An elevator can carry a maximum load of `1800 kg` (elevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The friction force opposite the motion is `4000 N`.What is minimum power delivered by the motor to the elevator?

Options

A. 22 kW
B. 44 kW
C. 66 kW
D. 88 kW

Answer

B

Explanation

Here, m=1800 kg
Frictional force, f=4000 N
Uniform speed, `v=2 m s^(-1)`
Downward force on elevator is
`F= mg+f=(1800 kg xx 10 ms^(-2)) + 4000 N=22000 N`
The motor must supply enough power to balance this force. Hence,
`P=Fv = (22000 N)(2 ms^(-1))`
`=44000 W = 44 xx 10^3 W =44 kW`