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An elevator of total mass (elevator+ pas...

An elevator of total mass (elevator+ passenger) 3600 kg in moving up with a constant speed of 2 `ms^-1` . A frictional force of 1300 M opposes its motion. Determine the minimum power delivered by the motor to the elevator.

Text Solution

Verified by Experts

We are given:
m = 1800 kg, Frictional force, F = 6000 N
v = 8 m/s.
Downward force on elevator,
`F = mg + F=1800 xx 10 + 6000 = 24000 N`
To balance this force necessary power is
`P = F.v = 24000xx8 = 192000 W. `
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