Potential Energy: Study Guide (Physics C: Mechanics 3.3)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 3.3 called Potential Energy. This guide is the calculus-based one. It builds each potential energy from an integral of force, gets forces back as derivatives, and uses the shape of U(x) to find and classify equilibrium positions. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise. This topic follows Topic 3.2, Work.

Potential energy belongs to a system

Kinetic energy belongs to one moving object. Potential energy is different: it is energy stored in the arrangement of a system of two or more objects that push or pull on each other.

Three conditions:

1. A system of two or more objects. A book on its own has no gravitational potential energy. The book–Earth system has it.
2. Conservative interactions only. The objects must interact through conservative forces, such as gravity or an ideal spring force. A force is conservative if the work it does between two points does not depend on the path. Equivalently, it does zero work around any closed path. Friction fails this test: a longer path means more negative work, so friction has no potential energy.
3. A scalar that depends on position. U has no direction. It depends only on where the objects are relative to each other, for example the separation r of two planets or the stretch Δx of a spring.

Defining U with an integral

Topic 3.2 gave the work done by a force as an integral. For a conservative force exerted between objects inside the system, the change in potential energy is defined as

ΔU = −W_c = −∫ F_c · dr (from the initial to the final position)

The minus sign matters. When a conservative force does positive work (gravity on a falling ball), the system’s potential energy decreases. The energy has not vanished: Topic 3.4 shows it becomes kinetic energy.

Choosing the zero. The definition only fixes changes. You pick where U = 0 to make the analysis easier: the floor, the lowest point of a swing, the relaxed length of a spring, or infinite separation for planets. Every ΔU, and every physical prediction, comes out the same for any choice. Only the values of U itself change.

Force from potential energy: F_x = −dU/dx

In one dimension, ΔU = −∫F_x dx. Differentiate both sides to reverse the integral:

F_x = −dU/dx

Read it as a rule about slopes: the conservative force is minus the slope of the U(x) graph.

• Where U(x) slopes upward (dU/dx > 0), F_x is negative: the force points to −x.
• Where U(x) slopes downward, F_x is positive.
• Either way, the force points in the direction of decreasing potential energy, like a ball rolling downhill on the graph.
• Where the graph is flat (dU/dx = 0), there is no conservative force. That is an equilibrium position.

Equilibrium and stability on a U(x) graph

At an equilibrium position x_eq, dU/dx = 0. Now nudge the object a little and see which way the force acts.

• Stable equilibrium: after a small displacement, the force points back towards x_eq. This happens at a local minimum of U(x). Test: d²U/dx² > 0.
• Unstable equilibrium: after a small displacement, the force points away from x_eq, in the same direction as the displacement. This happens at a local maximum of U(x). Test: d²U/dx² < 0.

If the second derivative is zero, look at the graph shape on both sides instead.

Elastic potential energy of an ideal spring

Take +x along the spring, with x = Δx measured from the relaxed length. The spring force on the object is F_x = −kx (Topic 2.8). Choose U_s = 0 at the relaxed length and integrate:

U_s(x) = −∫₀ˣ (−kx′) dx′ = ½kx², usually written U_s = ½k(Δx)².

The energy is the same for a stretch and a compression of the same size, because Δx is squared. It also grows as the square of the deformation: double the stretch, four times the energy. Check the slope: −dU_s/dx = −kx, which is Hooke’s law again.

The system here is the spring and the object it is attached to. The spring stores energy because it changes shape reversibly.

Gravitational potential energy

General form. Two roughly spherical masses m₁ and m₂ attract with a force of size Gm₁m₂/r². Take r as the centre-to-centre separation. The force on m₂ points towards m₁, so its radial component is F_r = −Gm₁m₂/r². Choose U_g = 0 when the objects are infinitely far apart and bring them in from infinity:

U_g(r) = −∫_∞ʳ (−Gm₁m₂/r′²) dr′ = −Gm₁m₂/r

U_g is negative at every finite separation and rises towards zero as r grows. Moving the objects apart increases U_g, as it should, because you must pull against the attraction.

Near a planet’s surface. Over small height changes, the field is nearly constant at g. For a mass m raised by Δy:

ΔU_g ≈ mgΔy

This is the limit of the general form. Raising m from r = R to R + h gives ΔU_g = GMm[1/R − 1/(R + h)] = GMmh / [R(R + h)]. When h is much smaller than R, R(R + h) ≈ R², and GM/R² = g, so ΔU_g ≈ mgh. Worked example 2 shows how good this approximation is.

Systems with three or more objects

Potential energy belongs to pairs of interacting objects. For a system of three or more, add the potential energy of each pair once:

U_total = U₁₂ + U₁₃ + U₂₃ + …

For three spheres: U_g = −G(m₁m₂/r₁₂ + m₁m₃/r₁₃ + m₂m₃/r₂₃). Three objects give three pairs; four objects give six. Different interaction types add too. A block on a vertical spring has U_s (block–spring) and U_g (block–Earth), and the system’s potential energy is their sum.

Worked example 1: forces and equilibria from U(x)

Question. Take +x to the right. A small object moves along the x-axis in a system whose potential energy is U(x) = (2.0 J/m³)x³ − (6.0 J/m)x (Figure 1). Find (a) the force F_x(x), (b) the equilibrium positions and their stability, (c) the force at x = 2.0 m, and (d) the work done by this force as the object moves from x = −1.0 m to x = +1.0 m.

