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 energy density of wire,
(b) the elastic potential energy stored in the wire.

Explanation

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

### 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) Finding the Energy Density of the Wire

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

2. **Calculate the Stress:**
\[ \text{Stress} = Y \times \text{Strain} = (2.0 \times 10^{11} \, \text{N/m}^2) \times (5.0 \times 10^{-4}) = 10.0 \times 10^{7} \, \text{N/m}^2 = 1.0 \times 10^{8} \, \text{N/m}^2 \]

3. **Calculate the Force:**
\[ F = \text{Stress} \times A = (1.0 \times 10^{8} \, \text{N/m}^2) \times (2.0 \times 10^{-6} \, \text{m}^2) = 200 \, \text{N} \]

4. **Calculate the Spring Constant \( K \):**
Using Hooke's Law \( F = K \Delta L \):
\[ K = \frac{F}{\Delta L} = \frac{200 \, \text{N}}{2.0 \times 10^{-3} \, \text{m}} = 100,000 \, \text{N/m} = 10^5 \, \text{N/m} \]

5. **Calculate the Elastic Potential Energy:**
\[ U = \frac{1}{2} K (\Delta L)^2 = \frac{1}{2} (10^5 \, \text{N/m}) \times (2.0 \times 10^{-3} \, \text{m})^2 \]
\[ U = \frac{1}{2} (10^5) \times (4.0 \times 10^{-6}) = 0.2 \, \text{J} \]

6. **Calculate the Volume of the Wire:**
\[ V = A \times L = (2.0 \times 10^{-6} \, \text{m}^2) \times (4.0 \, \text{m}) = 8.0 \times 10^{-6} \, \text{m}^3 \]

7. **Calculate the Energy Density:**
\[ \text{Energy Density} = \frac{U}{V} = \frac{0.2 \, \text{J}}{8.0 \times 10^{-6} \, \text{m}^3} = 25,000 \, \text{J/m}^3 = 2.5 \times 10^4 \, \text{J/m}^3 \]

### (b) Elastic Potential Energy Stored in the Wire
The elastic potential energy has already been calculated as:
\[ U = 0.2 \, \text{J} \]

### Final Answers:
(a) Energy Density of the wire = \( 2.5 \times 10^4 \, \text{J/m}^3 \)

(b) Elastic Potential Energy stored in the wire = \( 0.2 \, \text{J} \)

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