Question

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?

Answer 1

Here, `m=1800kg`
frictional force, `f=4000N,`
uniform speed, `v=2.0m//s`
Downward force on elevator is
`F=mg+f=1800xx10+4000=22000N`
The motor must supply enough power to balance this force. As
`P=Fxxv`
`:. P=22000xx2=44000wat t`
`P=(44000)/(746)h.p. =59 h.p.`