Question

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

Answer 1

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Answer 2

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 :

$F_{net}=mg+f$

$F_{net}=1800\times 9.8+4000$

$F_{net}=21640\ N$

We have to find the minimum power delivered by the motor to the elevator.

Power, $P=F_{net}\times v$

$P=21640\times 2$

P = 43280 Watts