Work Done

This lesson explains Work Done for Cambridge IGCSE Physics 0625. It covers the Core knowledge required by every candidate. Focus on the cause-and-effect explanation and the exact quantities being compared. Read the explanation once, then attempt the worked method from a blank page.

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 example

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.

Original marking points

• 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.

These marking points belong to this original example. They are not an official Cambridge mark scheme.

Common errors and how to correct them

• 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).

How to apply this in an exam

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 may not earn credit; a three-line answer with the same slip still scores one. Two marks per paper hinge on this habit alone.

Where this skill matters

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.