The big idea: Run up a flight of stairs and you're soon panting; stroll up slowly and you're fine — you moved the same energy, but running delivers it faster.
Power
- That rate of transferring energy is power.
- Unit: the watt (W) — just 1 joule per second (J s⁻¹).
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Spot it — P = Fv at a steady speed: When something moves at a constant speed, the driving force balances the resistive (drag) force.
So the power is P = Fv, where F is the resistive force. Faster cruise ⇒ more power needed.
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Power has two given forms in the data booklet. Pick the one that matches what you know.
Which form to use
- Use P = ΔW/Δt when you know the energy transferred and the time.
- Use P = Fv when a force moves at a steady speed (for example a vehicle cruising against drag).
- power — energy transferred per second (W, watts = J s⁻¹)
- work done / energy transferred (J, joules)
- time taken (s)
- the force (N) — at constant speed, equal in size to the resistive force
- speed in the direction of the force (m s⁻¹)
Average vs instantaneous power: Average power = total energy ÷ total time (use ΔW/Δt over the whole interval).
Instantaneous power = the power right now (use Fv with the speed at that moment).
Define: 'instantaneous' means at one instant, not averaged over time.
A motor lifts a load, transferring 6000 J of energy in 4.0 s. Find its average power output.
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A car cruises at a steady 25 m s⁻¹ against a total resistive force of 480 N. Find the power its engine delivers.
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Efficiency
No machine turns all the energy it takes in into the job you want. Efficiency is the fraction that is useful. Use energies or powers, whichever you are given:
- efficiency (no unit)
- the energy (J) or power (W) that does the wanted job
- all the energy (J) or power (W) supplied
Useful power when lifting: Lifting at a steady speed v: the force equals the weight, so useful power = mgv.
Pumping water up a height h: useful power = (mass lifted per second) × g × h.
Two ropes each pulling with tension F at θ to the vertical lift at speed v: useful power = 2F cos θ × v (only the upward parts do work).
A motor that is 75 % efficient lifts a 40 kg load at a steady 0.50 m s⁻¹ (g = 9.8 N kg⁻¹). Find the power supplied to the motor.
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How this is tested — power and efficiency appear across both papers:
Paper 1A
- A quick average power (energy ÷ time).
- e.g. a spring releasing its stored energy.
Paper 2
- Determine a drag constant from power and speed at constant velocity, then state its SI unit.
- Or average power delivered to an object from rest.
The classic trap: At constant speed the engine force equals the resistive force, so P = Fv lets you find that force (or a drag constant) from the power and the speed.
Drag constant from power & speed: For many objects the drag force grows with speed as F = cv, where c is the drag constant.
At constant speed P = Fv = cv × v = cv², so c = P ÷ v².
Its unit comes from the units: W ÷ (m s⁻¹)² = kg s⁻¹ (work it out from N = kg m s⁻²).
A boat moves at a steady 5.0 m s⁻¹. Its engine delivers 750 W, and the drag force is F = cv. (a) Determine the drag constant c. (b) State its fundamental SI unit.
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