A spring, which is initially in its unstretched condition, is first stretched by a length x and then again by a further length x. The work done in the first case is `W_(1)` and in the second case is `W_(2)`.

Options

A. `W_(2) = W_(1)`
B. `W_(2) = 2W_(1)`
C. `W_(2) = 3W_(1)`
D. `W_(2) = 4W_(1)`

Answer

To solve the problem, we need to calculate the work done in stretching a spring twice, first by a length \( x \) and then by an additional length \( x \). We will use the formula for the potential energy stored in a spring, which is given by:

\[ PE = \frac{1}{2} k x^2 \]

where \( k \) is the spring constant and \( x \) is the extension of the spring.

### Step-by-step Solution:

1. **Calculate Work Done in the First Stretch (W1)**:
- Initially, the spring is unstretched. When it is stretched by a length \( x \), the work done \( W_1 \) is equal to the potential energy stored in the spring after the first stretch.
- Using the formula for potential energy:
\[ W_1 = \frac{1}{2} k x^2 \]

2. **Calculate Work Done in the Second Stretch (W2)**:
- After the first stretch, the spring is now at an extension of \( x \). When it is stretched further by an additional length \( x \), the total extension becomes \( 2x \).
- The work done \( W_2 \) in this case is equal to the change in potential energy when the spring is stretched from \( x \) to \( 2x \).
- The potential energy at \( 2x \) is:
\[ PE_{final} = \frac{1}{2} k (2x)^2 = \frac{1}{2} k (4x^2) = 2 k x^2 \]
- The potential energy at \( x \) is:
\[ PE_{initial} = \frac{1}{2} k x^2 \]
- Therefore, the work done \( W_2 \) is:
\[ W_2 = PE_{final} - PE_{initial} = 2 k x^2 - \frac{1}{2} k x^2 = \frac{4}{2} k x^2 - \frac{1}{2} k x^2 = \frac{3}{2} k x^2 \]

3. **Relate W1 and W2**:
- We already have \( W_1 = \frac{1}{2} k x^2 \).
- Now substituting \( W_1 \) into the equation for \( W_2 \):
\[ W_2 = 3 \left(\frac{1}{2} k x^2\right) = 3 W_1 \]

### Conclusion:
Thus, the relationship between the work done in the first and second stretches is:
\[ W_2 = 3 W_1 \]