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Question

An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is: (g=10 ms2)
  1. 34500
  2. 23000
  3. 20000
  4. 23500

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Solution

The correct option is A 34500
Total force required to move with constant speed,

Force = weight + friction

F=20000+3000

F=23000 N

Power, P=FV

=23000×1.5

P=34500 W

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