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Work and Mechanical Energy

AP Physics 1· difficulty 4/5

A 2.0 kg block, initially at rest on a frictionless surface, is pushed by a force whose magnitude varies with position as F(x)=10−2xF(x)=10-2x (in newtons, with xx in meters) for 0≤x≤5 m0\le x\le 5\,\text{m}. What is the block's kinetic energy at x=5 mx=5\,\text{m}?

  • A

    12.5 J12.5\,\text{J}

  • B

    25 J25\,\text{J}

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  • C

    10 J10\,\text{J}

  • D

    50 J50\,\text{J}

Explanation

Work equals area under the FF-vs-xx graph. The triangle has base 5 m and height 10 N, so W=12(5)(10)=25 JW=\tfrac{1}{2}(5)(10)=25\,\text{J}. By work-energy theorem K=25 JK=25\,\text{J}.

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