Two springs A and B are identical but A is harder than `B(k_A gt k_B)` . Let `W_A and W_B` represent the work done when the springs are stretched through the same distance and `W'_A and W'_B` are the work done when these are stretched by equal forces, then which of the following is true

Options

A. `W_A gt W_B and W'_A = W'_B`
B. `W_A gt W_B and W'_A lt W'_B`
C. `W_A gt W_B and W'_A gt W'_B`
D. `W_A lt W_B and W'_A lt W'_B`

Explanation

To solve the problem involving two springs A and B, we will analyze the work done on both springs under two different scenarios: stretching them through the same distance and stretching them with equal forces.

### Step 1: Work Done When Stretched Through the Same Distance

1. **Identify the Work Done Formula**: The work done on a spring when it is stretched by a distance \( x \) is given by the formula:
\[ W = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant.

2. **Calculate Work Done on Spring A**: For spring A, which has a spring constant \( k_A \):
\[ W_A = \frac{1}{2} k_A x^2 \]

3. **Calculate Work Done on Spring B**: For spring B, which has a spring constant \( k_B \):
\[ W_B = \frac{1}{2} k_B x^2 \]

4. **Compare Work Done**: Since it is given that \( k_A > k_B \):
\[ W_A > W_B \]
Thus, when both springs are stretched through the same distance, the work done on spring A is greater than that on spring B.

### Step 2: Work Done When Stretched by Equal Forces

1. **Identify the Force Applied**: When equal forces \( F \) are applied to both springs, we can express the force in terms of the spring constant and the extension:
\[ F = k_A x_A = k_B x_B \]
where \( x_A \) and \( x_B \) are the extensions of springs A and B, respectively.

2. **Express Extensions in Terms of Force**:
- For spring A:
\[ x_A = \frac{F}{k_A} \]
- For spring B:
\[ x_B = \frac{F}{k_B} \]

3. **Compare Extensions**: Since \( k_A > k_B \), it follows that:
\[ x_B > x_A \]
This means that spring B stretches more than spring A when the same force is applied.

4. **Calculate Work Done**:
- For spring A:
\[ W'_A = \frac{1}{2} k_A x_A^2 = \frac{1}{2} k_A \left(\frac{F}{k_A}\right)^2 = \frac{F^2}{2 k_A} \]
- For spring B:
\[ W'_B = \frac{1}{2} k_B x_B^2 = \frac{1}{2} k_B \left(\frac{F}{k_B}\right)^2 = \frac{F^2}{2 k_B} \]

5. **Compare Work Done**: Since \( k_B < k_A \):
\[ W'_B > W'_A \]
Thus, when equal forces are applied, the work done on spring B is greater than that on spring A.

### Conclusion

- When stretched through the same distance: \( W_A > W_B \)
- When stretched by equal forces: \( W'_B > W'_A \)