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

Explanation

The downward force on the elevator is
`F = mg + F_f = (1800 xx 10) + 4000 = 22000 N`.
The motor must supply enough power to balance this force.
Hence, `P= vecF . vecv = 22000 xx 2 = 44000 W. `