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

Explanation

Weight of (elevator + passenger) `=mg=1800xx10N=18000N`
Frictional force `=4000N`
Total downward force on the elevator is
`(18000+4000)N=22000N`
Clearly, the motor must have enough power to balance this force.
Now power, `P=Fv=22,000Nxx2ms^-1=44,000W`
`=44000/746hp=58.98hp`