Calculate work done by forces at any angle, understand energy conservation, and connect power to force and velocity.
AQA A-Level Physics · Unit 4: MechanicsWork is done when a force causes a displacement in the direction of the force. When the force and displacement are not parallel, only the component of force along the displacement does work.
| Symbol | Quantity | Unit |
|---|---|---|
| W | Work done | J (joules) |
| F | Force applied | N |
| d | Displacement | m |
| θ | Angle between force and displacement | ° |
Special cases:
On a force-displacement graph, the area under the curve equals the work done. This is especially useful for springs, where the force is not constant.
| Symbol | Quantity | Unit |
|---|---|---|
| KE | Kinetic energy | J |
| m | Mass | kg |
| v | Speed | m s⁻¹ |
| GPE | Gravitational potential energy | J |
| g | Gravitational field strength | N kg⁻¹ (= m s⁻²) |
| h | Height above reference level | m |
The work-energy theorem connects work done to change in kinetic energy:
The net work done on an object equals its change in kinetic energy. This can be derived from F = ma and the suvat equation v² = u² + 2as: multiplying both sides by ½m gives ½mv² − ½mu² = Fs = W.
Gravitational potential energy is measured relative to a chosen reference level (often the ground). Only changes in GPE matter — the absolute value depends on your reference.
For a system where only gravity does work (no friction, no air resistance):
When friction or air resistance is present, mechanical energy is not conserved — some is converted to thermal energy (heat). In this case:
Or equivalently: the decrease in mechanical energy equals the work done against resistive forces.
Conservation of energy is one of the most powerful tools in physics — it avoids needing to calculate forces and accelerations at every instant, giving answers directly in terms of energy.
Power is the rate of doing work, or the rate of energy transfer.
| Symbol | Quantity | Unit |
|---|---|---|
| P | Power | W (watts) |
| W | Work done | J |
| t | Time | s |
| F | Driving force | N |
| v | Speed | m s⁻¹ |
The formula P = Fv is derived: P = W/t = (Fd)/t = F(d/t) = Fv. It applies when the force is constant and parallel to the velocity.
At terminal velocity, the driving force equals the resistive forces, so the net force is zero and acceleration is zero. Using P = Fv: if the engine power is constant, terminal velocity is reached when all engine power is dissipated by resistive forces.
Efficiency is always between 0% and 100%. It tells you what fraction of the input energy is actually transferred to the intended useful form. The rest is wasted as heat, sound, etc.
Q1. A force acts perpendicular to the direction of motion. How much work does it do?
Q2. A 2 kg ball is thrown upward at 10 m s⁻¹. Using energy conservation, find the maximum height reached. (g = 9.81 m s⁻²)
Q3. A motor lifts a 500 N load through 8.0 m in 5.0 s. If the motor is 75% efficient, what input power is required?
Q4. A car engine produces 60 kW and the car has a drag force of 1200 N. What is the maximum (terminal) speed?
Q5. Which energy transfer occurs when friction acts on a sliding block?
Challenge Q1. A ski slope is 500 m long and descends 120 m vertically. A 70 kg skier starts from rest. If the friction force is 85 N throughout, calculate the speed at the bottom of the slope.
Challenge Q2. Derive the equation KE = ½mv² starting from Newton's second law (F = ma) and the kinematic equation v² = u² + 2as. Show clearly each step.
Challenge Q3. A vehicle of mass 1500 kg accelerates from 0 to 30 m s⁻¹ in 12 s against a constant drag force of 600 N. Assuming constant driving force, calculate (a) the driving force and (b) the power at maximum speed.