Question

Question: A spring whose stiffness is 800N/m has a relaxed length of 0.66m. If the length of the spring changes from 0.55m to 0.96m. What is the change in potential energy of the spring?

Short answer

Answer

The change in potential energy is 31.16 J.

Step-by-step solution

Given information in the problem 

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 .

 Step 2: The formula of the potential energy

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:

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

Calculation of the change in the potential energy of the spring

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

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

Thus, the change in potential energy is 31.16J.