Work Done

Work done is the bridge between forces and energy in Cambridge IGCSE 0625. It is a Core topic, so every candidate meets it, usually as a quick 2-3 mark calculation. Examiners test it because students mix up force, work and power under time pressure.

What does “work done” mean in physics?

Work is done when a force moves an object in the direction of the force. The work done equals the energy transferred. In words: work done = force × distance moved in the direction of the force. In symbols: $W = Fd = \Delta E$. The unit is the joule (J); 1 J of work moves a force of 1 N through 1 m.

Quantity	Symbol	Unit
Work done / energy transferred	$W$ (or $\Delta E$)	J
Force	$F$	N
Distance moved (in direction of force)	$d$	m

Two consequences earn marks. First, no movement means no work: holding a heavy bag still does zero work on it, however tired you feel. Second, the distance must be in the direction of the force. Lifting a 50 N box 2 m up while walking 10 m across the room transfers $50 \times 2 = 100\ \text{J}$ against gravity, not $50 \times 10$.

How is work linked to energy stores?

$W = \Delta E$ means every work calculation is an energy calculation. Lifting an object does work against gravity, filling its gravitational potential store: $Fd$ here equals $mg\Delta h$, because the lifting force equals the weight $mg$. Dragging a crate against friction does work that fills the internal (thermal) store of the surfaces. Stating which store the energy enters is a common 1-mark add-on.

Worked Exam Question

A worker pushes a crate 8.0 m across a warehouse floor with a constant horizontal force of 150 N.

(a) Calculate the work done on the crate. [2] (b) The crate moves at constant speed. State the size of the friction force and explain what happens to the energy transferred. [2]

Solution (a). Equation: $W = Fd$. Substitute: $W = 150 \times 8.0$. Answer: $W = \textbf{1200 J}$.

Solution (b). Friction = 150 N (constant speed, so resultant force is zero). The 1200 J is transferred to the internal (thermal) store of the floor and crate by heating; it is dissipated.

Mark scheme

• M1: $W = Fd$ stated or used with $150 \times 8.0$.
• A1: 1200 J with unit.
• B1: friction = 150 N.
• B1: energy dissipated / transferred to internal (thermal) store of surroundings.

Common Mistakes

• Multiplying force by time instead of distance. Fix: $W = Fd$; time belongs in the power equation, not this one.
• Using the wrong distance. Only distance in the force’s direction counts. Fix: for lifting, use vertical height; ignore horizontal walking.
• Leaving the unit off or writing N/m. Fix: work and energy are always joules (J).
• Saying friction “destroys” the energy. Fix: the energy is dissipated to the internal store of the surroundings, so conservation of energy still holds.
• Forgetting weight $= mg$ when lifting. A 12 kg load needs $F = 12 \times 9.8 = 117.6\ \text{N}$, not 12 N. Fix: convert mass to weight with $g = 9.8\ \text{N/kg}$ (some papers say 10, so read the question).

Exam Technique Tip

Write the equation, substitution and answer as three separate lines every time. Cambridge mark schemes award M1 for a correct substitution even when the arithmetic fails, but only if the examiner can see it. A one-line answer that is numerically wrong scores zero; a three-line answer with the same slip still scores one. Two marks per paper hinge on this habit alone.

How This Is Examined

Work done sits on every written paper. Papers 1 and 2 (MCQ) ask single-step $W = Fd$ questions or “in which case is no work done?” conceptual picks. Papers 3 and 4 embed it in multi-part questions: calculate weight, then work done lifting, then power, with each part feeding the next. Paper 6 can ask for work calculated from a measured force and distance in a friction or ramp experiment. Core and Extended use the same equation; Extended versions simply chain it with $E_k = \dfrac{1}{2}mv^2$ or efficiency. Drill the three-line layout until it is automatic.