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: (A)

$W_P > W_Q ; W_Q > W_P$

Answer 2

The correct answer is A:$W_P>W_Q,W_Q>W_P$
Here, $K_P > K_Q$
Case (a): Elongation (x) in each spring is same.
$W_P=\frac{1}{2} K_P x^2,W_Q=\frac{1}{2}K_Qx^2$
$\therefore \, \, \, \, \, \, W_P > W_Q$
Case (b) : Force of elongation is same.
So, $x_1=\frac{F}{K_p}$ and $x_2=\frac{F}{K_Q}$
$W_P=\frac{1}{2}K_P x_2^1=\frac{1}{2} \frac{F^2}{K_P}$
$W_Q=\frac{1}{2}K_Q x_2^2=\frac{1}{2} \frac{F^2}{K_Q}$
$\therefore \, \, \, \, \, \, W_P < W_Q$