Since the speed is constant the motor has to give equal and opposite upward force.
∴F=3000N+2000×10 =23000N
P=FV=23000×1.5 =34500W
An electric lift with a maximum load of 2000kg (lift + passengers) is moving up with a constant speed of 1.5ms−1. The frictional force opposing the motion is 3000N. The minimum power delivered by the motor to the lift in watts is : (g=10ms−2)
Held on 30 Apr 2022 · Verified 9 Jul 2026.
20000
34500
23500
23000
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