⚡Potential Energy
Stored Energy
Potential energy (PE) is stored energy associated with position or configuration. There are two types in AP Physics 1:
Gravitational PE: energy stored due to height: U_g = mgh
where h is the height above the chosen reference point (usually the ground or the lowest point of motion).
Elastic/Spring PE: energy stored in a compressed/stretched spring: U_s = ½kx²
where k is the spring constant (N/m) and x is the deformation from equilibrium.
Both forms are measured in Joules. Both can be converted to kinetic energy.
Reference Points
For gravitational PE, you get to choose your reference point (h = 0). The actual value of PE depends on your choice, but differences in PE do not.
For example: a 5 kg book on a 2 m table. - If h = 0 at the floor: U_g = 5×10×2 = 100 J - If h = 0 at the table: U_g = 0 J - If h = 0 at 10 m ceiling: U_g = 5×10×(−8) = −400 J
This doesn't matter! What matters in energy problems is the CHANGE in PE, not the absolute value.
Think About It
A spring with k = 200 N/m is compressed 0.1 m. How much energy is stored? If you compress it twice as much (0.2 m), how does the stored energy change?
✏️ Worked Example
Problem: A 3 kg ball is held 5 m above the ground. Find its gravitational PE. (g = 10 m/s²)
📐 Key Equations
Potential energy
U_g = mghU_s = (1)/(2)kx²Δ U = -W_conservative⚠️ Common Mistakes
Misconception: Using a fixed reference point (e.g., "the ground") when the problem is easier with a different choice.
✓ Correct thinking: You are free to set h = 0 anywhere. Choose the reference point to make the arithmetic simplest — often the lowest point of the motion.
Why: The physics only depends on ΔPE (change in PE), not the absolute value. A smart choice of reference eliminates unnecessary terms.
Misconception: Thinking the spring PE formula uses the total length of the spring, not the compression/extension.
✓ Correct thinking: U_s = ½kx² uses x = deformation from the natural (equilibrium) length — the amount the spring is stretched or compressed, not its total length.
Why: A spring at its natural length stores zero elastic PE regardless of how long it is. Only displacement from equilibrium stores energy.
Misconception: Doubling spring compression doubles the stored energy.
✓ Correct thinking: U_s = ½kx². Doubling x multiplies x² by 4, so the stored energy quadruples.
Why: Like KE ∝ v², spring PE has a quadratic (squared) dependence. This is a recurring pattern worth memorizing.
📝 Practice Problems
Try these problems. Check your answer when ready.
A 4 kg book sits on a shelf 2.5 m above the floor. Find its gravitational PE. (g = 10 m/s²)
U_g = mghA spring with k = 400 N/m is compressed 0.05 m. Find the elastic PE stored.
U_s = (1)/(2)kx²A spring (k = 500 N/m) is compressed 0.1 m, then compressed an additional 0.1 m (total 0.2 m). How does the stored PE compare?
A 2 kg object falls from a shelf 3 m high (reference h = 0 at floor) to a table 1 m high. Find (a) U_g at each position and (b) the change in U_g.
U_g = mghA spring launcher (k = 800 N/m) is compressed 0.15 m and launches a 0.2 kg ball. Assuming all spring PE converts to KE, find the launch speed.
(1)/(2)kx² = (1)/(2)mv²On the Moon (g = 1.6 m/s²), a 70 kg astronaut climbs 10 m. How much gravitational PE is gained? Compare to climbing the same height on Earth.
U_g = mghA spring (k = 1000 N/m) is attached to a wall and compressed by a 3 kg block. If released, what compression x is needed so the block reaches a speed of 5 m/s on a frictionless surface?
(1)/(2)kx² = (1)/(2)mv²Finished reading through this lesson?