Work done by a force

CIE A-Level Physics· 9702 AS & A Level Physics Syllabus 2022-2024 Section 5.1· 15 min read

1. Definition and Formula for Work Done★★☆☆☆⏱ 4 min

Work done is defined as the product of the force component in the direction of displacement, multiplied by the magnitude of displacement. It is a scalar quantity, even though both force and displacement are vectors.

📘 Definition

Work done by a constant force

$W = F s cos θ$

Energy transferred when a force acts on an object that moves through a displacement

Example:

A 10 N force pulling a box 2 m along a horizontal surface does 20 J of work.

W = F s \cos \theta

📐 Worked Example

Calculate the work done by a 25 N constant force acting at 30° to the horizontal, pulling an object 4.0 m horizontally across a frictionless surface.

1
Identify known values: $F = 25$ N, $s = 4.0$ m, $θ = 3 0^{\circ}$
2
Substitute into the work done formula:
3
$W = (25)(4.0)\cos(30^\circ)$
4
Calculate: $cos (3 0^{\circ}) \approx 0.866$ , so:
5
$W = 100 \times 0.866 = 86.6 \approx 87 \text{ J}$

2. Positive, Negative and Zero Work Done★★☆☆☆⏱ 4 min

The value of $cos θ$ changes with the angle between force and displacement, so work can be positive, negative or zero depending on this angle:

If 0 ≤ θ < 90°: $cos θ > 0$ , work done is positive (force does work on the object, adds energy to it)
If 90° < θ ≤ 180°: $cos θ < 0$ , work done is negative (force opposes motion, removes energy from the object)
If θ = 90°: $cos θ = 0$ , work done is zero (force is perpendicular to displacement, no energy transferred)

📐 Worked Example

A moving box slides 3 m across a rough floor and is slowed by a frictional force of 12 N. What is the work done by friction?

1
Friction acts opposite to the direction of displacement, so $θ = 18 0^{\circ}$
2
Substitute into the work done formula:
3
$W = F s \cos(180^\circ) = (12)(3)(-1) = -36 \text{ J}$
4
The negative sign confirms friction removes energy from the box, as expected.

✓ Quick check

Check your understanding of work signs:

A person holds a 10 kg bag stationary above their head for 1 minute. What is the work done by the person on the bag?
- 0 J
- 98 J
- 980 J
Reveal answer
0 J —
Correct! There is no displacement, so work done is zero regardless of the force applied.

3. Work Done from Force-Displacement Graphs★★★☆☆⏱ 5 min

For a force that changes with displacement (for example, the force needed to stretch a spring), we cannot use the constant force formula directly. Instead, the total work done is equal to the area under the force-displacement (F-s) graph.

📘 Definition

Area under F-s graph

The total work done by the force as the object moves between two displacements on the graph

Example:

A triangular area for a stretching spring gives work = $\frac{1}{2} F x$ , matching the elastic potential energy formula.

📐 Worked Example

A spring is stretched from 0 to 0.5 m. The force increases linearly from 0 N to 20 N at maximum extension. Calculate the total work done to stretch the spring.

1
The F-s graph forms a triangle with base equal to extension 0.5 m and height equal to maximum force 20 N.
2
Area of the triangle equals total work done:
3
$W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.5 \times 20 = 5.0 \text{ J}$

4. Net Work Done by Multiple Forces★★★☆☆⏱ 5 min

When multiple forces act on an object at the same time, the net work done on the object is the sum of the work done by each individual force. You can also calculate the net resultant force first, then apply the work done formula to the net force.

Method 1: Calculate work done by each force separately, add all values (keeping positive/negative signs) to get net work
Method 2: Find the net resultant force in the direction of displacement, multiply by displacement to get net work

📐 Worked Example

A 5 kg box is pulled 2 m horizontally by a 20 N force, against a 5 N frictional force. Find the net work done on the box.

1
Method 1: Sum work from individual forces
2
Pulling force: $θ = 0^{\circ}$ , $W_{pull} = 20 \times 2 \times cos 0 = + 40$ J
3
Friction: $θ = 18 0^{\circ}$ , $W_{friction} = 5 \times 2 \times cos 180 = - 10$ J
4
Net work $= 40 - 10 = 30$ J
5
Method 2: Use net force
6
$F_{\text{net}} = 20 - 5 = 15 \text{ N}, \quad W_{\text{net}} = 15 \times 2 = 30 \text{ J}$
7
Both methods give the same result, as expected.

5. Common Pitfalls

Wrong move:

Forgetting to multiply by $cos θ$ and just calculating $W = F s$ when force is at an angle.

Why:

Only the component of force in the direction of displacement contributes to work done, so this gives an overestimated value.

Correct move:

Always identify the angle between the force vector and displacement vector, and include the $cos θ$ term in the calculation.

Wrong move:

Ignoring the sign of work done and always giving a positive value.

Why:

Negative work indicates energy is removed from the object, which is an important detail exam questions expect.

Correct move:

Always check the direction of the force relative to displacement, and keep the negative sign for forces opposing motion.

Wrong move:

Calculating the area above the force-displacement graph instead of under it.

Why:

Total work done is defined as the area between the F-s curve and the displacement (x) axis, so this gives an incorrect result.

Correct move:

Always find the area bounded by the force curve, the x-axis, and the start/end displacement lines.

Wrong move:

Claiming work done is non-zero when a force acts on a stationary object.

Why:

Work requires displacement: if $s = 0$ , then $W = 0$ regardless of how large the force is.

Correct move:

Always confirm the object moves through a displacement before calculating work done.

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Note
Constant force work	$W = F s cos θ$	Unit: joules (J), 1 J = 1 N·m
$θ = 0^{\circ}$ (same direction)	$W = + F s$	Positive work = energy added to object
$θ = 9 0^{\circ}$ (perpendicular)	$W = 0$	No work done, no energy transferred
$θ = 18 0^{\circ}$ (opposite direction)	$W = - F s$	Negative work = energy removed from object
Varying force	$W =$ Area under F-s graph	Count all areas, keep sign for negative force
Net work from multiple forces	$W_{net} = \sum W_{i} = F_{net} s$	Keep signs when summing individual work

7. Frequently Asked

When is work done equal to zero?

Work done is zero if the force is perpendicular to displacement ((\cos\theta = 0)) or the object has no displacement, regardless of how large the force is.

Why can work done be negative?

Negative work means the force opposes the motion of the object, so energy is transferred away from the moving object, rather than to it.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 11
MCQ on negative work done by friction
2023 · 22
Calculate work done at an angle to displacement
2024 · 13
Find work from F-displacement graph area

Going deeper

syllabusOfficial CIE 9702 Syllabus

What's Next

Understanding work done is the foundation for the entire work, energy and power unit. The work-energy principle, which states that the net work done on an object equals its change in kinetic energy, follows directly from the definition of work we covered here. Work done also connects to gravitational potential energy, elastic potential energy, and power calculations, all core topics for CIE A-Level Physics exams. Many longer structured exam questions combine work done with force resolution and energy conservation, so mastering this sub-topic is critical for scoring full marks.