8.1 Potential Energy of a System

In Work, we saw that the work done on an object by the constant gravitational force, near the surface of Earth, over any displacement is a function only of the difference in the positions of the end-points of the displacement. This property allows us to define a different kind of energy for the system than its kinetic energy, which is called potential energy. We consider various properties and types of potential energy in the following subsections.

Potential Energy Basics

In Motion in Two and Three Dimensions, we analyzed the motion of a projectile, like kicking a football in Figure 8.2. For this example, let’s ignore friction and air resistance and begin the analysis immediately after the kick, at the start of the projectile motion of the ball. As the football rises, the work done by the gravitational force on the football is negative, because the ball’s displacement is positive vertically and the force due to gravity is negative vertically. We also noted that the ball slowed down until it reached its highest point in the motion, thereby decreasing the ball’s kinetic energy. This loss in kinetic energy translates to a gain in gravitational potential energy of the football-Earth system.

As the football falls toward Earth, the work done on the football is now positive, because the displacement and the gravitational force both point vertically downward. The ball also speeds up, which indicates an increase in kinetic energy. Therefore, energy is converted from gravitational potential energy back into kinetic energy.

An illustration of a football’s trajectory and energy. The kicker kicks the ball, doing work on it and giving it maximum kinetic energy. The potential energy is minimum. This is point one. On the way up, at point two, the kinetic energy of the ball decreases and its potential energy decreases. At the highest point, point three, the kinetic energy of the ball is at its minimum and its potential energy is maximum. As the ball descends, point four, the kinetic energy increases and the potential energy decreases. The receiver catches the ball at the same height above the ground as it was kicked, at point five. The kinetic energy equals maximum, potential energy is minimum. — **Figure 8.2** As a football starts its descent toward the wide receiver, gravitational potential energy is converted back into kinetic energy.

Based on this scenario, we can define the difference of potential energy from point A to point B as the negative of the work done:

$Δ U_{A B} = U_{B} - U_{A} = − W_{A B} .$

(8.1)

This formula explicitly states a potential energy difference, not just an absolute potential energy. Therefore, we need to define potential energy at a given position in such a way as to state standard values of potential energy on their own, rather than potential energy differences. We do this by rewriting the potential energy function in terms of an arbitrary constant,

$Δ U = U (\vec{r}) - U ({\vec{r}}_{0}) .$

(8.2)

The choice of the potential energy at a starting location of $r → sub 0$ is made out of convenience in the given problem. Most importantly, whatever choice is made should be stated and kept consistent throughout the given problem. There are some well-accepted choices of initial potential energy. For example, the lowest height in a problem is usually defined as zero potential energy, or if an object is in space, the farthest point away from the system is often defined as zero potential energy. Then, the potential energy, with respect to zero at $r → sub 0,$ is just $U (\vec{r}) .$

As long as there is no friction or air resistance, the change in kinetic energy of the football equals negative of the change in gravitational potential energy of the football. This can be generalized to any potential energy:

$Δ K_{A B} = - Δ U_{A B} .$

(8.3)

Let’s look at a specific example, choosing zero potential energy for gravitational potential energy at convenient points.

Example 1 Basic Properties of Potential Energy

A particle moves along the x-axis under the action of a force given by $F equals − a x squared$ , where $a equals 3 N/m squared$ . (a) What is the difference in its potential energy as it moves from $x sub A equals 1 m$ to $x sub B equals 2 m$ ? (b) What is the particle’s potential energy at $x equals 1 m$ with respect to a given 0.5 J of potential energy at $x equals 0$ ?

Strategy (a) The difference in potential energy is the negative of the work done, as defined by Equation 8.1. The work is defined in the previous chapter as the dot product of the force with the distance. Since the particle is moving forward in the x-direction, the dot product simplifies to a multiplication ( $\hat{i} \cdot \hat{i} = 1$ ). To find the total work done, we need to integrate the function between the given limits. After integration, we can state the work or the change in potential energy. (b) The potential energy function, with respect to zero at $x equals 0$ , is the indefinite integral encountered in part (a), with the constant of integration determined from Equation 8.3. Then, we substitute the x-value into the function of potential energy to calculate the potential energy at $x equals 1 m.$

