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Step-by-Step Solution
1. Understanding the Problem
We have two springs, P and Q, whose spring constants are respectively $K_P$ and $K_Q$, with $K_P > K_Q$. We examine two cases:
(a) Both springs are stretched by the same amount.
(b) Both springs are stretched by the same force.
We need to compare the work done by these two springs in both cases.
2. Work Done by a Spring
The work done by a spring, when stretched or compressed from its natural length by an extension $x$, is given by:
$$ W = \frac{1}{2} K x^2. $$
3. Case (a): Same Extension
Let the common extension in both springs be $x$. Then:
Work done by spring P:
$$W_P = \frac{1}{2} K_P x^2.$$
Work done by spring Q:
$$W_Q = \frac{1}{2} K_Q x^2.$$
Given $K_P > K_Q$, we compare:
$$W_P = \frac{1}{2} K_P x^2 > \frac{1}{2} K_Q x^2 = W_Q.$$
Thus, in case (a),
$$W_P > W_Q.$$
4. Case (b): Same Force
Let the common force applied on both springs be $F$. The extension of a spring under force $F$ is given by:
$$ x = \frac{F}{K}. $$
Hence:
Extension in spring P:
$$x_1 = \frac{F}{K_P}.$$
Extension in spring Q:
$$x_2 = \frac{F}{K_Q}.$$
Because $K_P > K_Q$, we have $x_1 < x_2$. Now, the work done by each spring is:
$$W_P = \frac{1}{2} K_P (x_1)^2 = \frac{1}{2} K_P \left(\frac{F}{K_P}\right)^2 = \frac{F^2}{2 K_P},$$
$$W_Q = \frac{1}{2} K_Q (x_2)^2 = \frac{1}{2} K_Q \left(\frac{F}{K_Q}\right)^2 = \frac{F^2}{2 K_Q}.$$
Since $K_P > K_Q$, we get:
$$\frac{1}{K_P} < \frac{1}{K_Q} \implies \frac{F^2}{2 K_P} < \frac{F^2}{2 K_Q}.$$
Therefore,
$$W_P < W_Q.$$
5. Conclusion
From the above analysis:
Case (a): $W_P > W_Q$ (same extension)
Case (b): $W_P < W_Q$ (same force)
This matches the correct answer: $W_P > W_Q$ in case (a) and $W_Q > W_P$ in case (b).
