If the maximum load carried by an elevator is 1400 kg ( 600 kg - Passengers + 800 kg - elevator), which is moving up with a uniform speed of 3 m s⁻¹ and the frictional force acting on it is 2000 N, then the maximum power used by the motor is __________ kW(g=10 m / s²)

Solution

Physics sir · 3 steps

Uniform speed means zero net force, so the motor must supply weight plus friction; then multiply by speed.

1. Forces on the lift. Moving up, the elevator is pulled down by its full weight mg and also by friction f, which opposes the upward motion. The speed is uniform, so there is no acceleration and the motor's pull must exactly balance both.
2. Find the pull. With the maximum load, mg = 1400 × 10 = 14000 N. Adding friction, F = 14000 + 2000 = 16000 N.

   P = Fv = (mg + f) v = 16000 × 3 = 48000 W

3. So. The maximum power used by the motor is 48000 W = 48 kW.

The common slip: 42. Leaving out the 2000 N friction force uses only mg = 14000 N and gives 42 kW instead of 48.

The idea to keep. For anything moving at constant velocity, find the force the engine must supply to cancel every opposing force, then use P = Fv. Friction always adds to the load when it opposes the motion.

Answer from the NTA final answer key. Solution by PrepWiser.