Unit 3 · Topic 3.3

3.3 Potential Energy

Potential energy is energy stored in the arrangement of a system of objects that interact through conservative forces, like gravity or a spring. The force and the potential energy are two views of the same thing: $F_x = -\frac{dU}{dx}$ . On a potential energy graph, the slope gives the force, minimums are stable equilibrium points and maximums are unstable ones.

Key terms

potential energy
conservative force
potential energy graph
stable equilibrium
unstable equilibrium
zero of potential energy

Potential energy belongs to a system

Potential energy, U, belongs to a system of two or more interacting objects, not to one object. A raised book has gravitational potential energy only as part of the book–Earth system. If your system is the book alone, gravity is an outside force doing work instead.

Only conservative forces have a potential energy. The change in potential energy is the negative of the work the conservative force does inside the system:

$\Delta U = -\int_a^b \vec{F}\cdot d\vec{r}$

So if gravity does positive work (an object falls), the system's potential energy drops. Only changes in U are physical; you choose where U = 0, and the choice doesn't affect any answer as long as you stay consistent. With three or more objects, add up the potential energy of every interacting pair.

The potential energies you need

System	Potential energy	Usual zero
ideal spring	$U_s = \frac{1}{2}k(\Delta x)^2$	relaxed length
object and Earth, near the surface	$\Delta U_g = mg\Delta y$	any convenient height
two spherical masses, any distance	$U_G = -\frac{Gm_1m_2}{r}$	infinitely far apart

Where the formulas come from

Spring: the spring force is −kx, so $\Delta U = -\int_0^x (-kx)\,dx = \frac{1}{2}kx^2$ . It's positive whether the spring is stretched or compressed.

Gravity between spheres: the force on m from M is $-\frac{GMm}{r^2}$ (toward M). Setting U = 0 at infinity, $U_G(r) = -\int_\infty^r \left(-\frac{GMm}{r'^2}\right)dr' = -\frac{GMm}{r}$ . It's negative everywhere and rises toward zero as the objects move apart, because you'd have to add energy to separate them. Near Earth's surface, its change over a small height h is very close to mgh, which is where $\Delta U_g = mg\Delta y$ comes from.

Force from potential energy

Going the other way, the force is the negative slope of the potential energy curve:

$F_x = -\frac{dU}{dx}$

The force always points "downhill" on a U(x) graph, toward lower potential energy, like a ball rolling in a valley. Where the graph is steepest, the force is largest.

Where the slope is zero, the force is zero: an equilibrium point. At a minimum of U, a small nudge produces a force back toward the point, so it's a stable equilibrium (think of a ball in a bowl). At a maximum, a nudge produces a force away from it, so it's an unstable equilibrium (a ball balanced on a hilltop).

If you draw the system's total mechanical energy E as a horizontal line on the U(x) graph, the kinetic energy at any x is the gap E − U. The object can only be where U ≤ E. Where the line meets the curve, K = 0 and the object turns around; these are turning points.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Reading forces and equilibria from U(x)
A particle's potential energy is $U(x) = 2x^3 - 6x$ (U in joules, x in meters). Find the force as a function of x, the equilibrium positions, whether each is stable or unstable, and the force at x = 2.0 m.
Show the solutionHide the solution
1. Step 1: $F(x) = -\frac{dU}{dx} = -(6x^2 - 6) = 6 - 6x^2$ N.
2. Step 2: Equilibrium where F = 0: $6x^2 = 6$ , so x = 1 m and x = −1 m.
3. Step 3: Classify with the curvature, $\frac{d^2U}{dx^2} = 12x$ . At x = 1 it's positive, so U has a minimum (U = −4 J): stable. At x = −1 it's negative, a maximum (U = +4 J): unstable.
4. Step 4: At x = 2.0 m: F = 6 − 24 = −18 N, pointing in the −x direction, back toward the stable point at x = 1.
Answer: $F = 6 - 6x^2$ N; stable at x = 1 m, unstable at x = −1 m; F(2.0) = −18 N.
Example 2Calculator allowed
When mgh isn't good enough (classic trap)
A 1000 kg satellite is moved from Earth's surface (radius $R = 6.37 \times 10^6$ m) to a distance 2R from Earth's center. Earth's mass is $5.97 \times 10^{24}$ kg. Find the change in gravitational potential energy, and compare with mgh using h = R.
Show the solutionHide the solution
1. Step 1: $\Delta U_G = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right) = \frac{GMm}{2R}$ .
2. Step 2: $\Delta U_G = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)}{2(6.37 \times 10^6)} \approx 3.13 \times 10^{10}$ J.
3. Step 3: mgh with g = 9.8 m/s² and h = R gives $(1000)(9.8)(6.37 \times 10^6) \approx 6.24 \times 10^{10}$ J, about twice as much. (It's exactly twice if you use $g = \frac{GM}{R^2}$ , because then mgR equals $\frac{GMm}{R}$ .)
4. Step 4: mgh assumes g stays 9.8 N/kg all the way up, but g falls off with distance. Use $-\frac{GMm}{r}$ whenever the height change isn't small compared with the planet's radius.
Answer: $\Delta U_G \approx 3.1 \times 10^{10}$ J; mgh would overestimate it by about a factor of 2.

Common mistakes

Giving potential energy to one object. It belongs to a system of interacting objects.
Forgetting the minus sign in $F_x = -\frac{dU}{dx}$ . The force points toward lower U.
Using mgh for large changes in height, such as launching to orbit. Use $U_G = -\frac{Gm_1m_2}{r}$ .
Thinking negative gravitational potential energy is impossible or an error. It's negative because of where zero was chosen.

On the exam

Expect a U(x) graph with questions about the force's direction, equilibrium points and turning points. Mark the total energy as a horizontal line before answering.
Derivations may ask you to get a force from a given U(x), or U(x) from a given force by integrating. Watch the sign in each direction.

Connected topics

Videos

Potential energy and conservative forces (part 1) | AP Physics | Khan Academy
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Check yourself

4 questions on 3.3 Potential Energy. Pick an answer to see if you got it, and why.

A particle moves along the x-axis under a single conservative force. The potential energy of the system is $U(x) = 3x^2 - 2x^3$ , with U in joules and x in meters.

Described situation

Question 1 of 4Calculator allowed

What is the force on the particle at x = 2.0 m?

Question 2 of 4Calculator allowed

Where are the particle's equilibrium positions?

Question 3 of 4Calculator allowed

Which statement about these equilibrium positions is correct?

Question 4 of 4Calculator allowed

A spring with k = 500 N/m is compressed 0.10 m from its relaxed length. How much elastic potential energy is stored?

0 of 4 answered