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  4. Two similar springs P and Q have spring constants KP and KQ, such that KP>KQ. They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs WP and WQ are related as, in case (a) and case (b) respectively

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Q. 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

AIPMTAIPMT 2015Work, Energy and Power

Solution:

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