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Question

An elevator of total mass (elevator + passenger) 1800kg is moving up with a constant speed of 2ms-1 . A frictional force of 2000N is opposing its motion. The minimum power delivered by the motor to the elevator is [Take g=10ms-2]


A

36 kW

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B

4 kW

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C

40 kW

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D

-40 kW

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Solution

The correct option is C

40 kW


Step 1: Given data:
Mass of the elevator + Passenger
m=1800kg
Frictional force f=2000N
Constant speed v=2m/s

Step 2: Formula

Since the elevator is moving upward, there is a downward force acting on the elevator which is given by,
F=mg+f
F=1800×10+2000
F=20000N

The minimum power delivered by the elevator is given by,
P=F.v
P=20000×2
P=40000W
P=40kW

The correct option is c) 40kW


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