An elevator in a building can carry a maximum | vedclass

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(C) The total mass of the persons is $M_p = 10 	imes 68 \; kg = 680 \; kg$.The total mass of the elevator system is $M = M_p + M_{elevator} = 680 \;

Vedclass · Accepted Answer

$66000$

Vedclass · Answer

(C) The total mass of the persons is $M_p = 10 	imes 68 \; kg = 680 \; kg$.The total mass of the elevator system is $M = M_p + M_{elevator} = 680 \; kg + 920 \; kg = 1600 \; kg$.The total downward force due to gravity is $F_g = M 	imes g = 1600 \; kg 	imes 10 \; m/s^{2} = 16000 \; N$.The frictional force opposing the upward motion is $f = 6000 \; N$.Since the elevator moves with a constant speed,the acceleration is zero. Therefore,the tension $T$ in the cable must balance the total downward force:$T = F_g + f = 16000 \; N + 6000 \; N = 22000 \; N$.The power $P$ delivered by the motor is given by $P = T 	imes v$,where $v$ is the velocity.$P = 22000 \; N 	imes 3 \; m/s = 66000 \; W$.

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