Chapter 5 · Work · Lesson 3.1
Work Done
Connect force, displacement and energy.
A force can do positive, negative or zero work. Explore how its direction changes the energy transfer.
Only the parallel force component contributes.
Lesson Overview
Work is energy transferred when a force acts through a displacement.
A 10 N force moving a box 3 m in the same direction does 30 J of work. If the force is at an angle, use its component parallel to the displacement.
Learning Objectives
- Define work as an energy transfer
- Resolve the parallel component of a force
- Calculate W = Fs cos θ in joules
- Explain positive, zero and negative work
- Distinguish individual work from net work
- Apply work ideas to pulling and lifting
Key Concepts
Interactive Work Lab
Change one input. Predict the sign before reading the result.
Force and displacement arrows use separate scales. Other forces are not shown.
Parallel force: 173.2 N
This force does positive work. Add the work of all other forces before deciding whether the kinetic energy changes.
Key Formulae
- Constant force
- W = Fs cos θF is force magnitude (N); s is displacement magnitude (m). θ is the angle between them.
- Parallel component
- F∥ = F cos θOnly the component along the displacement contributes to work.
- Unit
- 1 J = 1 N·mOne newton acting through one metre in the force's direction transfers one joule.
- Net work
- Wnet = ΣWi = ΔKAdd the signed work of all forces on the object to find its change in kinetic energy.
- Work by gravity
- Wg = −mgΔhNear Earth's surface with constant g; upward motion gives negative gravitational work.
- Variable force
- W = ∫ F∥ dsAdd force × tiny displacement along the path. Continue in the graphs and calculus lessons.
Practice Questions
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.
One force is not the whole story
A rope can do positive work while friction does an equal amount of negative work. The net work is zero, so the object's kinetic energy does not change.
Use W = Fs cos θ for a constant force and the displacement of its point of application; the lab treats the object as a translating particle.
Direction decides the sign.
Positive work
0° ≤ θ < 90°The force component points along the displacement.
Zero work
θ = 90°, F = 0 or s = 0A perpendicular force—or no force or displacement—does no work.
Negative work
90° < θ ≤ 180°The force component opposes the displacement.
Misconception check
In an ideal circular orbit, gravity is radial and the instantaneous displacement is tangential. Does gravity do work along the orbit?
Choose an answer to test the idea.
Worked Examples
Choose a challenge. Name the force, resolve it, calculate and check the sign.
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 200 N rope pulls a crate 5.0 m horizontally 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 moves 25 m horizontally along level ground, pulled by a 90 N rope at 35° above the horizontal. Friction exerts 40 N backwards. Find the rope work, friction work and 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.
W(rope) ≈ 1.8 kJ, W(friction) = −1.0 kJ, net ≈ +0.84 kJ
ChallengingA 2.0 kg block is pushed parallel to a frictionless 30° slope at constant speed for 3.0 m. Compare the pushing work with 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.
W(push) ≈ +29 J and W(gravity) ≈ −29 J, so net work on the block is zero. For the block–Earth system, the external pushing work increases gravitational potential energy by ≈29 J.
Work in the World
Lifting a load
A crane lifting steadily does positive work against gravity. The load’s speed stays constant because gravity does equal negative work.
Try Crane Loader →Pulling at an angle
For horizontal motion, only the rope’s horizontal component contributes to its work. Increasing the angle at fixed F and s decreases this contribution.
Compare the angles →Friction and energy
Sliding friction usually does negative work on a moving object. Mechanical energy can be transferred into thermal energy in the object and its surroundings.
Review friction →Exam diagrams for this topic2 figures to inspect and practiseQuestions, hints and marking points in one compact 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 Level
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.
Guided questions 5 parts
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.
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.
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.
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.
Check answer 4 marks
- volume of wire = 2.46 × 10⁻⁷ × 2.50 = 6.2 × 10⁻⁷ m³
- 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 powerIB
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.
Guided questions 5 parts
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.
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.
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.
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
- fraction raising the cyclist = 1.76 × 10⁴ / 2.59 × 10⁴ = 68%
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.
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
Three checks before you move on.
- Work is energy transferred by a force through displacement.
- The angle is measured between the force and displacement vectors.
- Resistive forces usually do negative work on the moving object.
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