Work done by a constant force: positive, negative and zero work
Work W = Fd cos θ: the force component along the displacement times the displacement. It can be positive, negative or zero; its unit is the joule.
Work in physics
In daily speech, studying or holding a heavy bag is 'work'. In physics, work is done only when a force moves a body. Let a constant force F act at an angle θ to the displacement d. Only the part of the force along the displacement, F cos θ, does work. So W = Fd cos θ. In vector form, W = F⃗·d⃗ (the dot product). Work is a scalar. Its SI unit is the joule: 1 J is the work done when a force of 1 N moves a body 1 m along the force. Push hard against a wall: it does not move, so no work is done, though you feel tired.
Positive, negative and zero work
| Angle θ between F and d | cos θ | Work done | Example |
|---|---|---|---|
| θ = 0° (force along the motion) | 1 | Positive, W = Fd | The engine force on a car when you press the accelerator |
| 0° < θ < 90° | Positive | Positive, less than Fd | Pulling a suitcase by a slanting handle |
| θ = 90° | 0 | Zero | Gravity on a ball rolling on a level floor; the pull of the string on a stone whirled in a circle |
| 90° < θ ≤ 180° | Negative | Negative; W = −Fd at 180° | Brakes on a moving car; friction on a sliding box |
Work done by gravity: lifting and lowering
In both cases the body is moved without acceleration. When a body moves along a level path, gravity does zero work on it.
Worked example: pulling at an angle
A child pulls a toy cart with a force of 10 N along a string at 60° to the ground. The cart moves 5 m along the ground. Find the work done by the pull.
- Given: F = 10 N, d = 5 m, θ = 60°, cos 60° = 0.5.
- W = Fd cos θ = 10 N × 5 m × 0.5.
- W = 25 J.
25 J. Only half the pull (F cos 60°) acts along the motion
Worked example: carrying a cement bag upstairs
A worker carries a 25 kg cement bag up to a floor 6 m higher, at steady speed. Find the work done by the worker against gravity, and the work done by gravity. Take g = 9.8 m s⁻².
- Force needed = weight = mg = 25 kg × 9.8 m s⁻² = 245 N.
- Work by the worker = mgh = 245 N × 6 m = 1470 J.
- Gravity acts down while the bag moves up, so the work by gravity = −1470 J.
Work by the worker = +1470 J (about 1.5 kJ); work by gravity = −1470 J
Worked example: work from force and displacement vectors
A force F⃗ = (3î + 4ĵ) N moves a particle through d⃗ = (2î − ĵ) m. Find the work done.
- W = F⃗·d⃗ = Fx dx + Fy dy.
- W = (3 × 2) + (4 × (−1)) = 6 − 4.
- W = 2 J.
2 J
A railway porter does work against gravity while lifting a bag onto his head, but gravity does no work on it along the level platform. Scooter brakes do negative work and take away the scooter's energy of motion.
2 marks: “When is the work done by a force zero? Give two examples.” When there is no displacement, or when the force is at right angles to the displacement (θ = 90°). Examples: pushing a fixed wall; gravity on a body moving along a level road; the string's pull on a stone whirled in a circle.
2 marks: “A force of 20 N moves a box 3 m along a floor while acting at 60° to the floor. Find the work done.” W = Fd cos θ = 20 × 3 × cos 60° = 60 × 0.5 = 30 J.
1 marks: “Is work a scalar or a vector? Give its SI unit.” A scalar; the joule (J).
Work needs movement along the force: W = Fd cos θ, with cos 0° = 1, cos 90° = 0, cos 180° = −1.