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

• A. \(k_B : k_A\)
• B. \(k_A : k_B\)
• C. \(k_A k_B : 1\)
• D. \(\sqrt{k_B} : \sqrt{k_A}\)
• E. \(\sqrt{k_A} : \sqrt{k_B}\)

Hint:

For same force, softer spring (smaller \(k\)) stores more energy.

Answer 1

The Correct Option is A

Understanding the 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: Final Answer. \[ \boxed{k_B : k_A} \]