Question

If the energy stored in spring $\text{A is}5\text{J}$, then the energy stored in spring $\text{B}$, under the same stretching force $\text{F}$ will be:

• A. $10\text{J}$
• B. $5\text{J}$
• C. $2.5\text{J}$
• D. $1.25\text{J}$

Answer 1

The correct option is A $10\text{J}$

$E_A=5\text{J}$ = Energy stored in spring '$\text{A}$'
$E_B=$ Energy stored in spring '$\text{B}$'
$x_A=$ Displacement in spring '$\text{A}$'
$x_B=$ Displacement in spring '$\text{B}$'

From the basic spring force relation:
$F=kx,E=\frac{k{x}^2}{2}=\frac{F^2}{2k}$
where $x=$ Displacement produced in spring w.r.t it's natural length

$E_B=\frac{1}{2}{k}_B{x}_B^2=\frac{1}{2}\times k\times {\left(\frac{F}{k}\right)}^2=\frac{F^2}{2k}...\left(i\right)$

$E_A=\frac{1}{2}{k}_A{x}_A^2=\frac{1}{2}.2k.{\left(\frac{F}{2k}\right)}^2=\frac{F^2}{4k}...\left(ii\right)$

Dividing Eq.($i$) and Eq.($ii$)

$\frac{E_A}{E_B}=\frac{\frac{F^2}{4k}}{\frac{F^2}{2k}}=\frac{1}{2}$
$\Rightarrow \frac{5}{E_B}=\frac{1}{2}$
$\therefore E_B=10\text{J}$

Hence option A is the correct answer.