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}}=2{{k}_{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} $