Power and efficiency flashcards

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Power
The rate of doing work, or the rate of transferring energy. A scalar, unit W.
The watt
1 W = 1 J s⁻¹ (= 1 kg m² s⁻³ in SI base units).
The kilowatt-hour
The energy transferred by a power of 1 kW in 1 hour: 1 kWh = 3.6 × 10⁶ J. A unit of energy, not power.
Efficiency
Useful energy output ÷ total energy input (or useful power output ÷ total power input). No unit; less than 1.
Useful energy output
The energy transferred in the form the device is designed to give, e.g. the gain in GPE for a lift.
Wasted energy
Input energy not transferred usefully, usually dissipated as thermal energy in the surroundings.
Power from energy and from work
P = E/t and P = W/t (Yes)
Efficiency (two forms)
useful energy output ÷ total energy input; useful power output ÷ total power input (Yes)
Power of a force F moving at constant velocity v
P = Fv (from P = W/t = FΔs/t, with Δs/t = v). (No — derive it)
Power to lift a mass m at constant speed v
P = mgv (P = Fv with tension F = mg). (No — derive it)
kWh to J
Multiply by 3.6 × 10⁶ (1000 W × 3600 s). (No — learn it)
Gradient of an energy transferred–time graph
The power.
Area under a power–time graph
The energy transferred (work done).
Sankey diagram
Energy-flow diagram with arrow widths proportional to the energies (or powers); useful output goes straight on, wasted output bends away.
How do the arrow widths of a Sankey diagram show energy conservation?
The output widths add up to the input width: input = useful + wasted.
Why does a vehicle at constant speed still need power?
Resistive forces still act; the driving force must balance them, and it does work at a rate Fv.
Why does a vehicle have a top speed?
Resistive force, and so the power needed, rises with speed; when the power needed equals the maximum power output, it cannot accelerate further.
Running upstairs: which measurements, and with what?
Mass (scales), vertical height (metre rule on one step × number of steps), time (stopwatch, repeated and averaged).
Running upstairs: why is mgh/t an underestimate of a person's power?
It ignores the kinetic energy gained and the energy wasted as thermal energy in the muscles.
Why can efficiency never exceed 100%?
Energy is conserved, so the useful output cannot exceed the input; some energy is always dissipated.
A hoist raises 620 kg through 15 m at constant speed in 25 s. Useful power?
mgΔh/t = 620 × 9.81 × 15 ÷ 25 = 3.6 kW (3649 W).
A boat moves at a constant 30 m s⁻¹ against a 450 N resistive force. Power of the engine's driving force?
Driving force = 450 N; P = Fv = 450 × 30 = 13.5 kW.
Convert 0.75 kWh to joules.
0.75 × 3.6 × 10⁶ = 2.7 × 10⁶ J.
A motor is 60% efficient with a useful output of 1.8 kW. Input and wasted power?
Input = 1.8 ÷ 0.60 = 3.0 kW; wasted = 3.0 − 1.8 = 1.2 kW.
A winch uses 2500 J to raise 64 kg through 3.2 m. Efficiency?
Useful = 64 × 9.81 × 3.2 = 2009 J; efficiency = 2009 ÷ 2500 = 0.80 (80%).
A 52 kg student runs up stairs of vertical height 3.0 m in 2.4 s. Useful power?
52 × 9.81 × 3.0 ÷ 2.4 = 640 W (638 W).
In P = Fv, which force is F?
The force doing the work (e.g. the driving force), not the resultant force — at constant speed the resultant is zero.
Efficiency answer above 1
The ratio is upside down or the useful output is wrong. Efficiency = useful ÷ total input.
Units before substituting into P = E/t
Convert kW → W, minutes or hours → s, kWh → J (× 3.6 × 10⁶).
Efficiency as a ratio or a percentage
Either 0.77 or 77% — never “0.77%”. Match the form the question asks for.

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