The big idea: Work is the energy transferred when a force moves something a distance.
No movement → no work. Push a wall and it doesn't budge → you do zero work on it.
Unit: the joule (J) — the same unit as all energy.
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).
When the force points along the direction of motion, work = force × distance. 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) |
Worked example — work at an angle
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?
Solution
- Start with the given formula:
- Put in the numbers (F = 25, s = 12, θ = 30°):
- Work it out — keep the unit:
Final answer
W ≈ 260 J. (Only the part of the pull along the snow does work, so the cos 30° shrinks it.)
Never wonder what to study next
Get a personalized daily plan based on your exam date, progress, and weak areas. We'll tell you exactly what to review each day.
How this is tested: Force–distance graphs are a classic Paper 1A / Paper 2 task.
- State it: the area under a force–distance graph = the work done. - Calculate it: read the area off the graph, then use that work to find a final speed (the work becomes kinetic energy).
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: E_k = 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.
- kinetic energy (J)
- mass (kg)
- speed (m s⁻¹)
IB-style question — (a) what the area represents
A trolley of mass 2.0 kg starts from rest. A constant net force of 9.0 N acts on it as it moves 4.0 m, shown on a force–distance graph. State what the area under the graph represents.
Solution
- The area under a force–distance graph is the work done by the force.
- Here that area is:
Final answer
The area represents the work done by the net force on the trolley — here 36 J.
IB-style question — (b) the final speed
The same trolley (mass 2.0 kg, starts from rest, 36 J of work done on it). Find its final speed.
Solution
- From rest, all the work becomes kinetic energy — start with the given formula:
- Set the kinetic energy equal to the work and put in the numbers (Ek = 36, m = 2.0):
- Rearrange and solve — keep the unit:
Final answer
v = 6.0 m s⁻¹ — the work from the graph area (36 J) became kinetic energy.