Question

The potential energy of a long spring when stretched by $2\mathrm{cm}$ is U. If the spring is stretched by $8\mathrm{cm}$ the potential energy stored in it is:-

• A. $4\mathrm{U}$
• B. $8\mathrm{U}$
• C. $16\mathrm{U}$
• D. $\frac{\mathrm{U}}{4}$

Answer 1

The correct option is C

$16\mathrm{U}$

Step 1: Given

Length of first stretch of spring, $x_1=2cm$

Length of second stretch of spring, $x_2=8cm$

Potential energy stored in spring for first stretch = $U$

Step 2: Formula Used

The general formula for the potential energy stored in the spring is given as:

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

Where $\text{U}$ is potential energy stored in spring for the stretch $x_1$ and $k$ is the spring constant.

$\frac{\mathrm{U}\text{'}}{\mathrm{U}}=\frac{{\left({\mathrm{x}}_2\right)}^2}{{\left({\mathrm{x}}_1\right)}^2}$

Where $U$ is potential energy stored in spring for first stretch, $U\text{'}$ is the potential energy stored in spring for second stretch, $x_1$ is the length of first stretch, and $x_2$is length of second stretch.

Step 3: Potential energy stored

$\mathrm{Potential}\mathrm{energy}\propto {\left(\mathrm{extension}\right)}^2$

By substituting the given values in the formula, we get

$\frac{\mathrm{U}\text{'}}{\mathrm{U}}=\frac{{\left(8\right)}^2}{{\left(2\right)}^2}$

$$\begin{gathered} \frac{\mathrm{U}\text{'}}{\mathrm{U}}=\frac{64}{4} \\ \mathrm{U}\text{'}=16\mathrm{U} \end{gathered}$$

Hence, the total potential energy stored in the spring is $16\mathrm{U}$. So the correct option is C $16\mathrm{U}$.