Question

A spring has a spring constant of 450 N/m. How much must this spring be stretched to

store 49 J of potential energy?

Answer 1

The potential energy of the spring is given as follows:

$$E = \dfrac{kx^2}{2}$$

where $k = 450N/m$ is the spring constant, and $x$ the spring elongation. Expressing $x$ and substituting $E = 49J$, obtain:

$$x = \sqrt{\dfrac{2E}{k}} = \sqrt{\dfrac{2\cdot 49}{450}} \approx 0.47m$$

Answer. 0.47 m.