Question

An elevator can carry a maximum load of 1800kg, is moving up with a constant speed of 2m/s. The frictional force opposing the motion is 4000N. Determine the minimum power delivered by the motor to the elevator in watts?

Answer 1

Acceleration due to gravity `g=10m/s^2`
So total force need to apply =`(1800*10+4000)=22000N`
Power delivered by the motor `P=force*velocity=22000N*2m/s`=
`44000`Watt