An ideal spring of negligible mass is 12.00cm long when nothing is attached to it. When you hang a 3.15kg weight from it, you measure its length to be 13.40cm . If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume it continues to obey Hooke’s law.

The given data can be listed below as,

• The initial natural length of a spring is, $l_1=12.0cm$.
• The mass of a attached weight is, $m=3.15kg$.
• The final measured length of the spring after the weight attached is, $l_2=13.40cm$.
• The amount of elastic potential energy has to be stored in the spring is, $U=10.0J$.

Whenever a mechanical spring is subjected to a specific external load, there would be a change in the spring's length. The value of the elastic potential energy stored in the spring can be obtained with the help of the change in length of the spring.

According to Hooke’s law, the expression to calculate the spring constant can be expressed as,

$$\begin{gathered} F=k∆l \\ mg=k\left(l_2-l_1\right) \\ k=\left[\frac{mg}{\left(l_2-l_1\right)}\right] \end{gathered}$$

Here, F is the weight force of the attached weight, k is the spring constant and g is the gravitational acceleration whose value is$9.81m/s^2$ .

Substitute all the known values in the expression.

$$\begin{gathered} k=\left[\frac{\left(3.15kg\right)\left(9.81m/s^2\right)}{\left(13.40cm-12.0cm\right)\left({\displaystyle \frac{{10}^{-2}m}{1cm}}\right)}\right] \\ =2207.25kg/s^2 \\ =\left(2207.25kg/s^2\right)\times \left(\frac{1N/m}{1kg/s^2}\right) \\ =2207.25N/m \end{gathered}$$

The expression to calculate the total length of the spring can be expressed as,

$$\begin{gathered} U=\frac{1}{2}k{\left(∆l\right)}^2 \\ =\frac{1}{2}k\left[{\left(l\text{'}\right)}^2-{\left(l_1\right)}^2\right] \end{gathered}$$

Here, $l\text{'}$ is the total length of the spring.

Substitute all the known values in the expression.

$$\begin{gathered} \left(10.0J\right)=\frac{1}{2}\left(2207.25N/m\right)\left[{\left(l\text{'}\right)}^2-{\left(12.0cm\times \frac{{10}^{-2}m}{1cm}\right)}^2\right] \\ \left[{\left(l\text{'}\right)}^2-{\left(12.0cm\times \frac{{10}^{-2}m}{1cm}\right)}^2\right]=\left[\frac{2\left(10.0J\right)}{\left(2207.25N/m\right)}\right] \\ l\text{'}\approx \left(0.1532m\right)\left(\frac{{10}^{2}cm}{1m}\right) \\ \approx 15.32cm \end{gathered}$$

Thus, the total length of the spring is 15.32 cm .