9 min readConcept sheetSection A MCQ (1 mark); Section B very short answer (2 marks), short answer (3 marks) or long answer (5 marks)

The work energy relation

The work done on a body equals its change in kinetic energy. Work done against gravity is stored as potential energy.

RelationWork done by the net force = change in kinetic energy: W = ½mv² − ½mu²

LiftingWork done against gravity in lifting a mass m through h = mgh = gain in potential energy

StoppingTo stop a moving body, work equal to its KE must be done against it, for example by brakes

Work changes energy

When you do work on a body, you give it energy. Push a trolley at rest and it speeds up: your work becomes its kinetic energy. Lift a bucket, and your work is stored as potential energy. When a body does work on something else, it loses energy. This link is called the work energy relation.

Derivation: KE = ½mv² from work

1
A constant force F acts on a body of mass m at rest (u = 0). It moves a distance s and reaches speed v.
2
Work done W = F × s.
3
By the second law, F = ma.
4
From v² = u² + 2as with u = 0, the distance s = v²/2a.
5
So W = ma × v²/2a = ½mv².
6★
So, the work done to take a body from rest to speed v is ½mv². This is its kinetic energy.

★ key step

Derivation: PE = mgh from work

1
To lift a body of mass m slowly, you apply an upward force equal to its weight, F = mg.
2
The body rises through a height h, along the force.
3
Work done W = F × h = mgh.
4★
This work is stored in the body. So, its potential energy at height h is PE = mgh.

★ key step

Worked example: work to speed up a cart

Problem

A 50 kg cart speeds up from 2 m s⁻¹ to 6 m s⁻¹. Find the work done on it.

Working

Initial KE = ½ × 50 × 2² = 100 J.
Final KE = ½ × 50 × 6² = 900 J.
Work done = change in KE = 900 J − 100 J = 800 J.

Answer

800 J

Worked example: braking distance

Problem

A 1000 kg car moving at 20 m s⁻¹ is stopped by brakes that give a steady force of 5000 N. How far does it travel before it stops?

Working

KE of the car = ½ × 1000 × 20² = 200 000 J.
The brakes must do this much work against it: F × s = 200 000 J.
s = 200 000 J ÷ 5000 N = 40 m.

Answer

40 m. At double the speed the KE is four times, so the braking distance would be 160 m

In real life

This is why speed limits near schools are low. A vehicle's braking distance grows with its kinetic energy. At twice the speed it needs about four times the distance to stop.

Exam angle

3 marks: “Show that the kinetic energy of a body of mass m moving with speed v is ½mv².” Start from rest; a constant force F over a distance s gives speed v. W = Fs, F = ma and s = v²/2a (from v² = u² + 2as with u = 0). So W = ma × v²/2a = ½mv², which is the KE.

2 marks: “How much work is needed to increase the speed of a 2 kg ball from 3 m s⁻¹ to 5 m s⁻¹?” W = change in KE = ½ × 2 × 5² − ½ × 2 × 3² = 25 − 9 = 16 J.

2 marks: “A body of mass 4 kg is lifted through 5 m. Find the work done against gravity and its gain in potential energy. (g = 10 m s⁻²)” W = mgh = 4 × 10 × 5 = 200 J. The gain in potential energy is also 200 J.

Remember it

Work in, energy up; work out, energy down.