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Power

AP Physics 1· difficulty 2/5

An elevator lifts a 400 kg400~\text{kg} load at constant 2.0 m/s2.0~\text{m/s}. Using g≈10 m/s2g \approx 10~\text{m/s}^2, what is the average power provided by the cable?

  • A

    800 W800~\text{W}

  • B

    4000 W4000~\text{W}

  • C

    8000 W8000~\text{W}

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  • D

    40,000 W40{,}000~\text{W}

Explanation

At constant speed, cable tension equals mgmg. Power P=Fv=mgv=(400)(10)(2.0)=8000 WP = F v = m g v = (400)(10)(2.0) = 8000~\text{W}.

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