STANDARD Physics Power in Mechanics MCQ Question
An elevator which can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 m s⁻¹. The frictional force opposing the motion is 4000 N. What is minimum power delivered by the motor to the elevator?
22 kW
44 kW
66 kW
88 kW
Correct Answer
Detailed Explanation
The total force opposing the elevator is the sum of gravitational force and frictional force: F = mg + f = (1800 kg × 10 m s⁻²) + 4000 N = 22000 N. The power required is P = Fv = 22000 N × 2 m s⁻¹ = 44000 W = 44 kW.
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