2022 Work, Energy and PowerPower Medium Question 1 of 11 in 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 : ($g=10\ ms^{-2}$)

A 23500
B 23000
C 20000
D 34500

The answer is D.

Solution

The lift moves with constant velocity, so a = 0. Tension (force by motor) $T = W + f = 2000 \times 10 + 3000 = 23000$ N Power $= Tv = 23000 \times 1.5 = 34500$ W

Concept power delivered by a lift motor

Power in past papers

11 questions from this chapter have appeared across 9 exam years.