Question

A spring of natural length ${}^{\prime }{x}^{\prime }\text{m}$ is compressed to half of its length. If the force constant $k$ of spring is $6\text{N/m}$, then it's potential energy ($\text{joule}$) is:

• A. $3x^2$
• B. $\frac{2}{3}{x}^2$
  
• C. $\frac{3}{4}{x}^2$
  
• D. $6x^2$
  

Answer 1

The correct option is C $\frac{3}{4}{x}^2$


Potential Energy of spring $=\frac{1}{2}k{x}^2$
$\Rightarrow x=\text{extension or compression in spring w.r.t natural length of spring}$

Since, the spring is compressed to half of its length.
$x=\left(\frac{x}{2}\right)$ ;where $x$ is natural length
$\therefore$Potential Energy $=\frac{1}{2}\times \left(6\right)\times {\left(\frac{x}{2}\right)}^2$
$=\left(\frac{3}{4}{x}^2\right)\text{joule}$

Hence option C is the correct answer.