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

Question

Dr. Abubacker · Accepted Answer

(d) Given$k_{p} = 2k_{Q}$
By stretched spring gain get energy
$E = \frac{1}{2}kx^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because F = kx)$
OR $E = \frac{1}{2}\ \frac{F^{2}}{K}$
$\because For\ spring\ P$
$E_{P} = \frac{1}{2}\ \frac{F^{2}}{K_{P}}$
$\because For\ spring\ Q$
$E_{Q} = \frac{1}{2}\ \frac{F^{2}}{K^{Q}}$
$\frac{E_{P}}{E_{Q}} = \frac{K_{Q}}{K_{P}} = \frac{1}{2}$
$E_{P} = \frac{E_{Q}}{2} = \frac{E}{2}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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