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

An elevator can carry a maximum load of 1800 kg (elevator + passengers

1.	An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2ms^(-1). The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.
Answer» Solution :The downward FORCE on the elevator is `F= MG+ F_(f)= (1800xx10)+ 4000= 22000N` The motor must supply enough power to balance this force. HENCE, `P= F*v= 22000xx2= 44000 W = 59 hp`.