In the Cambridge IGCSE 0625 syllabus students must understand that mechanical or electrical work done is equal to the energy transferred. In other words, work is the amount of energy moved from one system to another by a force (mechanical) or by an electric current (electrical).
Core (scalar) form – used when the force and displacement are parallel:
\$W = F\,s\$
General form – when the force makes an angle θ with the displacement:
\$W = F\,s\,\cos\theta\$
\$W = VQ \qquad\text{or}\qquad W = V I t\$
Because \$V\times Q\$ (or \$V I t\$) has the unit joule, the result is the amount of energy transferred by the circuit.
The numerical value of work done on (or by) a system is exactly the amount of energy transferred to (or from) that system:
\$\boxed{\text{Work} = \text{Energy transferred}}\$
Consequently the joule (J) is the SI unit for both work and energy.
| Quantity | Symbol | SI Unit | Definition |
|---|---|---|---|
| Force | F | newton (N) | 1 N = 1 kg·m·s⁻² |
| Displacement | s | metre (m) | – |
| Work / Energy | W | joule (J) | 1 J = 1 N·m = 1 kg·m²·s⁻² |
| Potential Difference | V | volt (V) | 1 V = 1 J·C⁻¹ |
| Charge | Q | coulomb (C) | 1 C = 1 A·s |
| Current | I | ampere (A) | – |
| Time | t | second (s) | – |
Mechanical work (parallel force)
A student pushes a 15 kg crate across a smooth floor with a constant horizontal force of 60 N over a distance of 5 m.
Solution:
\$W = F\,s = 60\;\text{N}\times5\;\text{m}=300\;\text{J}\$
The work is positive, so 300 J of kinetic energy is transferred to the crate.
Mechanical work (angled force)
A 10 N force pulls a sled up a 30° slope for 8 m.
Solution:
\$W = F\,s\,\cos30^\circ = 10\;\text{N}\times8\;\text{m}\times0.866 = 69.3\;\text{J}\$
Positive work adds 69.3 J of energy to the sled.
Electrical work
A 12 V lamp draws 2 A for 3 minutes.
Solution:
\$t = 3\;\text{min}=180\;\text{s}\$
\$W = V I t = 12\;\text{V}\times2\;\text{A}\times180\;\text{s}=4320\;\text{J}\$
The battery transfers 4320 J of energy to the lamp.
Negative work (braking)
A car of mass 800 kg traveling at 20 m s⁻¹ is brought to rest by a constant braking force over 40 m.
Solution (using \$W = \Delta KE\$):
\$\Delta KE = \tfrac12 m vf^2 - \tfrac12 m vi^2 = 0 - \tfrac12(800)(20)^2 = -160\,000\;\text{J}\$
The brakes do –160 kJ of work, removing that amount of kinetic energy from the car.
The correct statement is: work is done if the force has a component along the displacement. If the component is zero (force ⟂ displacement) no work is done.
Energy can also be transferred as heat, sound, light, etc. Only the part transferred by a force (mechanical) or by moving charge (electrical) is called work.
\$V I\$ is power (watts). Work (or energy) is obtained by multiplying power by the time the power acts: \$W = V I t\$.
Work is a scalar; its sign indicates whether energy is added to (+) or removed from (–) the system.
Power is the rate at which work is done (or energy is transferred):
\$P = \frac{W}{t}\$

Your generous donation helps us continue providing free Cambridge IGCSE & A-Level resources, past papers, syllabus notes, revision questions, and high-quality online tutoring to students across Kenya.