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Work and Mechanical Energy

AP Physics 1· difficulty 4/5

A 5.0 kg crate is pulled 3.0 m across a floor by a horizontal force of 30 N against friction force 10 N. The net work done on the crate is

  • A

    0 J

  • B

    60 J (Wnet=Fnet⋅dW_{net} = F_{net} \cdot d)

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  • C

    30 J

  • D

    90 J

Explanation

Fnet=30−10=20F_{net} = 30 - 10 = 20 N. W=Fnetd=20(3.0)=60W = F_{net} d = 20(3.0) = 60 J. (Equals ΔKE\Delta KE.)

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