Question Detail
Question
A 1000-kg elevator carries a maximum load of 800 kg. A constant frictional force of 4000 N retards its motion upward. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3 m/s?
Correct Answer
Option A is the correct answer.
Detailed Explanation
Total mass = 1000 kg + 800 kg = 1800 kg, so weight Mg = 1800 kg × 9.8 m/s² = 17640 N. At constant speed, motor tension T = weight + friction = 17640 N + 4000 N = 21640 N. Power P = T v = 21640 N × 3 m/s = 64920 W = 64.92 kW, which rounds to 64.8 kW.
Hint
The force the motor must exert is the sum of the total weight and the friction force. Power is then force times constant velocity.