Solution

The work done by the given force as the particle moves from coordinate x to $x + d x$ in one dimension is
$d W = \vec{F} \cdot d \vec{r} = F d x = − a x^{2} d x .$
Substituting this expression into Equation 8.1, we obtain
$Δ U = − W = \int_{x_{1}}^{x_{2}} a x^{2} d x = \frac{1}{3} (3 {N/m}^{2}) {x^{3} |}_{1 m}^{2 m} = 7 J .$
The indefinite integral for the potential energy function in part (a) is
$U (x) = \frac{1}{3} a x^{3} + const .,$
and we want the constant to be determined by
$U open parenthesis 0 close parenthesis equals 0.5 J.$
Thus, the potential energy with respect to zero at $x equals 0$ is just
$U (x) = \frac{1}{3} a x^{3} + 0.5 J .$
Therefore, the potential energy at $x equals 1 m$ is
$U (1 m) = \frac{1}{3} (3 {N/m}^{2}) {(1 m)}^{3} + 0.5 J = 1.5 J .$

Significance In this one-dimensional example, any function we can integrate, independent of path, is conservative. Notice how we applied the definition of potential energy difference to determine the potential energy function with respect to zero at a chosen point. Also notice that the potential energy, as determined in part (b), at $x equals 1 m$ is $U open parenthesis 1 m close parenthesis equals 1 J$ and at $x equals 2 m$ is $U open parenthesis 2 m close parenthesis equals 8 J$ ; their difference is the result in part (a).

Systems of Several Particles

In general, a system of interest could consist of several particles. The difference in the potential energy of the system is the negative of the work done by gravitational or elastic forces, which, as we will see in the next section, are conservative forces. The potential energy difference depends only on the initial and final positions of the particles, and on some parameters that characterize the interaction (like mass for gravity or the spring constant for a Hooke’s law force).

It is important to remember that potential energy is a property of the interactions between objects in a chosen system, and not just a property of each object. This is especially true for electric forces, although in the examples of potential energy we consider below, parts of the system are either so big (like Earth, compared to an object on its surface) or so small (like a massless spring), that the changes those parts undergo are negligible when included in the system.

Types of Potential Energy

For each type of interaction present in a system, you can label a corresponding type of potential energy. The total potential energy of the system is the sum of the potential energies of all the types. (This follows from the additive property of the dot product in the expression for the work done.) Let’s look at some specific examples of types of potential energy discussed in Work. First, we consider each of these forces when acting separately, and then when both act together.

Gravitational potential energy near Earth’s surface

The system of interest consists of our planet, Earth, and one or more particles near its surface (or bodies small enough to be considered as particles, compared to Earth). The gravitational force on each particle (or body) is just its weight mg near the surface of Earth, acting vertically down. According to Newton’s third law, each particle exerts a force on Earth of equal magnitude but in the opposite direction. Newton’s second law tells us that the magnitude of the acceleration produced by each of these forces on Earth is mg divided by Earth’s mass. Since the ratio of the mass of any ordinary object to the mass of Earth is vanishingly small, the motion of Earth can be completely neglected. Therefore, we consider this system to be a group of single-particle systems, subject to the uniform gravitational force of Earth.

In Work, the work done on a body by Earth’s uniform gravitational force, near its surface, depended on the mass of the body, the acceleration due to gravity, and the difference in height the body traversed, as given by Equation 7.4. By definition, this work is the negative of the difference in the gravitational potential energy, so that difference is

$Δ U = - W_{A B} = m g (y_{B} - y_{A}) .$

(8.4)

You can see from this that the gravitational potential energy function, near Earth’s surface, is

$U (y) = m g y + const .$

(8.5)

You can choose the value of the constant, as described in the discussion of Equation 8.2; however, for solving most problems, the most convenient constant to choose is zero for when $y equals 0,$ which is the lowest vertical position in the problem.

Example 2 Gravitational Potential Energy of a Hiker

The summit of Great Blue Hill in Milton, MA, is 147 m above its base and has an elevation above sea level of 195 m (Figure 8.3). (Its Native American name, Massachusett, was adopted by settlers for naming the Bay Colony and state near its location.) A 75-kg hiker ascends from the base to the summit. What is the gravitational potential energy of the hiker-Earth system with respect to zero gravitational potential energy at base height, when the hiker is (a) at the base of the hill, (b) at the summit, and (c) at sea level, afterward?

Figure 8.3 Sketch of the profile of Great Blue Hill, Milton, MA. The altitudes of the three levels are indicated.

Strategy First, we need to pick an origin for the y-axis and then determine the value of the constant that makes the potential energy zero at the height of the base. Then, we can determine the potential energies from Equation 8.5, based on the relationship between the zero potential energy height and the height at which the hiker is located.

