Question

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?

Answer 1

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,

$U=\frac{1}{2}kx^2$

where x is the distance compressed,

Substituting the known values,

$\begin{gathered} 35=\frac{1}{2}\times82\times x^2 \\ x^2=\frac{35\times2}{82} \\ x^2=0.854 \\ x=0.92\text{ m} \end{gathered}$

Thus, the spring is compressed to 0.92 m.