An elevator can carry a maximum load of 1800 kg is moving up with a constant speed of 2m/s the frictional force opposing the motio n is 4000n what is the mimimum power delivered by the motor to the elevator
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Answer:
Power, P = 43280 watts
Explanation:
It is given that,
Mass of load, m = 1800 kg
Speed of elevator, v = 2 m/s
Frictional force, f = 4000 N
The weight of load is acting vertically downward and also the frictional force will act in downward direction. So, net force is given by :
We have to find the minimum power delivered by the motor to the elevator.
Power,
P = 43280 Watts
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