Potential Energy: Study Guide (Physics 1 3.3)

This guide is for the algebra-based Physics 1 course. Everything here uses algebra and graphs. If you are taking the calculus-based Physics C: Mechanics course, it has its own separate guide.

Potential energy belongs to a system

In Topic 3.1 you met kinetic energy, the energy of motion. Potential energy is different. It is energy stored in the arrangement of objects that pull or push on each other.

Three conditions must hold:

1. The system has two or more objects (or one object that can change its shape, such as a spring, whose parts push on each other).
2. Those objects interact with each other.
3. The interaction is through a conservative force, such as gravity or an ideal spring force.

A conservative force is one whose work depends only on the start and end positions, not on the path. Carry a box upstairs by the lift or by the stairs: gravity does the same work on it either way. Friction is not conservative. Drag a box the long way round a room and friction does more (negative) work than on the short path. There is no “friction potential energy”.

So potential energy is always a property of a system. A ball on its own has kinetic energy only. The ball–Earth system has gravitational potential energy, because it depends on the separation of the ball and Earth. A block–spring system has elastic potential energy, because it depends on how far the spring is stretched.

Scalar, and set by position

Potential energy is a scalar. It has a size and can be positive or negative, but it has no direction. Its value is fixed by the positions of the objects in the system, not by how fast they move or how they got there.

For a conservative force inside the system, the link to Topic 3.2 is:

ΔU = −W (work done by that conservative force on the objects inside the system)

When gravity does +20 J of work on a falling ball, the ball–Earth system’s potential energy falls by 20 J. That energy has not vanished. In Topic 3.4 you will follow it into kinetic energy.

Choosing the zero

Where U = 0 is your decision. Choose whatever makes the problem simplest: the floor, the lowest point of a swing, the relaxed length of a spring. Changing the zero shifts every value of U by the same amount, so every change ΔU stays the same. Only changes in potential energy affect what happens physically.

State your zero, just as you state your axis in kinematics: “U_g = 0 at the floor”.

Elastic potential energy of an ideal spring

An ideal spring obeys Hooke’s law (Topic 2.8): the force it exerts has size kΔx, where Δx is the stretch or compression measured from its relaxed (equilibrium) length. Its potential energy is

U_s = ½kΔx²

• k is the spring constant in N/m; Δx is in m; U_s is in joules (J).
• The zero is at the relaxed length. That is the natural choice and the one the formula assumes.
• Δx is squared, so a spring stores the same energy when stretched 3 cm as when compressed 3 cm. U_s is never negative.
• Functional dependence: U_s ∝ Δx². Double the stretch and U_s becomes 2² = 4 times as big. Triple it: 9 times.

Where does the ½ come from? In Topic 3.2 you found work from the area under a force–displacement graph. The force needed to stretch the spring rises in a straight line from 0 to kΔx. The area under that line is a triangle: ½ × Δx × kΔx = ½kΔx². That work is what the spring stores.

Gravitational potential energy

The general form

For two roughly spherical masses (a planet and a moon, a star and a planet, or Earth and a satellite) with centres a distance r apart:

U_g = −Gm₁m₂ / r

• r is measured centre to centre, not from the surface.
• The zero is at infinite separation. Every finite separation gives a negative value.
• As r increases, U_g gets less negative, so it increases. Pulling two masses apart takes energy; letting them fall together releases it.
• Functional dependence: U_g ∝ 1/r. Double the separation and U_g becomes half as big (half as negative). Compare the force, which goes as 1/r².

Near a planet’s surface

Close to the surface, the gravitational field is nearly constant, with strength g. Then the change in potential energy of an object–planet system when the object rises a height Δy is

ΔU_g = mgΔy

Here Δy is positive for upward. This is a short cut, and it works only while Δy is tiny compared with the planet’s radius. Worked example 3 tests where it breaks down.

Graphs of potential energy

Three features to be able to sketch without notes:

• Spring: a symmetric U-shape (parabola) with its lowest point at Δx = 0.
• General gravity: a curve below the r-axis, steep close in, flattening towards zero far away. It never crosses zero.
• Near-surface gravity: U_g against height y is a straight line with slope mg. That is what the curve in Figure 1(b) looks like over a very short stretch near r = R.

Systems with more than two objects

Potential energy is shared between pairs of interacting objects. For three or more objects, add up the potential energy of every pair. Three objects A, B and C have three pairs: AB, BC and AC.

Example. Three small moons lie in a straight line: A (mass m), then B (mass 2m) a distance d further on, then C (mass m) a further d beyond B. The pairs are:

• AB: separation d, so U = −G(m)(2m)/d = −2Gm²/d
• BC: separation d, so U = −2Gm²/d
• AC: separation 2d, so U = −G(m)(m)/(2d) = −0.5Gm²/d

U_total = −4.5Gm²/d

The outer pair AC is easy to miss because B sits between them, but A and C still attract each other. If every spacing doubles, every pair term halves, so U_total halves too: −2.25Gm²/d. That is an increase, because the value is less negative.

