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AS & A Level · Practice questions · Work done by a force

Only the useful component.

Six original Cambridge-style questions on the work done by constant forces, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Resolve along the motion, times distance.

01
Recall
[2 marks]

A box is pushed 4 m across a level floor by a horizontal force of 25 N. Find the work done by the push, and state the work done by the weight of the box.

  • Push: \(25 \times 4 = 100\) J. ✓
  • Weight: 0 J, because it is perpendicular to the motion. ✓
02
Calculation
[2 marks]

A rope with tension 60 N inclined at \(25^\circ\) above the horizontal pulls a crate 8 m along a horizontal floor. Find the work done by the tension.

  • \(W = 60 \times 8 \times \cos 25^\circ\). ✓
  • \(W = 435\) J. ✓
03
Calculation
[3 marks]

A crane lifts a load of mass 200 kg through 15 m at constant speed, then lowers it back to its starting point at constant speed. Find the work done by the crane cable on the way up, and the total work done by gravity over the whole trip.

  • Up: \(T = 2000\) N, so work \(= 2000 \times 15 = 30\,000\) J. ✓
  • Gravity does \(-30\,000\) J going up and \(+30\,000\) J coming down. ✓
  • Total work by gravity \(= 0\): it depends only on the overall change in height. ✓
04
Analysis
[5 marks]

A crate of mass 20 kg is pulled at constant speed through 10 m across rough horizontal ground by a rope inclined at \(30^\circ\) above the horizontal. The coefficient of friction is 0.4. Find the tension and the work done by the rope.

  • \(R = 200 - T\sin 30^\circ = 200 - 0.5T\). ✓
  • \(T\cos 30^\circ = 0.4(200 - 0.5T)\), so \(T(0.8660 + 0.2) = 80\). ✓
  • \(T = 75.0\) N. ✓
  • Work \(= 75.05 \times 10 \times \cos 30^\circ = 650\) J. ✓
05
Analysis
[4 marks]

A particle of mass 3 kg slides 6 m down a line of greatest slope of a rough plane inclined at \(30^\circ\), \(\mu = 0.2\). Find (a) the work done by the weight, (b) the work done against friction, (c) the net work done on the particle.

  • (a) \(30 \times 6\sin 30^\circ = 90\) J. ✓
  • (b) \(0.2 \times 30\cos 30^\circ \times 6 = 31.2\) J. ✓
  • (c) \(90 - 31.18 = 58.8\) J (the normal reaction does no work). ✓
06
Analysis
[5 marks]

A sledge of mass 15 kg is pulled 50 m up a line of greatest slope of a hill inclined at \(10^\circ\) to the horizontal, at constant speed, by a rope parallel to the slope. The coefficient of friction is 0.1. Find the work done by the rope.

  • Work against gravity \(= 150 \times 50\sin 10^\circ = 1302\) J. ✓
  • \(R = 150\cos 10^\circ = 147.7\) N; work against friction \(= 0.1 \times 147.7 \times 50 = 738.6\) J. ✓
  • Constant speed: work by the rope \(= 1302 + 738.6 = 2040\) J (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.