Topic 5 of 8 · Form 5

Work, energy and power

Power = rate of doing work = Fv (W). At maximum speed, the driving force equals the resistance.

Notes for this topic 1 file, in order

Work done by a constant force = Fs cos θ (J).
Kinetic energy = ½mv²; gravitational potential energy = mgh.
Work–energy principle: work done by all forces = change in kinetic energy.
Conservation of mechanical energy when only gravity does work: KE + PE = constant.
Work done against friction = friction × distance (energy lost).

Worked example: A 2 kg particle slides 5 m down a rough slope at 30° from rest; friction is 4 N. Find its speed at the bottom (g = 9.8).

PE lost = 2 × 9.8 × 5 sin 30° = 49 J; work against friction = 4 × 5 = 20 J

KE gained = 29 J → ½ × 2 × v² = 29 → v = 5.39 m s⁻¹

Power = rate of doing work = Fv (W). At maximum speed, the driving force equals the resistance.

Worked example: A car's engine works at 30 kW against a constant resistance of 1 200 N. Find its maximum speed on level ground.

At maximum speed, driving force = 1 200 N → v = 30 000 ÷ 1 200 = 25 m s⁻¹

Hooke's law T = λx/l (λ = modulus of elasticity, l = natural length). Elastic potential energy = λx²/(2l).

Exam tip: In energy problems, list all energy forms at the start and end, then account for work against resistance.

Find the kinetic energy of a 1 500 kg car at 20 m s⁻¹.
A pump raises 200 kg of water through 15 m each minute. Find the useful power (g = 9.8).
A string of natural length 0.5 m and modulus 20 N is stretched 0.1 m. Find the tension.

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