A.3 · IB Physics
Work, power and efficiency
An electric motor winds a cable and lifts a load at constant speed. Set the input power and the efficiency: the useful power sets the lift speed through P = Fv, and the energy flow diagram shows where every joule goes.
Predict first
Start from the default settings (50 kg, 60% efficient, 1000 W). If you raise the Mass of load m slider from 50 to 100 kg, what happens to the lift speed?
Useful power ηP_in = mgv is fixed at 600 W. Doubling m doubles the force mg, so v halves, from 1.22 to 0.612 m s⁻¹.
Check it in the model: Move the Mass of load m slider from 50 to 100 and compare Lift speed v in the readout.
useful wasted load height / input
Puseful = ηPin = Fv = mgv
v = ηPin ÷ mg ; t = h ÷ v
Ein = Pint ; gain in Ep = mgh
g = 9.8 m s⁻²
Try this
- Press Typical motor, note the lift speed, then press Double the load. Explain the new speed using P = Fv.
- Lift the same load through the same height with η = 20% and then η = 90%. Compare the useful energy gained with the energy wasted.
- Set η = 25%. Read the wasted and useful power in the energy flow diagram and show that wasted = 3 × useful. Which input power gives a 1.00 m s⁻¹ lift of 100 kg?
Common error: Using the motor’s input power in P = Fv: only the useful output ηPin lifts the load, and the wasted energy does not vanish, so input = useful + wasted at every instant.