Answer:
The spring would compress by approximately [tex]0.39\; {\rm m}[/tex].
The elastic potential energy stored in the spring would be approximately [tex]19\; {\rm J}[/tex].
(Assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].)
Explanation:
Assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex], the weight of this block ([tex]m = 10\; {\rm kg}[/tex]) would be:
[tex]\begin{aligned} (\text{weight}) &= m\, g \\ &= (10\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1}}) \\ &= 98.1\; {\rm N} \end{aligned}[/tex].
The spring compresses under the weight of this block. Dividing the external force [tex]F[/tex] by the spring constant [tex]k[/tex] would give the displacement [tex]x[/tex] from equilibrium:
[tex]\begin{aligned}(\text{displacement}) &= \frac{(\text{external force})}{(\text{spring constant})}\end{aligned}[/tex].
[tex]\begin{aligned}x &= \frac{F}{k} \\ &= \frac{98.1\; {\rm N}}{250\; {\rm N \cdot m^{-1}}} \approx 0.392\; {\rm m}\end{aligned}[/tex].
When a spring of spring constant [tex]k[/tex] is compressed by a displacement of [tex]x[/tex], the elastic potential energy [tex](\text{EPE})[/tex] stored in this spring would be:
[tex]\begin{aligned}(\text{EPE}) &= \frac{1}{2}\, (\text{spring constant})\, (\text{displacement})^{2}\end{aligned}[/tex].
For the spring in this question:
[tex]\begin{aligned}(\text{EPE}) &= \frac{1}{2}\, k\, x^{2} \\ &= \frac{1}{2}\, (250\; {\rm N\cdot m^{-1}})\, (0.392\; {\rm m})^{2} \\ &\approx 19\; {\rm J}\end{aligned}[/tex].
