Power and Efficiency

Work and energy tell you how much energy has been transferred, but not how fast. A marathon runner and a sprinter can do the same total work — climbing the same flight of stairs, say — yet the sprinter does it in a tenth of the time. That time difference is captured by the concept of power.

Efficiency, meanwhile, quantifies how much of the input energy is actually used for its intended purpose, as opposed to being wasted as heat, sound or friction. Together these two ideas are central to engineering, environmental physics and the energy sections of OCR Module 3.3.1.

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Definition of Power

Power is the rate of doing work, or equivalently, the rate of energy transfer.

P = W / t = ΔE / t

SI unit: watt (W) = J s⁻¹. One watt is the rate of doing 1 joule of work per second.

Scale	Typical power
Human at rest (basal metabolism)	80 W
Bright incandescent bulb	100 W
Person cycling hard	300 W
Kettle	2 kW
Sports car	150 kW
Wind turbine	2–3 MW
Nuclear reactor	1–3 GW
Total human civilisation	19 TW
Sun	3.8 × 10²⁶ W

Other units sometimes seen in the UK:

• Horsepower (hp): 1 hp ≈ 746 W (imperial horsepower).
• kilowatt-hour (kWh): a unit of energy (not power!), equal to 1 kW for 1 hour = 3.6 MJ. This is how your electricity bill is calculated.

Common Exam Mistake: Confusing kWh (energy) with kW (power). A 2 kW kettle run for 30 minutes consumes 1 kWh of energy.

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Power in Terms of Force and Velocity

For a constant force F acting along the direction of motion:

W = F s P = W / t = F × s/t = F v

So:

P = F v

This is extremely useful whenever a motor or engine is maintaining a constant speed against resistance.

Worked Example — Car at Constant Speed

A car travels at a steady 30 m s⁻¹ against a total resistive force of 700 N. Find the useful power output of the engine.

At constant speed, the driving force equals the resistive force (Newton I): F = 700 N.

P = Fv = 700 × 30 = 21 000 W = 21 kW

This is the useful output power delivered by the engine at the wheels. The actual fuel energy consumed is much higher because of engine inefficiencies — see below.

Worked Example — Cyclist

A cyclist rides uphill at 5.0 m s⁻¹ on a 5° slope. She and her bike together have a mass of 75 kg. Ignoring drag and friction, find the power she must generate.

Force needed up the slope = mg sin θ = 75 × 9.81 × sin 5° = 735.8 × 0.0872 = 64.1 N

P = Fv = 64.1 × 5.0 = 321 W

This is close to the maximum sustained output of an average amateur cyclist. Tour de France riders sustain around 400 W for long periods, and over 1000 W in sprints.

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Worked Example — Lift Motor

A lift of total mass 800 kg ascends at a steady 2.5 m s⁻¹. Ignoring friction, find the useful power of the motor.

At constant velocity the lift is in equilibrium, so the upward tension from the cable equals the weight:

T = mg = 800 × 9.81 = 7848 N P = T × v = 7848 × 2.5 = 19 620 W ≈ 19.6 kW

If the lift has to accelerate at 1.5 m s⁻² starting from rest, at the moment when v = 2.0 m s⁻¹, the net upward force must provide the acceleration:

• Net force = ma = 800 × 1.5 = 1200 N
• Tension T = 1200 + 7848 = 9048 N
• P = T × v = 9048 × 2.0 = 18.1 kW

Notice that power varies with velocity during acceleration — this is not a steady-state problem.

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Worked Example — Pump Lifting Water

A pump raises 15 kg of water per second through a height of 8.0 m. Find the useful power output of the pump.

Work done per second = rate of GPE gain = mass-per-second × g × h = 15 × 9.81 × 8.0 = 1177 W ≈ 1.18 kW

This calculation is typical of hydroelectric and irrigation problems. If the pump's electrical input is 1.5 kW, its efficiency is 1177/1500 = 78%.

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Efficiency