The same idea covers mixed systems. A block hanging on a vertical spring near Earth, with the system block + spring + Earth, has two potential energy terms: U_s for the block–spring interaction and U_g for the block–Earth interaction. (We treat the spring as massless, so it has no gravitational term of its own.)

Worked example 1: spring energy and factors of change

Question. A toy launcher uses a spring with k = 250 N/m. (a) How much elastic potential energy is stored when it is compressed by 4.0 cm? (b) Predict the energy at 12 cm compression without substituting again. (c) What compression stores 0.80 J?

1. Convert: Δx = 4.0 cm = 0.040 m.
2. (a) U_s = ½kΔx² = ½ × 250 N/m × (0.040 m)² = 0.20 J.
3. (b) 12 cm is 3 times 4.0 cm. U_s ∝ Δx², so the energy is 3² = 9 times bigger: 9 × 0.20 J = 1.8 J.
4. (c) Rearrange: Δx = √(2U_s / k) = √(2 × 0.80 J ÷ 250 N/m) = 0.080 m (8.0 cm).

Check. 0.80 J is 4 times 0.20 J, so the compression should be √4 = 2 times bigger: 2 × 4.0 cm = 8.0 cm. It is. These three points are marked on Figure 1(a).

Worked example 2: the same shelf, two choices of zero

Question. A 2.5 kg bag of rice is lifted from the floor onto a shelf 1.60 m above the floor. A table top is 0.75 m above the floor. The system is bag + Earth, and +y is upward. Find the gravitational potential energy with the bag on the floor, on the table and on the shelf, (a) with U_g = 0 at the floor and (b) with U_g = 0 at the table top. (c) Find ΔU_g from floor to shelf in each case.

Near the surface, U_g = mgy, where y is the height above the chosen zero. Here mg = 2.5 kg × 9.8 N/kg = 24.5 N.

Position	(a) zero at floor	(b) zero at table
floor (y = 0 or −0.75 m)	0 J	24.5 × (−0.75) = −18 J
table (y = 0.75 m or 0)	24.5 × 0.75 = 18 J	0 J
shelf (y = 1.60 m or 0.85 m)	24.5 × 1.60 = 39 J	24.5 × 0.85 = 21 J

(c) Floor to shelf: (a) 39.2 J − 0 = 39 J; (b) 20.8 J − (−18.4 J) = 39 J.

Interpretation. The values in the two columns differ, and some are negative. The change is the same, 39 J, because mgΔy = 24.5 N × 1.60 m = 39 J in both. The choice of zero is bookkeeping, not physics.

Worked example 3: when mgΔy stops working

Question. Use Earth’s mass 5.97 × 10²⁴ kg and mean radius 6.37 × 10⁶ m. A 500 kg satellite is moved from Earth’s surface to a distance of 2R from Earth’s centre (one Earth radius above the surface). (a) Find U_g at the surface and at 2R. (b) Find ΔU_g. (c) Compare with mgΔy using g = 9.8 N/kg and Δy = R.

1. (a) At the surface, r = R: U_g = −GMm / R = −(6.67 × 10⁻¹¹)(5.97 × 10²⁴)(500) ÷ (6.37 × 10⁶) = −3.1 × 10¹⁰ J.
2. At r = 2R the separation doubles, so U_g halves: −1.6 × 10¹⁰ J.
3. (b) ΔU_g = (−1.56 × 10¹⁰ J) − (−3.13 × 10¹⁰ J) = +1.6 × 10¹⁰ J. Positive: the system gained potential energy as the satellite moved away.
4. (c) mgΔy = 500 kg × 9.8 N/kg × 6.37 × 10⁶ m = 3.1 × 10¹⁰ J, about twice the true change.

Why. mgΔy assumes the field stays at 9.8 N/kg all the way up. In fact the field weakens as 1/r², so less energy is needed. For a 1.0 km lift, the weakening of the field changes the answer by only about 0.02% (1 km ÷ 6370 km), which is why mgΔy is fine for anything happening near the ground. Use U_g = −Gm₁m₂/r whenever the distance moved is a sizeable fraction of the planet’s radius.

Common misconceptions

• “The ball has potential energy.” A single object has no potential energy. The ball–Earth system does. Name the system.
• “Potential energy is always positive.” It can be negative. U_g = −Gm₁m₂/r is always negative, and near-surface values are negative below your chosen zero (Worked example 2).
• “Changing the zero changes the answer.” It changes the values of U but never ΔU.
• “A compressed spring has negative potential energy.” U_s = ½kΔx² uses the square, so compression and stretch both give positive values.
• “Doubling the stretch doubles the energy.” It quadruples it (Worked example 1).
• “In U_g = −Gm₁m₂/r, r is the height above the surface.” r is the centre-to-centre distance.
• “mgΔy works at any height.” Only while Δy is much smaller than the planet’s radius (Worked example 3).
• “Friction stores potential energy.” Friction is not conservative. The energy it removes does not come back; Topic 3.4 tracks it as thermal energy.

Where this leads

Topic 3.4, Conservation of Energy, combines potential energy with kinetic energy to predict speeds and heights without forces or time. Try the practice questions now, then use the revision notes and the checklist to consolidate. You can also look back at Topic 3.2, Work, or return to the course roadmap.