A steel wire` 4.0m` in length is stretched through `2.0mm`.The cross -sectional area of the wire is `2.0 mm^(2)`.If young's modulus of steel is `2.0xx10^(11) N//m^(2)`
(a) the enrgy density of wire,
(b) the elastic potential energy stored in the wire.

Explanation

To solve the problem step by step, we will calculate the energy density of the wire and then the elastic potential energy stored in the wire.

### Given Data:
- Length of the wire, \( L = 4.0 \, \text{m} \)
- Elongation of the wire, \( \Delta L = 2.0 \, \text{mm} = 2.0 \times 10^{-3} \, \text{m} \)
- Cross-sectional area of the wire, \( A = 2.0 \, \text{mm}^2 = 2.0 \times 10^{-6} \, \text{m}^2 \)
- Young's modulus of steel, \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \)

### (a) Calculate the Energy Density of the Wire

1. **Calculate Strain**:
\[ \text{Strain} = \frac{\Delta L}{L} = \frac{2.0 \times 10^{-3}}{4.0} = 5.0 \times 10^{-4} \]

2. **Use the formula for Energy Density**:
The energy density \( U \) is given by:
\[ U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \]
From Young's modulus, we know:
\[ \text{Stress} = Y \times \text{Strain} \]
Therefore,
\[ U = \frac{1}{2} \times Y \times \text{Strain}^2 \]

3. **Substitute the values**:
\[ U = \frac{1}{2} \times (2.0 \times 10^{11}) \times (5.0 \times 10^{-4})^2 \]
\[ U = \frac{1}{2} \times (2.0 \times 10^{11}) \times (25.0 \times 10^{-8}) \]
\[ U = \frac{1}{2} \times (5.0 \times 10^{3}) = 2.5 \times 10^{4} \, \text{J/m}^3 \]

### (b) Calculate the Elastic Potential Energy Stored in the Wire

1. **Calculate the Volume of the Wire**:
The volume \( V \) of the wire can be calculated using:
\[ V = A \times L = (2.0 \times 10^{-6}) \times (4.0) = 8.0 \times 10^{-6} \, \text{m}^3 \]

2. **Calculate the Elastic Potential Energy**:
The elastic potential energy \( E \) stored in the wire is given by:
\[ E = U \times V \]
Substituting the values:
\[ E = (2.5 \times 10^{4}) \times (8.0 \times 10^{-6}) \]
\[ E = 0.2 \, \text{J} \]

### Final Answers:
- (a) The energy density of the wire is \( 2.5 \times 10^{4} \, \text{J/m}^3 \).
- (b) The elastic potential energy stored in the wire is \( 0.2 \, \text{J} \).