Question

What is the spring constant of a spring that stores$\mathbf{25}\mathbf{\text{\hspace{0.17em}J}}$of elastic potential energy when compressed by$\mathbf{7}\mathbf{.5}\mathbf{\text{\hspace{0.17em}cm}}$?

Short answer

Spring constant is,$k=8.9\times {10}^3\text{\hspace{0.17em}N/m}$ .

Step-by-step solution

Given

Elastic potential energy$U=25\text{\hspace{0.17em}J}$

Compressed length$x=7.5\text{\hspace{0.17em}cm}=7.5\times {10}^{-2}\text{\hspace{0.17em}m}$ .

To understand the concept 

The problem is based on the concept of elastic potential energy. It is energy stored as a result of applying a force to deform an elastic object. By using the concept of elastic potential energy, we can find the spring constant.

Formula:

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

Calculate the spring constant of the spring 

Elastic potential energy can be written as

$\begin{array}{c}U=\frac{1}{2}k{x}^2\\ k=\frac{2U}{x^2}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}k=\frac{2\times 25\text{\hspace{0.17em}J}}{{(7.5\times {10}^{-2}\text{\hspace{0.17em}m})}^2}\\ k=8.9\times {10}^3\text{\hspace{0.17em}N/m}\end{array}$

Hence the spring constant is, $k=8.9\times {10}^3\text{\hspace{0.17em}N/m}$.