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 $ms^{-2}$)

Question

Dr. Abubacker · Accepted Answer

[Solution] Force needed = mg + f = 20000 + 3000 = 23000 N; P = Fv = 23000 × 1.5 = 34500 W. Answer (4).

Q. 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 $ms^{-2}$)

Explanation

Reviewed by Dr. Abubacker

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