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Learning Objectives

By the end of this section, you will be able to do the following:

Explain gravitational potential energy in terms of work done against gravity
Show that the gravitational potential energy of an object of mass m at height h on Earth is given by PEg = mgh
Show how knowledge of potential energy as a function of position can be used to simplify calculations and explain physical phenomena

The information presented in this section supports the following AP® learning objectives and science practices:

2.E.1.1 The student is able to construct or interpret visual representations of the isolines of equal gravitational potential energy per unit mass, and identify each line as a gravitational equipotential.
4.C.1.1 The student is able to calculate the total energy of a system and justify the mathematical routines used in the calculation of component types of energy within the system whose sum is the total energy. (S.P. 1.4, 2.1, 2.2)
5.B.1.1 The student is able to set up a representation or model showing that a single object can only have kinetic energy and use information about that object to calculate its kinetic energy. (S.P. 1.4, 2.2)
5.B.1.2 The student is able to translate between a representation of a single object, which can only have kinetic energy, and a system that includes the object, which may have both kinetic and potential energies. (S.P. 1.5)

Work Done Against Gravity

Climbing stairs and lifting objects is work in both the scientific and everyday sense—it 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.

Let us calculate the work done in lifting an object of mass $m$ through a height $h$ , such as in Figure 7.5. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight $mg$ . The work done on the mass is then $W = Fd = mgh$ . We define this to be gravitational potential energy $({PE}_{g})$ 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 ${PE}_{g}$ gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word system? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s 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. Since 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’s surface, but this point is arbitrary; what is important is the difference 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.

Converting Between Potential Energy and Kinetic Energy

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 $mgh$ 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 ${PE}_{g}$ to $KE$ without explicitly considering the intermediate step of work (see Example 7.7). This shortcut makes it is easier to solve problems using energy, when possible, rather than explicitly using forces.

(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.

Figure 7.5 (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.

More precisely, we define the change in gravitational potential energy $Δ {PE}_{g}$ to be

7.27

Δ {PE}_{g} = mgh,

where, for simplicity, we denote the change in height by $h$ rather than the usual $Δ h$ . Note that $h$ 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

7.28

\begin{array}{l} mgh & = & (0.500 kg) ({9.80 m/s}^{2}) (1.00 m) \\ = & 4.90 kg \cdot m^{2} {/s}^{2} = 4.90 J. \end{array}

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, without directly considering the force of gravity that does the work.

Using Potential Energy to Simplify Calculations

The equation $Δ {PE}_{g} = mgh$ applies for any path that has a change in height $h$ , not just when the mass is lifted straight up (see Figure 7.6). It is much easier to calculate $mgh$ , 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 $h$ of a mass $m$ is accompanied by a change in gravitational potential energy $mgh$ , and we will avoid the equivalent but more difficult task of calculating work done by or against the gravitational force.

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.

Figure 7.6 The change in gravitational potential energy

(Δ {PE}_{g})

between points A and B is independent of the path.

Δ {PE}_{g} = mgh

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.

Example 7.6 The Force to Stop Falling

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 ${PE}_{g}$ is transformed into $KE$ 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

7.29

W = Fd cos θ = - Fd,

with a minus sign because the displacement while stopping and the force from floor are in opposite directions $(cos θ = cos 180º = - 1) .$ 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 $h$

7.30

KE = - Δ {PE}_{g} = - mgh .

The distance $d$ that the person’s knees bend is much smaller than the height $h$ of the fall, so the additional change in gravitational potential energy during the knee bend is ignored.

The work $W$ done on the person by the floor stops the person and brings the person’s kinetic energy to zero

7.31

W = - KE = mgh .

Combining this equation with the expression for $W$ gives

7.32

- Fd = mgh .

Recalling that $h$ is negative because the person fell down, the force on the knee joints is given by

7.33

F = - \frac{mgh}{d} = - \frac{(60.0 kg) (9.80 m {/s}^{2}) (- 3 . 00 m)}{5 . 00 \times 10^{- 3} m} = 3 . 53 \times 10^{5} N.

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 bending its hind legs in each jump (see Figure 7.7).

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.

Figure 7.7 The work done on the kangaroo by the ground reduces the kangaroo's 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)

Example 7.7 Finding the Speed of a Roller Coaster from Its Height

(a) What is the final speed of the roller coaster shown in Figure 7.8 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?

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’s 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.

Figure 7.8 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’s gravitational potential energy is converted to kinetic energy. If work done by friction is negligible, all

Δ {PE}_{g}

is converted to

KE

.

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 $h$ equals the gain in kinetic energy. This can be written in equation form as $- Δ {PE}_{g} = Δ KE$ . Using the equations for ${PE}_{g}$ and $KE$ , we can solve for the final speed $v$ , which is the desired quantity.

Solution for (a)

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

7.34

- Δ {PE}_{g} = Δ KE

becomes

7.35

mg ∣ h ∣ = \frac{1}{2} {mv}^{2} .

Solving for $v$ , we find that mass cancels and that

7.36

v = \sqrt{2 g ∣ h ∣} .

Substituting known values,

7.37

\begin{array}{l} v & = & \sqrt{2 (9 . 80 m {/s}^{2}) (20.0 m)} \\ = & 19 .8 m/s. \end{array}

Solution for (b)

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

7.38

mg ∣ h ∣ = \frac{1}{2} {mv}^{2} - \frac{1}{2} m {v_{0}}^{2} .

Rearranging gives

7.39

\frac{1}{2} {mv}^{2} = mg ∣ h ∣ + \frac{1}{2} m {v_{0}}^{2} .

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

7.40

v = \sqrt{2 g ∣ h ∣ + {v_{0}}^{2}} .

