Work is energy transferred when a force acts through a displacement.
Example
A 10 N force moving a box 3 m in the same direction does 30 J of work.
01
Core idea
Work connects a force to what moves.
Work is an energy transfer, not a stored substance. A push can add energy, a resistive force can remove it, and a perpendicular force can transfer none.
Choose the object or system first. Then compare each force direction with the displacement of that same object.
02
Formula toolkit
Name every quantity before substituting.
Constant forceW = F s cos θ
Work is the push (F) times the distance moved (s), counting only the part of the push that points along the motion — that is what cos θ does.
θ is the angle between force and displacement.
SignW > 0 · W = 0 · W < 0
Pushing along the motion adds energy (positive), pushing at right angles adds nothing (zero), and pushing against the motion takes energy away (negative).
With motion · perpendicular · against motion.
Unit1 J = 1 N·m
One joule is the energy used when a force of one newton moves something one metre.
Work and energy use joules, J.
03
Interactive lab
Change one input. Explain the result before changing another.
objectdisplacement sforce F
energy transferpositive
W = F s cos θ866 J
The force transfers energy into the object's motion.
Misconception check
A satellite moves in a circular orbit while gravity points toward the centre. Does gravity do work during a short tangential displacement?
Choose an answer to test the idea.
04
Worked examples
Model → calculate → interpret.
EasyA 50 N horizontal force pushes a box 4.0 m along a floor in the direction of the force. How much work is done?
W = Fs cosθ with θ = 0°, so cosθ = 1.
W = 50 × 4.0 = 200 J.
W = 200 J
MediumA rope pulls a crate 5.0 m with a force of 200 N at 30° above the horizontal. Find the work done by the rope.
The crate moves horizontally, so θ = 30°.
Use the force component parallel to the motion: F cos θ.
W = (200 N)(5.0 m) cos 30°.
W = 866 J (3 s.f.), positive because the rope helps the motion.
HardA sledge is pulled 25 m by a 90 N rope at 35° above the horizontal while friction exerts 40 N backwards. Find the work done by the rope, by friction, and the net work.
Rope: W = Fs cosθ = 90 × 25 × cos35° ≈ 1843 J.
Friction acts opposite the motion: W = −40 × 25 = −1000 J.
The vertical forces do no work; net W ≈ 1843 − 1000 ≈ 843 J.
ChallengingA 2.0 kg block is pushed up a frictionless 30° incline at constant speed for 3.0 m along the slope. Show that the pushing force's work equals the gain in gravitational potential energy.
Constant speed on a frictionless slope: push F = mg sin30° = 2.0 × 9.81 × 0.5 ≈ 9.8 N along the slope.
W(push) = 9.8 × 3.0 ≈ 29 J.
Height gained h = 3.0 sin30° = 1.5 m, so ΔUg = mgh = 2.0 × 9.81 × 1.5 ≈ 29 J — they match because kinetic energy never changes.
Both equal ≈ 29 J: with ΔK = 0, every joule of pushing work becomes potential energy
Leave with this
Three checks before you move on.
01Work is energy transferred by a force through displacement.
02The angle is measured between the force and displacement vectors.
03Resistive forces usually do negative work on the moving object.
Exam diagrams for this topic2 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.Open subsectionHide subsection
See it. Read it. Work it.
These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.
01
InspectRead the figure comment.
02
TraceFollow labels, arrows and axes.
03
AnswerWork one part at a time.
04
CheckReveal hints and marking points.
A Level
01Fig. 8.1Work, energy and power · Deformation of solidsA LevelOpen diagramClose diagram
Figure comment
Fig. 8.1A steel wire hangs vertically from a rigid support drawn as a hatched ceiling, and a block is attached to its lower end. An arrow starting at the centre of that block points vertically downwards and is labelled 45 N. A dimension line to the left of the wire marks its original length as 2.50 m, and a leader line to the wire labels it as a steel wire of diameter 0.56 mm. The figure is not to scale.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. The 45 N is the weight of the block, so it equals the tension only if the wire's own weight is ignored; 0.56 mm is a diameter, and 2.50 m is the length before loading.
aCalculate Calculate the cross-sectional area of the wire labelled in Fig. 8.1.
recall2 marks
Check answer · 2 marks
A = πd²/4 with d = 0.56 × 10⁻³ m
A = 2.5 × 10⁻⁷ m²
bDetermine Steel of this type breaks at a tensile stress of 8.0 × 10⁸ Pa. Determine the greatest load this wire could support, and the factor by which the 45 N load could be increased before the wire breaks.
