An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.20 ms⁻¹. What would be the minimum horsepower of the motor to be used?…

Question

An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.20 ms⁻¹. What would be the minimum horsepower of the motor to be used? [Take g = 9.8 ms⁻², 1 hp = 746 W]

Answer 1

An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.20 ms⁻¹. What would be the minimum horsepower of the motor to be used? [Take g = 9.8 ms⁻², 1 hp = 746 W]

0.76 hp
1.3 hp ✓
2.1 hp
0.54 hp

Explanation

At constant velocity the motor only needs to supply a force equal to the weight of the elevator.

Force = mg = 500 × 9.8 = 4900 N.

Power = force × velocity = 4900 × 0.20 = 980 W.

Convert to horsepower: 980 ÷ 746 ≈ 1.3 hp.

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