Solution

Let’s choose the origin for the y-axis at base height, where we also want the zero of potential energy to be. This choice makes the constant equal to zero and
$U (base) = U (0) = 0 .$
At the summit, $y equals 147 m$ , so
$U (summit) = U (147 m) = m g h = (75 \times 9.8 N) (147 m) = 108 kJ .$
At sea level, $y = (147 - 195) m = −48 m$ , so
$U (sea-level) = (75 \times 9.8 N) (−48 m) = −35.3 kJ .$

Significance Besides illustrating the use of Equation 8.4 and Equation 8.5, the values of gravitational potential energy we found are reasonable. The gravitational potential energy is higher at the summit than at the base, and lower at sea level than at the base. Gravity does work on you on your way up, too! It does negative work and not quite as much (in magnitude), as your muscles do. But it certainly does work. Similarly, your muscles do work on your way down, as negative work. The numerical values of the potential energies depend on the choice of zero of potential energy, but the physically meaningful differences of potential energy do not. [Note that since Equation 8.2 is a difference, the numerical values do not depend on the origin of coordinates.]

Elastic potential energy

In Work, we saw that the work done by a perfectly elastic spring, in one dimension, depends only on the spring constant and the squares of the displacements from the unstretched position, as given in Equation 7.5. This work involves only the properties of a Hooke’s law interaction and not the properties of real springs and whatever objects are attached to them. Therefore, we can define the difference of elastic potential energy for a spring force as the negative of the work done by the spring force in this equation, before we consider systems that embody this type of force. Thus,

$Δ U = − W_{A B} = \frac{1}{2} k (x_{B}^{2} - x_{A}^{2}),$

(8.6)

where the object travels from point A to point B. The potential energy function corresponding to this difference is

$U (x) = \frac{1}{2} k x^{2} + const .$

(8.7)

If the spring force is the only force acting, it is simplest to take the zero of potential energy at $x equals 0$ , when the spring is at its unstretched length. Then, the constant in Equation 8.7 is zero. (Other choices may be more convenient if other forces are acting.)

Example 3 Spring Potential Energy

A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm. (a) How much elastic potential energy does the spring contribute when its length is 23 cm? (b) How much more potential energy does it contribute if its length increases to 26 cm?

Strategy When the spring is at its unstretched length, it contributes nothing to the potential energy of the system, so we can use Equation 8.7 with the constant equal to zero. The value of x is the length minus the unstretched length. When the spring is expanded, the spring’s displacement or difference between its relaxed length and stretched length should be used for the x-value in calculating the potential energy of the spring.

Solution

The displacement of the spring is $x = 23 cm - 20 cm = 3 cm$ , so the contributed potential energy is $U = \frac{1}{2} k x^{2} = \frac{1}{2} (4 N/cm) {(3 cm)}^{2} = 0.18 J$ .
When the spring’s displacement is $x = 26 cm - 20 cm = 6 cm$ , the potential energy is $U = \frac{1}{2} k x^{2} = \frac{1}{2} (4 N/cm) {(6 cm)}^{2} = 0.72 J$ , which is a 0.54-J increase over the amount in part (a).

Significance Calculating the elastic potential energy and potential energy differences from Equation 8.7 involves solving for the potential energies based on the given lengths of the spring. Since U depends on $x^{2}$ , the potential energy for a compression (negative x) is the same as for an extension of equal magnitude.

Gravitational and elastic potential energy

A simple system embodying both gravitational and elastic types of potential energy is a one-dimensional, vertical mass-spring system. This consists of a massive particle (or block), hung from one end of a perfectly elastic, massless spring, the other end of which is fixed, as illustrated in Figure 8.4.

A vertical mass spring system is illustrated. The upper end of the spring is attached to the ceiling. A block of mass m is attached to the lower end. The spring is drawn at two positions. On the left, the mass is in the equilibrium position. To the right of this, the spring is drawn with the mass pulled down a distance y sub pull. This position of the mass is labeled as h equal to zero. A graph of y as a function of X is shown to the rightly the illustrations, with y equals zero aligned with the equilibrium position in the illustrations. The plot is sinusoidal, with the minimum y at x=0 and even with the lower mass position in the illustrations. — **Figure 8.4** A vertical mass-spring system, with the positive y-axis pointing upward. The mass is initially at an unstretched spring length, point A. Then it is released, expanding past point B to point C, where it comes to a stop.

First, let's consider the potential energy of the system. We need to define the constant in the potential energy function of Equation 8.5. Often, the ground is a suitable choice for when the gravitational potential energy is zero; however, in this case, the highest point or when $y equals 0$ is a convenient location for zero gravitational potential energy. Note that this choice is arbitrary, and the problem can be solved correctly even if another choice is picked.

We must also define the elastic potential energy of the system and the corresponding constant, as detailed in Equation 8.7. This is where the spring is unstretched, or at the $y equals 0$ position.

