An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1 …

An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 m s-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
  1. 20000
  2. 34500
  3. 23500
  4. 23000

Solution

Since the speed is constant the motor has to give equal and opposite upward force.

∴   F=3000N+2000×10 =23000 N

P=FV=23000×1.5 =34500 W

Asked in: NEET 2022 (Phase 1)

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