Gravitational and Elastic Potential Energy — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Definition of gravitational potential energy (near Earth's surface), gravitational potential energy change formula, elastic potential energy for ideal springs, conservation of mechanical energy with two potential energy types, and conceptual identification of potential energy as a system property.

You should already know: Work and the work-energy theorem for kinetic energy, Hooke's Law for ideal springs, the definition of a closed system for energy analysis.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Gravitational and Elastic Potential Energy?

Potential energy is stored energy in a closed system arising from the relative positions of interacting objects. Gravitational potential energy is stored due to the gravitational interaction between two masses; for AP Physics 1, we only work with gravitational potential energy near Earth’s surface, where gravity is approximately constant. Elastic potential energy is stored when an elastic object (such as an ideal spring) is stretched or compressed from its equilibrium position.

This subtopic is part of Unit 4: Energy in the AP Physics 1 CED, which accounts for 20–28% of total exam score; gravitational and elastic potential energy make up roughly one-third of that unit, or ~6–9% of the total exam. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with kinematics, forces, or momentum problems. By convention, we use (U) for potential energy, with (U_g) for gravitational and (U_s) (or (U_{el})) for elastic. Only changes in potential energy are physically meaningful, so we can choose any reference point to set (U=0).

2. Gravitational Potential Energy Near Earth's Surface

Gravitational potential energy describes stored energy from the separation between masses in a gravitational field. Near Earth's surface, gravitational force on a mass (m) is approximately constant: (F_g = mg), pointing downward. The change in gravitational potential energy of the mass-Earth system equals the negative of the work done by gravity when the mass changes position: $$\Delta U_g=-W_g$$ If the mass changes height by (\Delta y = y_{final} - y_{initial}) (with upward as positive), gravity does work (W_g = -mg\Delta y). Substituting into the potential energy change formula gives: $$\Delta U_g=mg\Delta y$$ If we choose a reference point where (U_g = 0), we can write (U_g = mgy), where (y) is the height of the mass relative to that reference. Intuition: increasing the height of the mass increases the stored gravitational potential energy, which makes sense: lifting a heavier object higher stores more energy that can be converted to motion when it falls.

Worked Example

A 62 kg hiker climbs from the base of a trail at 1200 m elevation to the summit at 2450 m elevation. What is the change in the gravitational potential energy of the hiker-Earth system? If the hiker sets (U_g = 0) at the base, what is (U_g) at the summit?

1. Identify known values: (m = 62\ \text{kg}), (g = 9.8\ \text{m/s}^2), (\Delta y = 2450\ \text{m} - 1200\ \text{m} = 1250\ \text{m}).
2. Use the gravitational potential energy change formula: (\Delta U_g = mg\Delta y).
3. Calculate the change: (\Delta U_g = (62)(9.8)(1250) = 759500\ \text{J} \approx 7.6 \times 10^5\ \text{J}).
4. If (U_{g,\text{base}} = 0), then (U_{g,\text{summit}} = U_{g,\text{base}} + \Delta U_g = 0 + 7.6 \times 10^5\ \text{J} = 7.6 \times 10^5\ \text{J}).

Exam tip: Always confirm that (\Delta y) is the change in vertical height, not horizontal distance or trail length—AP 1 questions often give trail length to test this distinction.

3. Elastic Potential Energy for Ideal Springs

Elastic potential energy is stored in an ideal spring (or other elastic material) when it is stretched or compressed from its equilibrium (relaxed) position. An ideal spring obeys Hooke's Law: (F_s = -kx), where (x) is displacement from equilibrium, and (k) is the spring constant (a measure of the spring's stiffness, units N/m).

To find the elastic potential energy, we again use the relation that change in potential energy equals negative work done by the conservative spring force. The work done by the spring moving from equilibrium ((x=0)) to displacement (x) is (W_s = -\frac{1}{2}kx^2), so: $$U_s=\frac{1}{2}k{x}^2$$ We set (U_s = 0) at equilibrium, which is the standard convention for AP 1. Notice that (x) is squared, so stretching a spring by 0.1 m stores the same amount of energy as compressing it by 0.1 m. Intuition: stiffer springs (higher (k)) store more energy for the same displacement, and energy grows with the square of displacement, so doubling displacement quadruples stored energy.

Worked Example

A 0.25 kg block is pressed against a horizontal spring with spring constant 120 N/m, compressing it 0.15 m from equilibrium. How much elastic potential energy is stored in the block-spring system? If the spring is compressed twice as far, by what factor does stored energy increase?

