Work, kinetic energy and gravitational potential energy

Work done

ΔW = FΔs, where Δs is the distance moved in the direction of the force. The unit is the joule (1 J = 1 N m). If the force is at angle θ to the motion, only the component F cos θ does work: ΔW = FΔs cos θ. A force at 90° to the motion does no work.

Kinetic energy

E_k = (1/2)mv². Because v is squared, doubling the speed quadruples the kinetic energy.

Gravitational potential energy

Near the Earth's surface, the change in gravitational potential energy when a mass m rises a height Δh is ΔE_grav = mgΔh, with g = 9.81 N/kg.

Worked example

A sledge is pulled 15 m along flat snow by a rope at 30° to the horizontal with a tension of 40 N. Find the work done by the rope.

1. Component along the motion = 40 cos 30° = 34.6 N
2. W = 34.6 x 15

Answer: W = 520 J (2 s.f.)

Check yourself

1. A 250 N force pushes a box 4.0 m in the direction of the force. Find the work done in J.
   Show answer

   1000 J. W = FΔs = 250 x 4.0 = 1000 J.

2. A rope at 25° to the horizontal pulls a cart 12 m along level ground with a force of 80 N. Find the work done in J, to 3 s.f.
   Show answer

   870 J. W = 80 x 12 x cos 25° = 870 J.

3. Find the kinetic energy of a 0.40 kg ball moving at 15 m/s, in J.
   Show answer

   45 J. E_k = 0.5 x 0.40 x 225 = 45 J.

4. Find the gain in gravitational potential energy when a 12 kg box is lifted 2.5 m, in J, to 3 s.f.
   Show answer

   294 J. ΔE = mgΔh = 12 x 9.81 x 2.5 = 294 J.

5. A 900 kg car has 180 kJ of kinetic energy. Find its speed in m/s.
   Show answer

   20 m/s. v = sqrt(2E_k / m) = sqrt(360000 / 900) = 20 m/s.

6. The speed of a car doubles. What happens to its kinetic energy?
   • It doubles
   • It halves
   • It stays the same
   • It becomes 4 times as large
   Show answer

   It becomes 4 times as large. E_k is proportional to v², and 2² = 4.

7. A satellite moves in a circle at constant speed; the gravitational force on it is always at 90° to its motion. How much work does this force do?
   • mgh per orbit
   • None
   • It depends on the speed
   • Equal to its kinetic energy
   Show answer

   None. cos 90° = 0, so a force perpendicular to the motion does no work.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson Edexcel IAL Physics Unit 1 (WPH11) question paper, January 2023: list of data (g = 9.81 m/s²)

Last updated 7 November 2026. Spotted a mistake? Tell us.