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)
Answer & explanation
Correct answer: option 3
Minimum power : Pmin = F.v
F = 2000 g + 3000 = 23000 N ; v = 1.5 ms–1
Pmin = 34500 W