Work done, kinetic energy and potential energy

Work done

The work done by a constant force F moving its point of application a distance s in the direction of the force is W = Fs (joules, J).

If the force is at angle θ to the motion, only the component F cos θ does work: W = Fs cos θ. A force perpendicular to the motion (such as a normal reaction) does no work.

Work against gravity and friction

• Raising a mass m through a vertical height h: work against gravity = mgh. On a slope, h = distance along the slope × sin α.
• Moving a distance s against friction F: work = Fs. With limiting friction, F = μR, so work = μRs.

Energies (you must know these)

• Kinetic energy KE = (1/2)mv².
• Gravitational potential energy PE = mgh, relative to a chosen level.

The spec says these two formulae are not in the formulae booklet, so learn them.

Worked example

A box of mass 10 kg is pulled 8 m across a rough horizontal floor by a horizontal rope. The coefficient of friction is 0.3. Find the work done against friction.

1. R = 10g = 98 N.
2. Friction = μR = 0.3 × 98 = 29.4 N.
3. Work = 29.4 × 8 = 235 J.

Answer: 235 J

Check yourself

1. Find the kinetic energy of a car of mass 1200 kg moving at 15 m/s.
   Show answer

   135000 J. (1/2)(1200)(15²) = 135 000 J.

2. Find the potential energy gained when a 3 kg mass is lifted 4 m vertically, to 3 significant figures.
   Show answer

   118 J. mgh = 3 × 9.8 × 4 = 118 J.

3. A force of 50 N acts at 30° above the horizontal on a sledge, which moves 12 m horizontally. Find the work done by the force, to 3 significant figures.
   Show answer

   520 J. 50 cos 30° × 12 = 520 J.

4. A 5 kg parcel is pushed 10 m up a slope inclined at 30° to the horizontal. Find the work done against gravity.
   Show answer

   245 J. Height gained = 10 sin 30° = 5 m. Work = 5 × 9.8 × 5 = 245 J.

5. A 4 kg box is pulled 6 m up a rough slope inclined at α, where sin α = 3/5. The coefficient of friction is 0.25. Find the work done against friction, to 3 significant figures.
   Show answer

   47.0 J. R = 4g cos α = 4(9.8)(0.8) = 31.36 N. Friction = 7.84 N. Work = 7.84 × 6 = 47.0 J.

6. A 2 kg particle speeds up from 4 m/s to 10 m/s. Find its increase in kinetic energy.
   Show answer

   84 J. (1/2)(2)(100 - 16) = 84 J.

7. A particle slides along a horizontal table. Which force does no work on it?
   • Friction
   • Its weight
   • A horizontal pulling force
   • Air resistance
   Show answer

   Its weight. The weight is perpendicular to the horizontal motion, so it does no work.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M2
• Pearson Edexcel IAL Mechanics M2 (WME02) question paper, June 2025: front-cover instructions on g and accuracy

Last updated 8 November 2026. Spotted a mistake? Tell us.