Two springs A and B are identical except that A is stiffer than B i.e., `k_A gt k_B`. If the two springs are stretched by the same force, then

Options

A. more work is done on B i.e., `W_B gt W_A`
B. more work is done on A i.e., `W_AgtW_B`
C. work done on A and B are equal
D. work done depends upon the way in which they are stretched

Answer

To solve the problem, we need to analyze the work done on two identical springs A and B when they are stretched by the same force. Given that spring A is stiffer than spring B (i.e., \( k_A > k_B \)), we can derive the relationship between the work done on each spring.

### Step-by-Step Solution:

1. **Understanding Hooke's Law**:
The force exerted by a spring is given by Hooke's law:
\[ F = k \cdot x \]
where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the extension of the spring.

2. **Work Done on a Spring**:
The work done \( W \) in stretching a spring is given by the formula:
\[ W = \frac{1}{2} k x^2 \]
However, since we are applying the same force \( F \) to both springs, we need to express \( x \) in terms of \( F \) and \( k \).

3. **Finding the Extension**:
Rearranging Hooke's law gives us:
\[ x = \frac{F}{k} \]
Substituting this into the work done formula gives:
\[ W = \frac{1}{2} k \left(\frac{F}{k}\right)^2 = \frac{F^2}{2k} \]

4. **Comparing Work Done on Springs A and B**:
- For spring A:
\[ W_A = \frac{F^2}{2k_A} \]
- For spring B:
\[ W_B = \frac{F^2}{2k_B} \]

5. **Analyzing the Relationship**:
Since \( k_A > k_B \), it follows that:
\[ \frac{1}{k_A} < \frac{1}{k_B} \]
Therefore, we can conclude:
\[ W_A < W_B \]
This indicates that more work is done on spring B than on spring A.

### Conclusion:
The work done on spring B is greater than the work done on spring A:
\[ W_B > W_A \]