# Work done by a force

> CIE A-Level Physics · Unit 5: Work, energy and power
> Source: https://www.owlsprep.com/study/cie-9702-u5-work-done-by-a-force/

This sub-topic covers the definition of work done by a constant force, calculation of work when force is at an angle to displacement, and finding work from force-displacement graphs. We also explain positive, negative and zero work.

**Prerequisites:** [Scalar and vector quantities](https://www.owlsprep.com/study/cie-9702-u1-scalar-vector-quantities/); [Resolution of forces](https://www.owlsprep.com/study/cie-9702-u2-forces-resolution/)

## Learning objectives

- Define work done by a constant force
- Calculate work done for forces at an angle to displacement
- Distinguish between positive, negative and zero work done
- Calculate work done from a force-displacement graph
- Find net work done by multiple forces

## Definition and Formula for Work Done

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.

**Work done by a constant force** — Energy transferred when a force acts on an object that moves through a displacement

*Notation:* $W = Fs \cos\theta$

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

$$W = F s \cos \theta$$

> **tip**
>
> Work done is energy transferred, so it has the same units as energy: joules (J). 1 J = 1 N·m (1 newton-metre).

**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, $\theta = 30^\circ$
2. Substitute into the work done formula:
3. $$W = (25)(4.0)\cos(30^\circ)$$
4. Calculate: $\cos(30^\circ) \approx 0.866$, so:
5. $$W = 100 \times 0.866 = 86.6 \approx 87 \text{ J}$$

## Positive, Negative and Zero Work Done

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

- If **0 ≤ θ < 90°**: $\cos\theta > 0$, work done is positive (force does work on the object, adds energy to it)
- If **90° < θ ≤ 180°**: $\cos\theta < 0$, work done is negative (force opposes motion, removes energy from the object)
- If **θ = 90°**: $\cos\theta = 0$, 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 $\theta = 180^\circ$
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.

**Check your understanding**

Check your understanding of work signs:

1. 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

   *Why:* Correct! There is no displacement, so work done is zero regardless of the force applied.

## Work Done from Force-Displacement Graphs

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.

> **info**
>
> This works for both constant and varying forces: for a constant force, the area under the graph is a rectangle, so area = $F \times s = Fs \cos\theta$, matching our original formula.

**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 = $\frac{1}{2}Fx$, 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. $$W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.5 \times 20 = 5.0 \text{ J}$$

## Net Work Done by Multiple Forces

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: $\theta = 0^\circ$, $W_{\text{pull}} = 20 \times 2 \times \cos 0 = +40$ J
3. Friction: $\theta = 180^\circ$, $W_{\text{friction}} = 5 \times 2 \times \cos 180 = -10$ 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.

## Common pitfalls

- **Wrong:** Forgetting to multiply by $\cos\theta$ and just calculating $W = Fs$ when force is at an angle.
  - Why it fails: Only the component of force in the direction of displacement contributes to work done, so this gives an overestimated value.
  - Correct: Always identify the angle between the force vector and displacement vector, and include the $\cos\theta$ term in the calculation.
- **Wrong:** Ignoring the sign of work done and always giving a positive value.
  - Why it fails: Negative work indicates energy is removed from the object, which is an important detail exam questions expect.
  - Correct: Always check the direction of the force relative to displacement, and keep the negative sign for forces opposing motion.
- **Wrong:** Calculating the area above the force-displacement graph instead of under it.
  - Why it fails: 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: Always find the area bounded by the force curve, the x-axis, and the start/end displacement lines.
- **Wrong:** Claiming work done is non-zero when a force acts on a stationary object.
  - Why it fails: Work requires displacement: if $s = 0$, then $W = 0$ regardless of how large the force is.
  - Correct: Always confirm the object moves through a displacement before calculating work done.

## Cheatsheet

| Concept | Formula/Rule | Key Note |
| --- | --- | --- |
| Constant force work | $W = Fs \cos\theta$ | Unit: joules (J), 1 J = 1 N·m |
| $\theta = 0^\circ$ (same direction) | $W = +Fs$ | Positive work = energy added to object |
| $\theta = 90^\circ$ (perpendicular) | $W = 0$ | No work done, no energy transferred |
| $\theta = 180^\circ$ (opposite direction) | $W = -Fs$ | 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 | $W_{\text{net}} = \sum W_i = F_{\text{net}} s$ | Keep signs when summing individual work |

## 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.

- [Power](https://www.owlsprep.com/study/cie-9702-u5-power/)
- [Gravitational potential energy and kinetic energy](https://www.owlsprep.com/study/cie-9702-u5-gravitational-potential-energy-and-kinetic/)
- [Conservation of Energy](https://www.owlsprep.com/study/cie-9702-u5-conservation-of-energy/)

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