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An elevator can carry a maximum load of ...

An elevator can carry a maximum load of `1800 kg` (elevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The friction force opposite the motion is `4000 N`.What is minimum power delivered by the motor to the elevator?

Text Solution

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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`.
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