Gravitational potential energy and kinetic energy

AS Level Physics · Work, energy and power. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Gravitational potential energy is the energy an object has because of its position in a gravitational field. Near the Earth's surface the field is uniform, so g is constant. To lift a mass m at steady speed you need a force equal to its weight mg, so the work done in raising it through a height Δh is mgΔh. Only the vertical height change matters, not the path taken.

Kinetic energy is the energy an object has because of its motion. It depends on the square of the speed.

If there are no resistive forces, the loss in potential energy equals the gain in kinetic energy. The mass cancels, so the speed does not depend on the mass.

Rules to remember

Common mistake

For a change in kinetic energy, students write ½m(v − u)2. Square each speed first and then subtract: ½mv2 − ½mu2.

Worked example

A ball of mass 0.40 kg is thrown vertically upwards at 12 m s−1. Ignoring air resistance, what is its kinetic energy when it is 5.0 m above the point of release? (g = 9.81 m s−2)

  1. Initial kinetic energy = ½mv2 = ½ × 0.40 × 122 = 28.8 J
  2. Gain in potential energy = mgΔh = 0.40 × 9.81 × 5.0 = 19.6 J
  3. Kinetic energy left = 28.8 − 19.6 = 9.2 J

Answer: 9.2 J

Practice questions

Try each one, then open the answer.

1. Which statement about gravitational potential energy near the Earth's surface is correct?

  1. A
    ΔEp = mgΔh applies for any height above the Earth
  2. B
    ΔEp = mgΔh applies only where g is effectively constant
  3. C
    potential energy is always positive
  4. D
    potential energy does not depend on mass
Show answer

Answer: B. The formula assumes a uniform field. Far from the Earth g changes, and the general expression involves −GMm/r.

2. A pendulum bob passes through its lowest point at 2.4 m s−1. Ignoring air resistance, how high above the lowest point does it rise? (g = 9.81 m s−2)

  1. A
    0.12 m
  2. B
    0.24 m
  3. C
    0.29 m
  4. D
    0.59 m
Show answer

Answer: C. ½mv2 = mgh, so h = v2/2g = 2.42 ÷ (2 × 9.81) = 0.29 m. The mass cancels.

3. A climber of mass 65 kg climbs from an altitude of 1200 m to an altitude of 1850 m. What is her gain in gravitational potential energy? (g = 9.81 m s−2)

  1. A
    6.4 × 102 J
  2. B
    4.2 × 104 J
  3. C
    4.1 × 105 J
  4. D
    1.2 × 106 J
Show answer

Answer: C. ΔEp = mgΔh = 65 × 9.81 × (1850 − 1200) = 65 × 9.81 × 650 = 4.1 × 105 J.

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