The big idea: Drag a heavy suitcase across the floor and you tire out — you're doing work: transferring energy by moving something with a force.
Two facts about work
- No movement, no work: hold that suitcase still and you do zero work on it (however much your arm aches).
- Work is measured in joules (J) — the same unit as all energy.
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Spot it on the graph: On a force–distance (F–x) graph the area under the line = the work done. A flat line → the area is just a rectangle (force × distance).
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When the force points along the direction of motion, work = force × distance.
When the force is at an angle
- If the force is at an angle to the motion, only the part along the motion does work.
- That's where the cos θ comes in.
- θ is the angle between the force and the direction it moves.
- work done (J)
- force applied (N)
- distance moved (m)
- angle between the force and the direction of motion (°)
When the force is along the motion: If the force pushes straight along the motion then θ = 0 and cos 0 = 1, so the equation is just W = Fs (force × distance). Most basic questions are this simple case.
| Angle θ | cos θ | Work done |
|---|---|---|
| 0° — force along the motion | 1 | W = Fs (the most you can get) |
| 60° — force at a slant | 0.5 | W = 0.5 Fs (only half counts) |
| 90° — force across the motion | 0 | W = 0 (no work done) |
| 180° — force against the motion (friction) | −1 | W = −Fs (the force takes energy out) |
A child pulls a sledge 12 m along flat snow with a rope. The rope tension is 25 N and the rope is at 30° above the ground. How much work does the tension do? (cos 30° = 0.87)
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Steeper ramp, same work: Push a box up a frictionless ramp to a height h. On a steep ramp you push harder (mg sin θ) over a shorter distance (h ÷ sin θ). Multiply them: W = mg sin θ × h ÷ sin θ = mgh.
The work depends only on the height, not on the angle of the ramp.
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How this is tested — force–distance graphs are a classic Paper 1A / Paper 2 task. They come in two flavours:
State it
- The area under a force–distance graph = the work done.
- Just name what the area means.
Calculate it
- Read the area (the work) off the graph.
- Turn that work into a final speed with ½mv².
The classic trap: The area gives you energy in joules, not the speed — you still have to put it into ½mv² to get the speed.
Work → kinetic energy → speed: Kinetic energy is the energy a moving object has: Ek = ½mv² (also given in the booklet). If an object starts from rest, the work done on it = its kinetic energy, so you can solve for the speed v.
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- kinetic energy (J)
- mass (kg)
- speed (m s⁻¹)
A 2.0 kg trolley starts from rest. A constant net force of 9.0 N acts on it over 4.0 m, shown on a force–distance graph. (a) State what the area under the graph represents, and find it. (b) Find the trolley's final speed.
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In the exam the force is often not constant. Split the area into shapes you know, then add them:
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An acceleration–distance graph: If the graph shows acceleration against distance, its area is work ÷ mass (because F = ma). Multiply the area by the mass to get the work.
A 900 kg car slows at a steady 4.0 m s⁻² over 25 m: area = 4.0 × 25 = 100, so the energy taken out = 900 × 100 = 9.0 × 104 J.