Question

Two springs $P$ and $Q$ of force constants $k_p$ and $k_Q$ $\left(k_Q=\frac{k_p}{2}\right)$ are stretched by applying forces of equal magnitude. If the energy stored in $Q$ is $E$, then the energy stored in $P$ is

• A. $E$
• B. $2E$
• C. $\frac{E}{8}$
• D. $\frac{E}{4}$
• E. $\frac{E}{2}$

Answer 1

Answer: (E)

Given: $k_p=2k_Q$
By stretching spring energy given $E=\frac{1}{2}k{x}^2$
$[\because F=kx]$
Or $E=\frac{1}{2}\frac{F^2}{k}$
For spring P $E_p=\frac{1}{2}\frac{F^2}{k_p}$
For spring Q $E_Q=\frac{1}{2}\frac{F^2}{k_Q}$
$\therefore$ $\frac{E_P}{E_Q}=\frac{k_Q}{k_P}=\frac{1}{2}$
Or $E_P=\frac{k_Q}{2}=\frac{E}{2}$