If we consider that the total energy of the system is conserved, then the energy at point A equals point C. The block is placed just on the spring so its initial kinetic energy is zero. By the setup of the problem discussed previously, both the gravitational potential energy and elastic potential energy are equal to zero. Therefore, the initial energy of the system is zero. When the block arrives at point C, its kinetic energy is zero. However, it now has both gravitational potential energy and elastic potential energy. Therefore, we can solve for the distance y that the block travels before coming to a stop:

$\begin{matrix} K_{A} & + & U_{A} = K_{C} + U_{C} \\ 0 & = & 0 + m g y_{C} + {\frac{1}{2} k (y_{C})}^{2} \\ y_{C} & = & \frac{- 2 m g}{k} \end{matrix}$

A photograph of a bungee jumper. — **Figure 8.5** A bungee jumper transforms gravitational potential energy at the start of the jump into elastic potential energy at the bottom of the jump.

Example 4 Potential Energy of a Vertical Mass-Spring System

A block weighing $1.2 N$ is hung from a spring with a spring constant of $6.0 N / m$ , as shown in Figure 8.4. (a) What is the maximum expansion of the spring, as seen at point C? (b) What is the total potential energy at point B, halfway between A and C? (c) What is the speed of the block at point B?

Strategy In part (a) we calculate the distance $y_{C}$ as discussed in the previous text. Then in part (b), we use half of the y value to calculate the potential energy at point B using equations Equation 8.4 and Equation 8.6. This energy must be equal to the kinetic energy, Equation 7.6, at point B since the initial energy of the system is zero. By calculating the kinetic energy at point B, we can now calculate the speed of the block at point B.

Solution

Since the total energy of the system is zero at point A as discussed previously, the maximum expansion of the spring is calculated to be:
$\begin{matrix} y_{C} & = & \frac{- 2 m g}{k} \\ y_{C} & = & \frac{- 2 (1.2 N)}{(6.0 N/m)} = -0.40 m \end{matrix}$
The position of $y_{B}$ is half of the position at $y_{C}$ or $y B = y C / 2 = - m g / k =$ . The total potential energy at point B would therefore be:
$\begin{matrix} U_{B} & = & m g (y_{C} / 2) + \frac{1}{2} k (y_{C} / 2)^{2} \\ U_{B} & = & m g (- m g / k) + (\frac{1}{2}) k (- m g / k)^{2} \\ U_{B} & = & - (m g)^{2} / (2 k) \\ U_{B} & = & (1.2 N) (-0.20 m) + {\frac{1}{2} (6 N/m) (- 0.20 m)}^{2} \\ U_{B} & = & - 0.12 J where we have used y_{C} = - 2 m g / k from part (a) \end{matrix}$
The mass of the block is the weight divided by gravity.

$m = \frac{F_{w}}{g} = \frac{1.2 N}{9.8 {m/s}^{2}} = 0.12 kg$
The kinetic energy at point B therefore is 0.12 J because the total energy is zero. Therefore, the speed of the block at point B is equal to
$\begin{matrix} K = & \frac{1}{2} m v^{2} \\ v = & \sqrt{\frac{2 K}{m}} = \sqrt{\frac{2 (0.12 J)}{(0.12 kg)}} = 1.4 m/s \end{matrix}$

Significance Even though the potential energy due to gravity is relative to a chosen zero location, the solutions to this problem would be the same if the zero energy points were chosen at different locations.

A sample chart of a variety of energies is shown in Table 8.1 to give you an idea about typical energy values associated with certain events. Some of these are calculated using kinetic energy, whereas others are calculated by using quantities found in a form of potential energy that may not have been discussed at this point.

**Table 8.1** Energy of Various Objects and Phenomena
Object/phenomenon	Energy in joules
Big Bang	$10 to the power 68$
Annual world energy use	$4.0 \times 10^{20}$
Large fusion bomb (9 megaton)	$3.8 \times 10^{16}$
Hiroshima-size fission bomb (10 kiloton)	$4.2 \times 10^{13}$
1 barrel crude oil	$5.9 \times 10^{9}$
1 metric ton TNT	$4.2 \times 10^{9}$
1 gallon of gasoline	$1.2 \times 10^{8}$
Daily adult food intake (recommended)	$1.2 \times 10^{7}$
1000-kg car at 90 km/h	$3.1 \times 10^{5}$
Tennis ball at 100 km/h	$22$
Mosquito $open parenthesis 10 to the power −2 g at 0.5 m/s close parenthesis$	$1.3 \times 10^{−6}$
Single electron in a TV tube beam	$4.0 \times 10^{−15}$
Energy to break one DNA strand	$10 to the power −19$
1 kg hydrogen (fusion to helium)	$6.4 \times 10^{14}$
1 kg uranium (nuclear fission)	$8.0 \times 10^{13}$

Summary

For a single-particle system, the difference of potential energy is the opposite of the work done by the forces acting on the particle as it moves from one position to another.
Since only differences of potential energy are physically meaningful, the zero of the potential energy function can be chosen at a convenient location.
The potential energies for Earth’s constant gravity, near its surface, and for a Hooke’s law force are linear and quadratic functions of position, respectively.

