If the maximum load carried by an elevator is | vedclass

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(C) The total mass of the elevator system is $M = 1400\,kg$.The elevator is moving upward at a uniform speed $v = 3\,m/s$. Since the speed is uniform,

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$48$

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(C) The total mass of the elevator system is $M = 1400\,kg$.The elevator is moving upward at a uniform speed $v = 3\,m/s$. Since the speed is uniform,the acceleration is zero.The motor must provide a force $F$ to overcome both the gravitational force and the frictional force.The gravitational force is $F_g = M 	imes g = 1400\,kg 	imes 10\,m/s^2 = 14000\,N$.The frictional force is $f = 2000\,N$.Therefore,the total force required by the motor is $F = F_g + f = 14000\,N + 2000\,N = 16000\,N$.The power $P$ used by the motor is given by $P = F 	imes v$.$P = 16000\,N 	imes 3\,m/s = 48000\,W$.Converting to kilowatts,$P = 48\,kW$.

If the maximum load carried by an elevator is $1400\,kg$ ($600\,kg$ passenger $+ 800\,kg$ elevator),which is moving up with a uniform speed of $3\,m/s$ and the frictional force acting on it is $2000\,N$,then the maximum power used by the motor is $...........\,kW$ (take $g = 10\,m/s^2$).

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