A uniform copper bar of density `rho` , length `L` , cross - sectional area `S` and young's modulus `Y` is moving horizontally on a frictionless surface with constant acceleration `a_(0) `. Find
(a) the stress at the center of the wire ,
(b) total elongation of the wire .

Explanation

To solve the problem step by step, we will find the stress at the center of the copper bar and then calculate the total elongation of the wire.

### Step 1: Understanding the Problem
We have a uniform copper bar with:
- Density: \( \rho \)
- Length: \( L \)
- Cross-sectional area: \( S \)
- Young's modulus: \( Y \)
- Acceleration: \( a_0 \)

The bar is moving horizontally on a frictionless surface with constant acceleration \( a_0 \).

### Step 2: Finding the Stress at the Center of the Wire
1. **Identify the forces acting on the bar**:
- The tension in the bar due to the acceleration can be considered. The total mass of the bar is \( m = \rho \cdot V = \rho \cdot (S \cdot L) \).

2. **Divide the bar**:
- Consider the left half of the bar, which has a length of \( \frac{L}{2} \) and a mass of \( \frac{m}{2} \).

3. **Calculate the tension (T)**:
- The tension at the center of the bar due to the acceleration is given by:
\[ T = \frac{m}{2} \cdot a_0 = \frac{\rho S L}{2} \cdot a_0 \]

4. **Calculate the stress**:
- Stress is defined as force per unit area. Therefore,
\[ \text{Stress} = \frac{T}{S} = \frac{\frac{\rho S L}{2} \cdot a_0}{S} = \frac{\rho L a_0}{2} \]

### Step 3: Finding the Total Elongation of the Wire
1. **Consider a small element of the wire**:
- Take a small segment of the wire of length \( dx \) at a distance \( x \) from one end.

2. **Calculate the tension at this segment**:
- The tension in this segment is given by:
\[ T(x) = \frac{m}{L} \cdot x \cdot a_0 = \frac{\rho S L}{L} \cdot x \cdot a_0 = \rho S a_0 x \]

3. **Calculate the elongation of the small segment \( dx \)**:
- The elongation \( dL \) of the small segment is given by:
\[ dL = \frac{T(x) \cdot dx}{Y \cdot S} = \frac{\rho S a_0 x \cdot dx}{Y \cdot S} = \frac{\rho a_0 x \cdot dx}{Y} \]

4. **Integrate to find the total elongation**:
- The total elongation \( \Delta L \) of the entire wire is obtained by integrating \( dL \) from \( 0 \) to \( L \):
\[ \Delta L = \int_0^L \frac{\rho a_0 x \cdot dx}{Y} = \frac{\rho a_0}{Y} \int_0^L x \cdot dx = \frac{\rho a_0}{Y} \cdot \left[\frac{x^2}{2}\right]_0^L = \frac{\rho a_0 L^2}{2Y} \]

### Final Answers
(a) The stress at the center of the wire is:
\[ \text{Stress} = \frac{\rho L a_0}{2} \]

(b) The total elongation of the wire is:
\[ \Delta L = \frac{\rho a_0 L^2}{2Y} \]