An elevator of total mass 1400 kg moves upward with uniform speed 3 m s⁻¹. A frictional force of 2000 N acts downward. Find the maximum power used by the motor in kW. Take g = 10 m s⁻².
Physics · Circular Motion
JEE Mains 10 April 2023 Shift 2 Evening Shift

Question

An elevator of total mass 1400 kg moves upward with uniform speed 3 m s⁻¹. A frictional force of 2000 N acts downward. Find the maximum power used by the motor in kW. Take g = 10 m s⁻².

G1 Physics · Accepted Answer

Since the elevator moves with uniform speed, acceleration is zero. Therefore the motor force must balance weight plus friction: F = mg + f = 1400×10 + 2000 = 16000 N Power: P = Fv = 16000×3 = 48000 W = 48 kW Final value: 48