An elastic spring of unstretched length `L` and force constant `K` is stretched by amoun t `x` .It is further stretched by another length `y` The work done in the second streaching is

Options

A. `(1)/(2)Ky^(2)`
B. `(1)/(2)K(x^(2) + y^(2))`
C. `(1)/(2)K(x+ y)^(2))`
D. `(1)/(2)Ky(2x+ y)`

Explanation

To find the work done in the second stretching of the elastic spring, we can follow these steps:

### Step 1: Understand the Work Done in Stretching a Spring
The work done \( W \) in stretching a spring is given by the formula:
\[ W = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the amount of stretch from the unstretched length.

### Step 2: Calculate Work Done for Initial Stretching
When the spring is initially stretched by an amount \( x \), the work done \( W_1 \) is:
\[ W_1 = \frac{1}{2} k x^2 \]

### Step 3: Calculate Work Done for Final Stretching
When the spring is further stretched by an additional amount \( y \), the total stretch becomes \( x + y \). The work done \( W_2 \) for this final stretch is:
\[ W_2 = \frac{1}{2} k (x + y)^2 \]

### Step 4: Calculate the Work Done in the Second Stretching
The work done in the second stretching (from \( x \) to \( x + y \)) is given by the difference between the work done in the final stretch and the work done in the initial stretch:
\[ W_{\text{second}} = W_2 - W_1 \]
Substituting the expressions we derived:
\[ W_{\text{second}} = \frac{1}{2} k (x + y)^2 - \frac{1}{2} k x^2 \]

### Step 5: Simplify the Expression
Now we simplify the expression:
\[ W_{\text{second}} = \frac{1}{2} k \left[ (x + y)^2 - x^2 \right] \]
Expanding \( (x + y)^2 \):
\[ W_{\text{second}} = \frac{1}{2} k \left[ x^2 + 2xy + y^2 - x^2 \right] \]
The \( x^2 \) terms cancel out:
\[ W_{\text{second}} = \frac{1}{2} k (2xy + y^2) \]

### Step 6: Factor Out Common Terms
Factoring out \( y \):
\[ W_{\text{second}} = \frac{1}{2} k y (2x + y) \]

### Final Answer
Thus, the work done in the second stretching is:
\[ W_{\text{second}} = \frac{1}{2} k y (2x + y) \]

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