Question

A spring of natural length $2x\text{m}$ is elongated such that its extension is twice its original length. If the force constant($k$) of spring is $8\text{N/m}$, the potential energy ($\text{J}$) stored in the spring is:

• A. $16x^2$
• B. $64x^2$
• C. $32x^2$
• D. $128x^2$

Answer 1

The correct option is B $64x^2$

Potential energy stored ($U$) in spring $=\frac{1}{2}k{x}^2$
where $x$ is the displacement produced in spring w.r.t it's natural length .

It is given that the spring is elongated to twice its original length

$\Rightarrow x^{\prime }=2\left(2x\right)=4x$
$\therefore$ Potential energy of spring $=\frac{1}{2}\times 8\times (4x{)}^2$
$=4\times 16x^2$

$\therefore U=\left(64x^2\right)\text{J}$

Hence option B is the correct answer