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OpenStudy (anonymous):
A 2200 kg elevator is attached to a 1500 kg counterweight. What power must the motor supply to raise the elevator at 0.5 m/s?
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OpenStudy (anonymous):
Substitute into the kinetic energy equation: E=1/2mv^2, E=1/2*700*0.5^2 E=87.5J Therefore, it will be 87.5W
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