Question:
When two springs A and B with force constants \(k_A\) and \(k_B\) are stretched by the same force, then the respective ratio of the work done on them is
When two springs A and B with force constants \(k_A\) and \(k_B\) are stretched by the same force, then the respective ratio of the work done on them is
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For same force, softer spring (smaller \(k\)) stores more energy.
Updated On: May 8, 2026
- \(k_B : k_A\)
- \(k_A : k_B\)
- \(k_A k_B : 1\)
- \(\sqrt{k_B} : \sqrt{k_A}\)
- \(\sqrt{k_A} : \sqrt{k_B}\)
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The Correct Option is A
Solution and Explanation
Concept:
Work done in stretching a spring:
\[
W = \frac{1}{2}kx^2
\]
Also, from Hooke’s law:
\[
F = kx \Rightarrow x = \frac{F}{k}
\]
Step 1: Substitute \(x = \frac{F}{k}\) into work formula. \[ W = \frac{1}{2}k \left(\frac{F}{k}\right)^2 \]
Step 2: Simplify expression. \[ W = \frac{1}{2}k \cdot \frac{F^2}{k^2} = \frac{F^2}{2k} \]
Step 3: Relation between work and spring constant. \[ W \propto \frac{1}{k} \]
Step 4: Take ratio. \[ \frac{W_A}{W_B} = \frac{1/k_A}{1/k_B} = \frac{k_B}{k_A} \]
Step 5: Conclusion. \[ \boxed{k_B : k_A} \]
Step 1: Substitute \(x = \frac{F}{k}\) into work formula. \[ W = \frac{1}{2}k \left(\frac{F}{k}\right)^2 \]
Step 2: Simplify expression. \[ W = \frac{1}{2}k \cdot \frac{F^2}{k^2} = \frac{F^2}{2k} \]
Step 3: Relation between work and spring constant. \[ W \propto \frac{1}{k} \]
Step 4: Take ratio. \[ \frac{W_A}{W_B} = \frac{1/k_A}{1/k_B} = \frac{k_B}{k_A} \]
Step 5: Conclusion. \[ \boxed{k_B : k_A} \]
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