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CGP EDU Academic Team
Published on: August 21, 2026
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 m s -2 )
Text Solution
Verified by ExpertsThe correct answer is:
Given that,
Mass on lift m = 2000
Friction force, f = 3000 N
velocity, v = 1.5 m/s
Now, minimum force needed to move lift upward
F up = 2000g + 3000
=23000N
Minimum power P min = 
= 34500W
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