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

The Correct Option is A

The problem involves understanding how the work done by two springs, P and Q, having different spring constants, relate under two different scenarios. Let's address each case step-by-step. 

1. Case (a): Stretching by the Same Amount
   • The work done by a spring when it is stretched or compressed is given by the formula: \(W = \frac{1}{2} k x^2\), where \( k \) is the spring constant and \( x \) is the displacement.
   • For spring P: \(W_P = \frac{1}{2} K_P x^2\)
   • For spring Q: \(W_Q = \frac{1}{2} K_Q x^2\)
   • Given \(K_P > K_Q\) and both springs are stretched by the same amount \( x \), it follows that: \(W_P > W_Q\)
   • Hence, in case (a), the work done by spring P is greater than that by spring Q: \(W_P > W_Q\).
2. Case (b): Applying the Same Force
   • When the same force \( F \) is applied, the displacement can be calculated as: \(F = k \times x \rightarrow x = \frac{F}{k}\)
   • So for spring P: The displacement is \(x_P = \frac{F}{K_P}\)
   • For spring Q: The displacement is \(x_Q = \frac{F}{K_Q}\)
   • The work done, in this case, becomes:
     • \(W_P = \frac{1}{2} K_P \left(\frac{F}{K_P}\right)^2 = \frac{F^2}{2K_P}\)
     • \(W_Q = \frac{F^2}{2K_Q}\)
   • Since \(K_P > K_Q\), it implies: \(\frac{F^2}{2K_P} < \frac{F^2}{2K_Q}\)
   • Thus, the work done by spring Q is greater than that by spring P when the same force is applied: \(W_Q > W_P\).
3. Conclusion: Based on the analysis above, the relationship of the work done in both scenarios is: \(W_P > W_Q\) and \(W_Q > W_P\) for cases (a) and (b) respectively.