1.

An elevator which can carry a maximum load of 1800 kg (elevator + passenger) 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 :Downward FORCE on the ELEVATOR is :
`F=mg+f=22000N`
`:.` Power SUPPLIED by motor to balance this force is:
`P=Fv=44000W`
`=74000/746=59`hp


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