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 ms-2)
- 23000
- 20000
- 34500
- 23500
Answer
(3) 34500
Solution
At constant speed, net force is zero:
F = mg + f = 2000 10 + 3000 = 23000 N
P = Fv
= 23000 1.5
= 34500 W
Answer: (3)