91Ó°ÊÓ

The stiffness of spring is 800 N/m , the relaxed length I_S of the spring is 0.66m , the compressed length I₁ is 0.55m and the expanded length I₂ is 0.96m .

The potential energy of ideal spring mass system is given by:

${\text{U}}_{\text{s}}\text{=}\frac{\text{1}}{\text{2}}{\text{K}}_{\text{s}}{\text{x}}^{\text{2}}$

Where, is the potential energy of ideal spring-mass system, is the spring constant and is the displacement

The change of potential energy is given by:

.$\mathrm{Δ}{\text{U =U}}_{\text{2}}{\text{-U}}_{\text{1}}$

Find the initial displacement of the spring.

$x_1=l_1-l_s$

Find the final displacement of the spring.

$x_2=l_2-l_s$

Substitute the values of in the equation of

$$\begin{gathered} x_1=l_1-l_s \\ =0.55-0.66 \\ =-0.11\text{m} \end{gathered}$$

Substitute the values of in the equation of

$$\begin{gathered} x_2=l_2-l_s \\ =0.96-0.66 \\ =0.3\text{m} \end{gathered}$$

Substitute the values of in potential energy of ideal spring-mass system to find the value of and

role="math" localid="1657105681284" $$\begin{gathered} U_1=\frac{1}{2}\left(800\right){\left(-0.11\right)}^2 \\ =4.84\text{J} \\ {U}_2=\frac{1}{2}\left(800\right){\left(0.3\right)}^2 \\ =36\text{J} \end{gathered}$$

Find the value of change of potential energy

$$\begin{gathered} \mathrm{Δ}U=U_2-U_1 \\ =36-4.84 \\ =31.16\text{J} \end{gathered}$$

Thus, the change in potential energy is 31.16J.