1. Identify known values: (k = 120\ \text{N/m}), (x = 0.15\ \text{m}), (U_s = 0) at (x=0).
2. Use the elastic potential energy formula: (U_s = \frac{1}{2}kx^2 = 0.5(120)(0.15)^2 = 1.35\ \text{J} \approx 1.4\ \text{J}).
3. For twice the compression, new displacement (x' = 2x), so (U_s' = \frac{1}{2}k(2x)^2 = 4\left(\frac{1}{2}kx^2\right) = 4U_s).
4. Stored energy increases by a factor of 4.

Exam tip: Never use the total length of the spring for (x)—(x) is always displacement from equilibrium (relaxed length), so you must subtract the relaxed length from the stretched/compressed length to get (x).

4. Conservation of Mechanical Energy with Multiple Potential Energies

Total mechanical energy of a closed system is the sum of kinetic energy, gravitational potential energy, and elastic potential energy: $$E_{mech}=K+U_g+U_s$$ If no non-conservative forces (friction, air resistance, external pushes) do net work on the system, total mechanical energy is conserved. This means the initial total mechanical energy equals the final total mechanical energy: $$K_i+U_{g,i}+U_{s,i}=K_f+U_{g,f}+U_{s,f}$$ This is one of the most useful problem-solving tools in AP Physics 1, because it lets you find speed or position without calculating acceleration or forces at every point along the motion. To simplify calculations, always choose a convenient zero reference for (U_g) (usually set (U_g = 0) at the lowest point of motion, so that term becomes zero in the final energy calculation).

Worked Example

A 0.50 kg ball is dropped from rest from a height of 2.0 m above a vertical relaxed spring. The spring compresses 0.25 m before the ball momentarily stops. What is the spring constant of the spring, assuming no air resistance?

1. Define the system as the ball, Earth, and spring (no non-conservative work, so energy is conserved). Set (U_g = 0) at the maximum compression point (where the ball stops), so (U_{g,f} = 0).
2. Identify initial and final energies: Initial state: (K_i = 0) (dropped from rest), (U_{s,i} = 0) (spring is relaxed), (U_{g,i} = mg(2.0\ \text{m} + 0.25\ \text{m}) = mg(2.25\ \text{m})) (total height change from initial to final is 2.25 m, including compression). Final state: (K_f = 0) (momentarily stopped), (U_{s,f} = \frac{1}{2}k(0.25\ \text{m})^2), (U_{g,f} = 0).
3. Substitute into conservation of energy: (0 + mg(2.25) + 0 = 0 + 0 + \frac{1}{2}k(0.25)^2).
4. Solve for (k): (k = \frac{2mg(2.25)}{(0.25)^2} = \frac{2(0.50)(9.8)(2.25)}{0.0625} \approx 350\ \text{N/m}).

Exam tip: Always sketch initial and final positions to confirm all displacement/height changes—students almost always forget to add the spring compression to the total height change in this type of problem.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the total length of a spring instead of displacement from equilibrium in the elastic potential energy formula. Why: Questions often give both relaxed and stretched length to test understanding, and students default to the larger number given. Correct move: Calculate (x) as (|\text{final length} - \text{relaxed equilibrium length}|) every time before plugging into (U_s = \frac{1}{2}kx^2).
• Wrong move: Forgetting to add the spring compression/extension to the total height change when solving energy problems with a falling object on a vertical spring. Why: Students stop at the height of the object above the relaxed spring, and ignore the additional distance the object falls while compressing the spring. Correct move: Always measure the total change in height from the initial position of the object to its final position, regardless of what the spring is doing.
• Wrong move: Dropping the negative sign for (U_g) when the object is below the chosen zero reference point. Why: Students assume potential energy can never be negative, forgetting only changes in potential energy are physically meaningful. Correct move: Keep the sign of (U_g) equal to the sign of (y) relative to your zero reference; the change in potential energy will still be correct.
• Wrong move: Claiming stretching a spring twice as far stores twice as much elastic potential energy. Why: Students confuse the linear force-displacement Hooke's Law relation with the quadratic energy-displacement relation. Correct move: Remember that energy depends on the square of displacement, so doubling displacement quadruples stored energy.
• Wrong move: Treating potential energy as a property of a single object instead of a system of interacting objects. Why: Textbooks often use shorthand like "the hiker's potential energy," leading to this mistake, which AP 1 explicitly tests. Correct move: Explicitly reference the system: gravitational potential energy belongs to the hiker-Earth system, elastic potential energy belongs to the spring-mass system.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A student stretches two springs. Spring 1 has a spring constant (2k) and is stretched by distance (x). Spring 2 has a spring constant (k) and is stretched by distance (3x). How does the elastic potential energy stored in Spring 1 compare to the elastic potential energy stored in Spring 2? A) (U_1 = \frac{2}{9} U_2) B) (U_1 = \frac{1}{3} U_2) C) (U_1 = \frac{2}{3} U_2) D) (U_1 = 2 U_2)