1. (a) F_x = −dU/dx = −(6.0x² − 6.0) = 6.0 − 6.0x² (N, with x in m).
2. (b) Equilibrium: dU/dx = 0, so 6.0x² = 6.0 and x = ±1.0 m. Second derivative: d²U/dx² = 12x.
   • At x = +1.0 m: d²U/dx² = +12 J/m² > 0, a minimum (U = −4.0 J). Stable.
   • At x = −1.0 m: d²U/dx² = −12 J/m² < 0, a maximum (U = +4.0 J). Unstable.
3. Check by nudging. At x = 1.1 m, F_x = 6.0 − 7.26 = −1.26 N, back towards 1.0 m. At x = 0.9 m, F_x = +1.14 N, also back towards 1.0 m. At x = −1.1 m, F_x = −1.26 N, further away from −1.0 m. Stable and unstable, as predicted.
4. (c) F_x(2.0) = 6.0 − 24 = −18 N. The graph slopes steeply upward at x = 2.0 m, so the force points to −x, downhill on the graph.
5. (d) W = −ΔU = −[U(1.0) − U(−1.0)] = −(−4.0 − 4.0) = +8.0 J.

Check (d) by integration. ∫₋₁¹ (6.0 − 6.0x²) dx = 12 − 4.0 = 8.0 J. Positive work and a fall in U agree: the object moved downhill on the graph.

Worked example 2: when does mgΔy stop working?

Question. Orvane is a fictional planet with mass M = 2.4 × 10²⁴ kg and radius R = 4.0 × 10⁶ m. A probe has mass m = 800 kg. Use G = 6.67 × 10⁻¹¹ N·m²/kg². (a) Find g at the surface and U_g of the probe–planet system at the surface, with U = 0 at infinite separation. (b) Find ΔU_g exactly, and with mgΔy, for lifting the probe 20 km. (c) Repeat (b) for 2000 km.

1. (a) g = GM/R² = (6.67 × 10⁻¹¹)(2.4 × 10²⁴) ÷ (4.0 × 10⁶)² = 10.0 m/s² (10.005 before rounding). U_g = −GMm/R = −(6.67 × 10⁻¹¹)(2.4 × 10²⁴)(800) ÷ (4.0 × 10⁶) = −3.2 × 10¹⁰ J.
2. (b) Exact: ΔU_g = GMm[1/R − 1/(R + h)] with h = 2.0 × 10⁴ m gives 1.593 × 10⁸ J. Approximate: mgh = 800 × 10.005 × 2.0 × 10⁴ = 1.601 × 10⁸ J. The approximation is only 0.50% too high.
3. (c) With h = 2.0 × 10⁶ m = R/2: exact ΔU_g = GMm[1/R − 1/(1.5R)] = GMm/(3R) = 1.07 × 10¹⁰ J. mgh = 1.60 × 10¹⁰ J, which is 50% too high.

Interpretation. mgΔy assumes the field stays at its surface value. Higher up, the real field is weaker, so the true energy needed is smaller. The ratio mgh ÷ exact equals (R + h)/R, so the error grows in proportion to h/R: 0.50% for h/R = 0.005, 50% for h/R = 0.5.

Worked example 3: a potential energy from a nonideal force

Question. Take +x in the direction of stretch, with x = 0 at the cord’s natural length. A stretched elastic cord exerts F_x = −(ax + bx²) for x ≥ 0, with a = 50 N/m and b = 200 N/m². (a) Derive U(x), choosing U = 0 at x = 0. (b) Find the energy stored at x = 0.30 m and compare it with an ideal spring of k = 50 N/m. (c) How much energy must be supplied to stretch the cord from 0.10 m to 0.30 m?

1. (a) U(x) = −∫₀ˣ F_x dx′ = ∫₀ˣ (ax′ + bx′²) dx′ = ½ax² + ⅓bx³. Check: −dU/dx = −(ax + bx²) = F_x. ✓
2. (b) U(0.30) = ½(50)(0.30)² + ⅓(200)(0.30)³ = 2.25 + 1.80 = 4.05 J ≈ 4.1 J. An ideal spring with k = 50 N/m stores ½(50)(0.30)² = 2.25 J. The cord stores 1.8 times as much, because it gets stiffer as it stretches.
3. (c) U(0.10) = 0.25 + 0.0667 = 0.317 J, so ΔU = 4.05 − 0.317 = 3.7 J.

Check the units. b has unit N/m², so ⅓bx³ has unit N·m = J. ✓ Note that ½k(Δx)² would give the wrong answer here: it applies only when F is proportional to x.

Common misconceptions

• “The ball has potential energy.” Potential energy belongs to the ball–Earth system. A system of one object can only have kinetic energy.
• Dropping the minus sign. ΔU = −W_c and F_x = −dU/dx. Positive work by a conservative force lowers U.
• “Equilibrium is where U = 0.” Equilibrium is where the slope dU/dx is zero. In Figure 1, U = 0 at x = 0, where F_x = +6.0 N.
• “Negative U means no energy.” The sign depends on the zero you chose. Only changes in U have meaning.
• Using mgΔy far from the surface. Use −Gm₁m₂/r when the height change is not small compared with the planet’s radius (Worked example 2).
• Using ½k(Δx)² for any elastic object. It holds only for an ideal spring with F ∝ Δx. Otherwise integrate the force (Worked example 3).
• Counting pairs twice. In a three-body system, U₁₂ and U₂₁ are the same pair. Count it once.
• Giving friction a potential energy. Friction is not conservative. Its work depends on the path.

Where this leads

Next, Topic 3.4, Conservation of Energy, combines U with kinetic energy: you will use U(x) graphs with a total energy line to find speeds and turning points. Later, the minimum of U at a stable equilibrium returns in simple harmonic motion, and −Gm₁m₂/r returns in orbits. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.