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

• A. $W_P>W_Q;W_Q>W_P$
• B. $W_P<W_Q;W_Q<W_P$
• C. $W_P=W_Q;W_P>W_Q$
• D. $W_P=W_Q;W_P=W_Q$

Answer 1

Answer: (B)

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 $x_2=\frac{F}{K_Q}$
$W_P=\frac{1}{2}{K}_P{x}_2^1=\frac{1}{2}\frac{F^2}{K_P}$
$W_Q=\frac{1}{2}{K}_Q{x}_2^2=\frac{1}{2}\frac{F^2}{K_Q}$
$\therefore W_P<W_Q$