Answer:
[tex]0.7\; {\rm m}[/tex].
Explanation:
If spring with spring constant [tex]k[/tex] is compressed by [tex]x[/tex] relative to the equilibrium, the elastic potential energy [tex]E[/tex] stored in that spring will be [tex]E = (1/2)\, k\, x^{2}[/tex].
For the spring in this question, it is given that [tex]E = 196\; {\rm J}[/tex] whereas [tex]k = 800\; {\rm N \cdot m^{-1}}[/tex]. The displacement [tex]x[/tex] needs to be found.
Note that [tex]\begin{aligned}E &= 196\; {\rm J} = 196\; {\rm N \cdot m} \end{aligned}[/tex] (one joule is the work done when a force of [tex]1\; {\rm N}[/tex] is applied over a distance of [tex]1\; {\rm m}[/tex].)
Rearrange the equation [tex]E = (1/2)\, k\, x^{2}[/tex] to find displacement [tex]x[/tex] in terms of potential energy [tex]E[/tex] and spring constant [tex]k[/tex]:
[tex]\begin{aligned} x^{2} = \frac{2\, E}{k}\end{aligned}[/tex].
[tex]\begin{aligned} x &= \sqrt{\frac{2\, E}{k}} \\ &= \sqrt{\frac{2 \times 196\; {\rm N \cdot m}}{800\; {\rm N\cdot m^{-1}}}} \\ &= \sqrt{0.49\; {\rm m^{2}}} \\ &= 0.7\; {\rm m} \end{aligned}[/tex].
In other words, the spring has been compressed by [tex]x = 0.7\; {\rm m}[/tex] from the equilibrium.