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Work and Mechanical Energy

AP Physics 1· difficulty 2/5

0 x stretch F

For a spring obeying Hooke's law, the elastic PE stored when stretched by xx equals

  • A

    2Fx2 F x

  • B

    12Fx\tfrac{1}{2} F x (triangle area)

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  • C

    Fx\sqrt{F x}

  • D

    FxF x (rectangle area)

Explanation

PE = area under FF-vs-xx for a spring = triangle of base xx and height F=kxF = k x, so Us=12kx2=12FxU_s = \tfrac{1}{2} k x^2 = \tfrac{1}{2} F x.

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