An elevator which can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms. 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 which can carry a maximum load of 1800 kg (elevator + pass

1.	An elevator which can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms. 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 :Downward force on the ELEVATOR is : `F=mg+f=22000N` `therefore "power SUPPLIED by motor to BALANCE this force is"`: `p=F_(V)=44000W=(44000)/(746)=59hp`