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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 m s^(-1)`. The frictional force opposing the motion is 4000 N. What is minimum power delivered by the motor to the elevator?

Text Solution

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The downword force on the elevator is
`F=mg+F_(f)=(1800xx10)+4000=22000N`
The motor must supply enough power to balance this force. Hence,
`P = Fv = 22000xx2`
`= 44000 W = 59 hp`.
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