If the maximum load carried by an elevator is 1400 kg (600 kg-Passengers + 800 kg-elevator), which is moving up with a uniform speed of 3 m s -1 and the frictional force acting on it is 2000 N, then the maximum power used by the motor is_____________kW. (g = 10 m s -2 )
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CHECK THE SOLUTION.
(48)
The maximum force acting is, F max =(14000 + 2000) N
= 16000 N
The formula for power is given by, P = F max v.
It is given, v = 3ms -1 .
P = 16000 N x 3 m s -1 = 48000 W = 48 kW
W = 48 kW
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