Worked Solution: Use the elastic potential energy formula (U_s = \frac{1}{2}kx^2) for both springs. For Spring 1: (U_1 = \frac{1}{2}(2k)(x)^2 = kx^2). For Spring 2: (U_2 = \frac{1}{2}(k)(3x)^2 = \frac{9}{2}kx^2). Take the ratio of the two energies: (\frac{U_1}{U_2} = \frac{kx^2}{\frac{9}{2}kx^2} = \frac{2}{9}), so (U_1 = \frac{2}{9}U_2). The correct answer is A.

Question 2 (Free Response)

A 0.40 kg toy car moves along a frictionless track that starts with a hill, then has a horizontal section ending in a spring bumper. The car starts from rest at the top of the hill, 1.5 m above the horizontal section. The spring bumper has a spring constant of 80 N/m. (a) Calculate the speed of the car when it reaches the horizontal section, before it hits the spring. (b) Calculate the maximum compression of the spring when the car momentarily stops. (c) A student claims that if the car started from a height twice as high, the maximum compression of the spring would also double. Do you agree with this claim? Justify your answer with physics reasoning.

Worked Solution: (a) Set (U_g = 0) at the horizontal section. Energy conservation: (mgh = \frac{1}{2}mv^2). The mass cancels, so (v = \sqrt{2gh} = \sqrt{2(9.8)(1.5)} \approx 5.4\ \text{m/s}). (b) At maximum compression, all gravitational potential energy is converted to elastic potential energy: (mgh = \frac{1}{2}kx^2). Solve for (x): (x = \sqrt{\frac{2mgh}{k}} = \sqrt{\frac{2(0.40)(9.8)(1.5)}{80}} = \sqrt{0.147} \approx 0.38\ \text{m}). (c) I do not agree with the claim. From the result in part (b), (x = \sqrt{\frac{2mgh}{k}}), so (x) is proportional to (\sqrt{h}), not (h). If (h) doubles, new (x' = \sqrt{\frac{2mg(2h)}{k}} = \sqrt{2} \cdot \sqrt{\frac{2mgh}{k}} = x\sqrt{2} \approx 1.41x). The maximum compression increases by a factor of (\sqrt{2}), not 2, so the claim is incorrect.

Question 3 (Application / Real-World Style)

A 75 kg bungee jumper jumps from a 50 m high platform. The bungee cord has an unstretched length of 20 m, and the jumper just touches the ground at the lowest point of the jump before bouncing back. Assuming the bungee cord behaves like an ideal spring and air resistance is negligible, what is the effective spring constant of the bungee cord?

Worked Solution: Set (U_g = 0) at the ground, so initial potential energy at the platform is (U_{g,i} = mg(50\ \text{m})), and (K_i = U_{s,i} = K_f = U_{g,f} = 0). The bungee cord is stretched by (x = 50\ \text{m} - 20\ \text{m} = 30\ \text{m}) at the lowest point, so final elastic potential energy is (U_{s,f} = \frac{1}{2}kx^2). Energy conservation gives (mg(50) = \frac{1}{2}k(30)^2). Solve for (k): (k = \frac{2(75)(9.8)(50)}{30^2} \approx 82\ \text{N/m}). This means the bungee cord has a stiffness of 82 N per meter of stretch, which is soft enough to stop the jumper gradually at the ground.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for all energy-based problem solving in the rest of AP Physics 1. Next, you will extend energy conservation to account for work done by non-conservative forces like friction, which converts mechanical energy to thermal energy, leading to the full, general work-energy theorem. Without mastering how to correctly calculate gravitational and elastic potential energy, you cannot solve energy problems for collisions, simple harmonic motion, or rotational systems that come later in the course. This topic also builds the core understanding of energy conservation as a fundamental law that applies to all physical systems in AP Physics 1 and beyond.

Work done by non-conservative forces Conservation of energy in collisions Simple harmonic motion of mass-spring systems Power and energy transfer rate