An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms ⁻¹. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is: (g = 10 ms ⁻²) (a) 23000 (b) 20000 (c) 34500 (d) 23500

Answer: (c). At constant speed, motor force F = mg + f. F = 2000 × 10 + 3000 = 23000 N P = Fv = 23000 × 1.5 = 34500 W So the answer is (c) 34500.

An electric lift with a maximum load of 2000 kg (lift + passengers)… | NEET 2022 (Phase 1)

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Power

An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms ${}^{-1}$ . The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is: (g = 10 ms ${}^{-2}$ )

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Answer: (c)

At constant speed, motor force $F = mg + f$ .
$F = 2000 \times 10 + 3000 = 23000\text{ N}$
$P = Fv = 23000 \times 1.5 = 34500\text{ W}$
So the answer is (c) 34500.

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