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Potential Energy

AP Physics 1· difficulty 1/5

A spring with k=400 N/mk = 400~\text{N/m} is compressed 0.05 m0.05~\text{m}. The stored elastic PE is

  • A

    10 J10~\text{J}

  • B

    0.50 J0.50~\text{J}

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  • C

    0.20 J0.20~\text{J}

  • D

    1.0 J1.0~\text{J}

Explanation

Us=12kx2=12(400)(0.0025)=0.50 JU_s = \tfrac{1}{2}k x^2 = \tfrac{1}{2}(400)(0.0025) = 0.50~\text{J}.

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