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A spring has a spring constant k of 82.0 N/m. How much must this spring becompressed to store 35.0 J of potential energy?

User Paul Leigh
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1 Answer

1 vote

Given data:

* The value of the potential energy given is,


U=35\text{ J}

* The spring constant of the spring is,


k=82\text{ N/m}

Solution:

The potential energy in terms of the compressed distance of the spring is,


U=(1)/(2)kx^2

where x is the distance compressed,

Substituting the known values,


\begin{gathered} 35=(1)/(2)*82* x^2 \\ x^2=(35*2)/(82) \\ x^2=0.854 \\ x=0.92\text{ m} \end{gathered}

Thus, the spring is compressed to 0.92 m.

User PokerFace
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