Question

An electric motor of power 200 W is switched on for 1 minute and 40 seconds. If 60% of the energy of the motor is useful, calculate

1. useful work done by the motor
2. load lifted by it through a vertical height of 10 m.

[g = 10 ms⁻²]

Answer 1

(a) Energy supplied to the motor = P × t

\[= 200\ \frac{\mathrm{J}}{\mathrm{s}} \times 100\ \mathrm{s} \]

= 20000 J

∴ Useful work done by the motor \[ = 20000\ \frac{\mathrm{J}}{\mathrm{s}} \times \frac{60}{100}\ \mathrm{s} \]

= 12000 J

(b) Useful work done by the motor in lifting load = mgh

= m × 10 m s⁻² × 10 m

= m × 100 m² s⁻²

∴ By the Law of Conservation of Energy,

m × 100 m² s⁻² = 12000 J

\[ \therefore\ m = \frac{12000}{100} \]

m = 120 kg