Two similar springs $P$ and $Q$ have spring constants $K_{P}$ and $K_{Q}$, such that $K_{P}K_{Q}$.

Question

Two similar springs $P$ and $Q$ have spring constants $K_{P}$ and $K_{Q}$, such that $K_{P}>K_{Q}$. They are stretched first by the same amount (case $a$ ), then by the same force (case $b$ ). The work done by the springs $W_{P}$ and $W_{Q}$ are related as, in case ( $a$ ) and case (b) respectively\

Professor PC Thomas Classes · Answer

(a): Here, $K_{p}>K_{Q}$ Case (a) : Elongation $(x)$ in each spring is same. $$ W_{P}=\frac{1}{2} K_{P} x^{2}, W_{Q}=\frac{1}{2} K_{Q} x^{2} 	herefore \quad W_{P}>W_{Q} $$ Case (b) : Force of elongation is same.\ So, $x_{1}=\frac{F}{K_{P}}$ and $x_{2}=\frac{F}{K_{Q}} ; W_{P}=\frac{1}{2} K_{P} x_{1}^{2}=\frac{1}{2} \frac{F^{2}}{K_{P}}$ $W_{Q}=\frac{1}{2} K_{Q} x_{2}^{2}=\frac{1}{2} \frac{F^{2}}{K_{Q}} \quad 	herefore W_{P}<W_{Q}$\