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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. |
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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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