A force does work when its point of application moves. Only the component of the force along the displacement counts, which is where the \(\cos\theta\) comes from.
Work done by a constant force \(= Fd\cos\theta\), where \(d\) is the distance moved and \(\theta\) the angle between the force and the direction of motion. Work is measured in joules (\(1\ \text{J} = 1\ \text{N m}\)). Work against gravity \(= mgh\); work against friction \(= Fd\) with \(F = \mu R\).
Fig. 1: a 60 N pull at 25° to the horizontal moves a crate 8 m. Only the 54.4 N component along the motion does work (angle drawn to scale)
Section 01
Work is force along the motion.
In the lab, keep the pulling force fixed and raise the angle of the rope from \(0^\circ\) towards \(90^\circ\). The work done per metre falls like \(\cos\theta\) and reaches zero when the rope is vertical: a force at right angles to the motion does no work.
\(W = Fd\cos\theta\): \(\theta = 0^\circ\) gives \(Fd\); \(\theta = 90^\circ\) gives 0 (the normal reaction on a level floor does no work).
Negative work: a force opposing the motion (\(\theta = 180^\circ\)) does work \(-Fd\). We say work is done AGAINST it, for example against friction or resistance.
Units: joules (J); 1 kJ \(= 1000\) J.
Vectors and scalar products are not needed: just resolve the force along the motion.
Worked example A rope at an angle (Fig. 1)
A crate is pulled 8 m along a horizontal floor by a rope with tension 60 N inclined at \(25^\circ\) above the horizontal. Find the work done by the tension.
Component of the tension along the motion \(= 60\cos 25^\circ = 54.38\) N.
\(W = 54.38 \times 8\).
Answer: 435 J (3 s.f.)
Your turn
A force of 40 N acts at \(60^\circ\) to the direction of motion of a particle that moves 5 m. Find the work done by the force.
\(W = 40 \times 5 \times \cos 60^\circ = 100\) J.
Examiner trap
Do not use the full force when it is at an angle: \(60 \times 8 = 480\) J is the work only if the rope is horizontal.
Section 02
Against gravity and friction.
Lifting a mass through a height \(h\) needs work \(mgh\) against gravity, whatever path it takes. Dragging against friction needs work \(\mu R \times d\). On a slope the vertical height is \(d\sin\alpha\).
Worked example Lifting at constant speed
A crane lifts a load of mass 200 kg through 15 m at constant speed. Find the work done by the crane.
Constant speed, so the tension equals the weight: \(T = 2000\) N.
\(W = 2000 \times 15 = 30\,000\) J.
Answer: 30 kJ, equal to the work done against gravity.
Worked example On a rough slope
A particle of mass 3 kg slides 6 m down a rough plane inclined at \(30^\circ\), \(\mu = 0.2\). Find the work done by the weight and the work done against friction.
Vertical drop \(= 6\sin 30^\circ = 3\) m, so work done by the weight \(= 30 \times 3 = 90\) J.
\(R = 30\cos 30^\circ = 25.98\) N, so friction \(= 0.2 \times 25.98 = 5.196\) N.
Work against friction \(= 5.196 \times 6 = 31.2\) J.
Answer: Weight: 90 J. Against friction: 31.2 J.
Examiner trap
Work done by the weight on a slope uses the VERTICAL height, \(mg \times d\sin\alpha\), not \(mg \times d\). The distance along the slope only multiplies the component \(mg\sin\alpha\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
A 40 N force moves its point of application 3 m in the direction of the force. Find the work done.
Quick check+10 XP
A particle moves along a level floor. How much work does the normal reaction do?
Quick check+10 XP
A 2 kg mass is raised vertically through 5 m. Find the work done against gravity.
Quick check+10 XP
A force of 50 N at \(60^\circ\) to the direction of motion acts over 4 m. Find the work done.
Section 03
Constant speed: work in = work out.
At constant speed the resultant force is zero, so the pulling force does exactly as much work as is done against friction, resistance and gravity. This gives a second way to check an answer.
Worked example Dragged at an angle on a rough floor
A crate of mass 20 kg is pulled at constant speed for 10 m across a rough floor (\(\mu = 0.4\)) by a rope at \(30^\circ\) above the horizontal. Find the tension and the work done by it.
\(T(\cos 30^\circ + 0.2) = 80\), so \(T = 75.05\) N.
Work \(= 75.05 \times 10 \times \cos 30^\circ = 650\) J (check: friction \(0.4 \times 162.5 = 65.0\) N over 10 m is also 650 J).
Answer: \(T = 75.0\) N; work done 650 J.
Worked example Up a slope
A sledge of mass 15 kg is pulled 50 m up a slope inclined at \(10^\circ\) at constant speed by a rope parallel to the slope. The coefficient of friction is 0.1. Find the work done by the rope.
Against gravity: \(150 \times 50\sin 10^\circ = 1302\) J.
Against friction: \(0.1 \times 150\cos 10^\circ \times 50 = 738.6\) J.
Constant speed, so work by the rope \(= 1302 + 738.6\).
Answer: 2040 J (3 s.f.)
Examiner trap
With the rope at an angle, the normal reaction is NOT \(mg\). Using \(\mu mg\) for friction in the first example gives 800 J instead of 650 J.
Stage 2 · Exam
Exam-style questions
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Exam style+20 XP
A rope with tension 60 N at \(25^\circ\) above the horizontal pulls a crate 8 m along level ground. Find the work done by the tension.
Exam style+20 XP
A 20 kg crate is pulled at constant speed for 10 m across a rough floor (\(\mu = 0.4\)) by a rope at \(30^\circ\) above the horizontal. Find the work done by the rope.
Exam style+20 XP
A 3 kg particle slides 6 m down a rough plane inclined at \(30^\circ\), \(\mu = 0.2\). Find the work done against friction.
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A crane raises a 200 kg load through 15 m with constant upward acceleration \(0.5\ \text{m s}^{-2}\). Find the work done by the cable.
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