91Ó°ÊÓ

Spring potential energy: When a spring is stretched or compressed by the application of force, the potential energy stored in the spring is,

${\text{PE}}_s=\frac{1}{2}k{x}^2$

Here, k is the force constant of the spring $\left(k=2.50×{10}^4\text{N}/\text{m}\right)$and x is the compression in the spring 12.0 cm.

When a child jumps on the spring, the spring potential energy in the spring gets converted into gravitational potential energy and the child attains some height. Mathematically,

$$\begin{gathered} {\text{PE}}_s={\text{PE}}_g \\ \frac{1}{2}k{x}^2=mgh \end{gathered}$$

Here, m is the mass of the child$\left(m=40.0\text{kg}\right)$, g is the acceleration due to gravity$\left(g=9.8\text{m}/{\text{s}}^2\right)$, and h is the maximum height attained by the child.

The expression for the maximum height attained by the child is,

$h=\frac{k{x}^2}{2mg}$

Putting all known values,

$$\begin{gathered} h=\frac{\left(2.50×{10}^4\text{N}/\text{m}\right)×{\left(12.0\text{cm}\right)}^2}{\left(40.0\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)} \\ =\frac{\left(2.50×{10}^4\text{N}/\text{m}\right)×{\left[\left(12.0\text{cm}\right)×\left(\frac{1\text{m}}{100\text{cm}}\right)\right]}^2}{2×\left(40.0\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)} \\ =0.46\text{m} \end{gathered}$$

Therefore, the child can jump up to the maximum height of $0.46\text{m}$.