routine3 marks
Check answer · 3 marks
maximum load = stress × area = 8.0 × 10⁸ × 2.46 × 10⁻⁷
= 2.0 × 10² N
factor = 197 / 45 = 4.4
cDeduce The wire is replaced by one of the same steel and the same original length but of diameter 1.12 mm, and the same 45 N load is hung from it. Deduce the factor by which the extension changes.
demanding3 marks
Check answer · 3 marks
doubling the diameter makes the cross-sectional area 4 times greater
for the same load the stress, and hence the strain, is one quarter of its previous value
the extension becomes one quarter of its previous value
dSuggest The density of steel is 7800 kg m⁻³. Suggest, with a supporting calculation, whether taking the tension in the wire to be 45 N along its whole length is justified.
weight of wire = 7800 × 6.2 × 10⁻⁷ × 9.81 = 0.047 N
this is about 0.1% of 45 N, and only the part of the wire below a given point adds to the tension there
so treating the tension as 45 N throughout introduces no significant error
Transfer challenge
A climbing rope of unstretched length 12 m and cross-sectional area 1.1 × 10⁻⁴ m² stretches by 0.16 m when a climber of weight 750 N hangs at rest from it. Calculate the Young modulus of the rope material.
Check answer · 3 marks
stress = 750 / 1.1 × 10⁻⁴ = 6.8 × 10⁶ Pa
strain = 0.16 / 12 = 0.013
E = stress / strain = 5.1 × 10⁸ Pa
IB
02Figure 4Work, energy and powerIBOpen diagramClose diagram
Figure comment
Figure 4Side view, drawn not to scale. A road climbs to the right from level ground, and the angle between the road and a dashed horizontal line drawn from the foot of the slope is marked 4.0°. The cyclist and bicycle are drawn as one wheeled body on the road, labelled total mass 78 kg. An arrow from the front of the body points up the slope and is labelled 5.5 m s⁻¹; a second arrow from the rear points down the slope and is labelled 25 N.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. The 25 N arrow points down the slope, so it is the resistance and not the drive; the 4.0° is measured from the dashed horizontal, so the weight component along the road uses sin, not cos.
aState State the resultant force acting on the cyclist and bicycle while they travel up the slope at the speed shown, and give a reason.
recall2 marks
Check answer · 2 marks
zero
the velocity is constant, so by Newton's first law the forward force, the weight component and the 25 N resistance balance
bDetermine Determine the gravitational potential energy gained by the cyclist and bicycle during one minute of riding at the speed shown.
routine3 marks
Check answer · 3 marks
distance along the road = 5.5 × 60 = 330 m
vertical rise = 330 × sin 4.0° = 23.0 m
ΔE_p = 78 × 9.81 × 23.0 = 1.76 × 10⁴ J
cDetermine Determine the work done against the resistive force over the same minute, and hence determine the percentage of the cyclist's useful work that goes into raising her and the bicycle.
demanding3 marks
Check answer · 3 marks
work against resistance = 25 × 330 = 8.25 × 10³ J
total useful work = 1.76 × 10⁴ + 0.825 × 10⁴ = 2.59 × 10⁴ J
dDiscuss Further on, the road steepens to 8.0° and the cyclist holds the same useful power output of 431 W. Discuss how her steady speed changes, supporting your answer with a calculation, and whether the resistive force would still be 25 N.
top of the paper4 marks
Check answer · 4 marks
component of weight down an 8.0° slope = 78 × 9.81 × sin 8.0° = 106 N
forward force needed = 106 + 25 = 131 N, so v = 431 / 131 = 3.3 m s⁻¹
the force needed has risen from 78.4 N only to 131 N, a factor of 1.7, so the speed falls by 1.7 and not by the factor of 2 the doubled angle might suggest — the 25 N resistance is unchanged while only the weight component doubles
air resistance is part of the 25 N and falls as the speed falls, so the true steady speed is a little above 3.3 m s⁻¹ and this figure is a lower bound
Transfer challenge
A lift of total mass 850 kg is raised vertically at a steady 1.2 m s⁻¹. Frictional forces on the lift total 400 N. Determine the useful power output of the motor.
Check answer · 4 marks
the motion is now vertical, so the whole weight opposes it: 850 × 9.81 = 8.34 × 10³ N
total upward force required = 8.34 × 10³ + 400 = 8.74 × 10³ N
P = Fv = 8.74 × 10³ × 1.2 = 1.0 × 10⁴ W
notes that the sine factor of the cyclist's slope has become 1, which is why so much more power is needed at a lower speed