Conceptual Questions

The kinetic energy of a system must always be positive or zero. Explain whether this is true for the potential energy of a system.

The potential energy of a system can be negative because its value is relative to a defined point.

The force exerted by a diving board is conservative, provided the internal friction is negligible. Assuming friction is negligible, describe changes in the potential energy of a diving board as a swimmer drives from it, starting just before the swimmer steps on the board until just after their feet leave it.

Describe the gravitational potential energy transfers and transformations for a javelin, starting from the point at which an athlete picks up the javelin and ending when the javelin is stuck into the ground after being thrown.

If the reference point of the ground is zero gravitational potential energy, the javelin first increases its gravitational potential energy, followed by a decrease in its gravitational potential energy as it is thrown until it hits the ground. The overall change in gravitational potential energy of the javelin is zero unless the center of mass of the javelin is lower than from where it is initially thrown, and therefore would have slightly less gravitational potential energy.

A couple of soccer balls of equal mass are kicked off the ground at the same speed but at different angles. Soccer ball A is kicked off at an angle slightly above the horizontal, whereas ball B is kicked slightly below the vertical. How do each of the following compare for ball A and ball B? (a) The initial kinetic energy and (b) the change in gravitational potential energy from the ground to the highest point? If the energy in part (a) differs from part (b), explain why there is a difference between the two energies.

What is the dominant factor that affects the speed of an object that started from rest down a frictionless incline if the only work done on the object is from gravitational forces?

the vertical height from the ground to the object

Two people observe a leaf falling from a tree. One person is standing on a ladder and the other is on the ground. If each person were to compare the energy of the leaf observed, would each person find the following to be the same or different for the leaf, from the point where it falls off the tree to when it hits the ground: (a) the kinetic energy of the leaf; (b) the change in gravitational potential energy; (c) the final gravitational potential energy?

Problems

Using values from Table 8.1, how many DNA molecules could be broken by the energy carried by a single electron in the beam of an old-fashioned TV tube? (These electrons were not dangerous in themselves, but they did create dangerous X-rays. Later-model tube TVs had shielding that absorbed X-rays before they escaped and exposed viewers.)

40,000

If the energy in fusion bombs were used to supply the energy needs of the world, how many of the 9-megaton variety would be needed for a year’s supply of energy (using data from this equation)?

A camera weighing 10 N falls from a small drone hovering $20 m$ overhead and enters free fall. What is the gravitational potential energy change of the camera from the drone to the ground if you take a reference point of (a) the ground being zero gravitational potential energy? (b) The drone being zero gravitational potential energy? What is the gravitational potential energy of the camera (c) before it falls from the drone and (d) after the camera lands on the ground if the reference point of zero gravitational potential energy is taken to be a second person looking out of a building $30 m$ from the ground?

$a. −200 J; b. −200 J; c. −100 J; d. −300 J$

Someone drops a $50 minus g$ pebble off of a docked cruise ship, $70.0 m$ above the water level. A person on a dock $3.0 m$ above the water level holds out a net to catch the pebble. (a) How much work is done on the pebble by gravity during the drop? (b) What is the change in the gravitational potential energy during the drop? If the gravitational potential energy is zero at the water line, what is the gravitational potential energy (c) when the pebble is dropped? (d) When it reaches the net? What if the gravitational potential energy was $30.0$ Joules at water level? (e) Find the answers to the same questions in (c) and (d).

A cat’s crinkle ball toy of mass $15 g$ is thrown straight up with an initial speed of $3 m/s$ . Assume in this problem that air drag is negligible. (a) What is the kinetic energy of the ball as it leaves the hand? (b) How much work is done by the gravitational force during the ball’s rise to its peak? (c) What is the change in the gravitational potential energy of the ball during the rise to its peak? (d) If the gravitational potential energy is taken to be zero at the point where it leaves your hand, what is the gravitational potential energy when it reaches the maximum height? (e) What if the gravitational potential energy is taken to be zero at the maximum height the ball reaches, what would the gravitational potential energy be when it leaves the hand? (f) What is the maximum height the ball reaches?

$a. 0.068 J; b. −0.068 J; c. 0.068 J; d. 0.068 J; e. −0.068 J; f. 46 cm$