An electric motor of efficiency 80% is used to lift a load of 500 N through a height of 6 m in 10 seconds. What is the power input to the motor?

The useful power output = (Force × distance)/time = (500 × 6)/10 = 300 W. Since efficiency = (power output/power input) × 100%, power input = power output/efficiency = 300/0.80 = 375 W. Option B is the output power (ignoring efficiency), and option D results from multiplying output power by efficiency instead of dividing.