energy to zero:

Combining this equation with the expression for [latex]W[/latex] gives

Recalling that [latex]h[/latex] is negative because the person fell down, the force on the knee joints is given by

Discussion

Such a large force (500 times more than the person’s 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 [link].)

(a) What is the final speed of the roller coaster shown in [link] 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?

Strategy

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 loss of gravitational potential energy from moving downward through a distance [latex]h[/latex] equals the gain in kinetic energy. This can be written in equation form as [latex]-\text{Δ}{\text{PE}}_{\text{g}}=\text{Δ}\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.

Solution for (a)

Here the initial kinetic energy is zero, so that [latex]\text{ΔKE}=\frac{1}{2}{\text{mv}}^{2}[/latex]. The equation for change in potential energy states that [latex]{\text{ΔPE}}_{\text{g}}=\text{mgh}[/latex]. Since [latex]h[/latex] is negative in this case, we will rewrite this as [latex]{\text{ΔPE}}_{\text{g}}=-\text{mg}\mid h\mid[/latex] to show the minus sign clearly. Thus,

becomes

Solving for [latex]v[/latex], we find that mass cancels and that

Substituting known values,

Solution for (b)

Again [latex]-{\text{ΔPE}}_{\text{g}}=\text{ΔKE}[/latex]. In this case there is initial kinetic energy, so [latex]\text{ΔKE}=\frac{1}{2}m{v}^{2}-\frac{1}{2}m{{v}_{0}}^{2}[/latex]. Thus,

Rearranging gives

This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass again cancels, and

This equation is very similar to the kinematics equation [latex]v=\sqrt{{{v}_{0}}^{2}+2\text{ad}}[/latex], but it is more general—the 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

Discussion and Implications

First, note that mass cancels. This is quite consistent with observations made in Falling Objects 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 any height along the way by simply using the appropriate value of [latex]h[/latex] at the point of interest.