This equation is very similar to the kinematics equation $v = \sqrt{{v_{0}}^{2} + 2 ad}$ , 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

7.41

\begin{array}{l} v & = & \sqrt{2 (9.80 {m/s}^{2}) (20 .0 m) + (5 .00 m/s)^{2}} \\ = & 20.4 m/s. \end{array}

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 h at the point of interest. While changes in the potential and kinetic energies depend only on h, changes in the potential and kinetic energies, expressed in terms of other quantities like time t or horizontal distance x, depend on constraints defined by how the roller coaster is constructed. The height h, for example, can be considered a function of x that essentially describes the design of the roller coaster.

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.

Making Connections: Take-Home Investigation—Converting Potential to Kinetic Energy

You 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 Figure 7.9). 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’s kinetic energy at the bottom is proportional to its potential energy at the release point.

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.

Figure 7.9 A marble rolls down a ruler, and its speed on the level surface is measured.

Newton’s Universal Law of Gravitation and Gravitational Potential Energy

Near the surface of Earth, where the gravitational force on an object of mass m is given by $F = m g$ , there is an associated gravitational potential energy, $Δ P E_{g} = m g h$ , where h is the height above some reference value (e.g., sea level), and the potential is defined to be zero at that reference height ( $h = 0$ ). In chapter 6 we learned that the magnitude of the gravitational force between two bodies having masses m and M, with a distance r between their centers of mass, is given by the equation $F = G \frac{m M}{r^{2}}$ . Again, in this case, an associated gravitational potential energy can be determined by

7.42

Δ P E_{g} = - G \frac{m M}{r},

where the potential energy approaches zero as r approaches infinity.

Conservation of energy principles can again be used to solve many problems of practical interest. Suppose you want to launch an object from Earth’s surface with just enough energy to escape Earth’s gravitational influence. At Earth’s surface, the total energy will be $E_{i} = \frac{1}{2} m v^{2} - G \frac{m M}{R}$ , where M and R are Earth’s mass and radius, respectively. The magnitude of the object’s velocity v will drop toward zero as r approaches infinity, leading to a final energy of $E_{f} = 0$ . Setting the two equal to each other and solving for v gives

7.43

v = \sqrt{\frac{2 G M}{R}} .

Substituting in values for G, M, and R gives $v = 11.2 \times 10^{3}$ m/s, which is the escape velocity for objects launched from Earth.

Gravitational Potential

The potential energy $Δ P E_{g}$ is proportional to the test mass m. To have a physical quantity that is independent of test mass, we define the gravitational potential $V_{g}$ to be the potential energy per unit mass. Near the surface of Earth, the gravitational potential is given by

7.44

V_{g} = g h.

The more general form for the potential due to an object of mass M, derived from Newton’s universal law of gravitation, is

7.45

V_{g} = - \frac{G M}{r},

where r is the distance from the object’s center of mass. Since the gravitational potential is a scalar quantity, the potential described as a function of location in three-dimensional space corresponds to a scalar field.

When we are interested in the influence of multiple masses on a test mass, the gravitational potential at any given point is simply the sum of the gravitational potentials of each individual object. Suppose that two point objects of mass M are located along the x-axis at $x = \pm a$ . The gravitational potential at any point in the x-y-plane is given by

7.46

V_{g} (x, y) = - \frac{G M}{\sqrt{(x - a) {}^{2}+ y^{2}}} - \frac{G M}{\sqrt{(x + a) {}^{2}+ y^{2}}} .

There are a couple of ways to visualize a function of the sort represented by the preceding equation. You can, for example, look at a 2-D plot for a specific x- or y-value. For example, if we want to look at the potential only along the x-axis, we can set $y = 0$ , which results in

7.47

V_{g} (x, y) = - \frac{G M}{| x - a |} - \frac{G M}{| x + a |} .

Enter this formula into a real or online graphing calculator. The origin is shown at the top of the graph, the locations of the two objects at $x = \pm a$ are noted along the x-axis, and the gravitational potential is plotted in arbitrary units. Keeping in mind that the potential represents the potential energy per unit mass of a test object, you can envision what would happen to such a test object located at some particular point along the x-axis. An object placed anywhere to the left of the origin will fall into the potential well (i.e., be drawn to the object at $x = - a$ ). An object placed to the right of the origin will be drawn to the object at $x = + a$ . An object located precisely at $x = 0$ will be in a state of unstable equilibrium.

Note also that details of the curve in the above plot provide information about the location and relative magnitude of the two masses. Even without knowing the function describing the potential, the location of the potential wells in the plot make clear the locations of the objects. The left-right symmetry of the plot also indicates that the masses of the two objects are equal.

Another way to visualize the potential is to draw a contour plot of the potential in a given plane. Figure 7.10 shows the gravitational potential energy of three objects.

This is a graph of a contour plot, showing isolines of gravitational potential for a collection of three point objects: object A (-1, 0), object B (0, 1), and object C (1, 0). The lines are most widely spaced for object B, and most closely spaced for object C.

Figure 7.10 A gravitational contour plot showing the gravitational potential of three masses.

The gravitational potential is constant along each of the lines, which are known as isolines. The potentials are in arbitrary units, with the outermost red line corresponding to a negative potential with relatively small magnitude. The innermost green lines correspond to negative potentials with relatively large magnitudes. The remaining lines correspond to equally spaced intermediate values of the potential. The locations of the three objects are clear from the contour plot, and the symmetry across the y-axis shows that their masses are not equal. Like the contour lines on a topographic map, the relative spacing between adjacent isolines represents how rapidly the potential changes with location. This information provides insight into the direction and magnitude of the gravitational force a test mass would experience at any particular point in the x-y-plane.