STANDARDPhysics-Work, Energy and Power

STANDARD Physics Power in Mechanics MCQ Question

Type: MCQ-numerical-Medium-Class 11

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?

A

22 kW

B

44 kW

C

66 kW

D

88 kW

Correct Answer

Option B

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.

Found an issue with this question?