Question #1841

A small domestic elevator has a mass of 454 kg and can ascend at a rate of 0.180 m/s. What is the average power that must be supplied for the elevator to move at this rate? answer in Watts.

Expert's answer

The power is equal to W = A/t

The work done by the elevator is equal to changing fo the potential energy:& mgΔH

thus W = mgΔH/t = mg v = 454[kg]*10[m/s2] * 0.18 [m/s] = 817.2 W.

The work done by the elevator is equal to changing fo the potential energy:& mgΔH

thus W = mgΔH/t = mg v = 454[kg]*10[m/s2] * 0.18 [m/s] = 817.2 W.

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