An electric motor lifts a crate of mass 40 kg vertically through a height of 6.0 m at a constant speed, taking 15 s to complete the lift. During this time, the motor is supplied with 3200 J of electrical energy. Take gravitational field strength g = 10 N/kg. (a) Calculate the us…

IGCSE Physics 0625: Work, Power and Efficiency of an Electric Motor

Part (a): Useful work done

Lifting at constant speed, the useful work done equals the work done against gravity, $W = Fd$ , where the force is the crate’s weight $mg$ and the distance is the height risen:

$W = mgh = 40 \times 10 \times 6.0$

$W = 2400\text{ J}$

Part (b): Efficiency of the motor

Efficiency compares the useful energy output with the total energy input:

$\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%$

$\text{efficiency} = \frac{2400}{3200} \times 100\%$

$\text{efficiency} = 75\%$

Part (c): Useful output power

Power is the useful energy transferred per unit time:

$P = \frac{E}{t} = \frac{2400\text{ J}}{15\text{ s}}$

$P = 160\text{ W}$

Part (d): Where the wasted energy goes

The $800\text{ J}$ ( $3200\text{ J} - 2400\text{ J}$ ) that is not usefully transferred is mostly dissipated as heat, due to friction in the motor’s moving parts (bearings and gears) and electrical resistance in its windings.

Final answers

(a) Useful work done $=$ $2400\text{ J}$
(b) Efficiency $=$ $75\%$
(c) Useful output power $=$ $160\text{ W}$
(d) Wasted energy is mostly transferred as heat