To solve the problem, we need to find the Young's modulus of the material of the wire and the potential energy stored in the wire. Let's go through the steps systematically.
### Step 1: Calculate the Stress
Stress (σ) is defined as the force (F) applied per unit area (A) of the wire.
1. **Calculate the force (F)**:
The force applied is due to the weight of the load:
\[
F = mg
\]
where \( m = 2 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \).
\[
F = 2 \times 9.81 = 19.62 \, \text{N} \approx 20 \, \text{N}
\]
2. **Calculate the cross-sectional area (A)**:
The area of the wire can be calculated using the formula for the area of a circle:
\[
A = \pi r^2
\]
The diameter of the wire is given as \( 0.3 \, \text{mm} \), so the radius \( r \) is:
\[
r = \frac{0.3 \, \text{mm}}{2} = 0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m} = 1.5 \times 10^{-4} \, \text{m}
\]
Now, substituting the radius into the area formula:
\[
A = \pi (1.5 \times 10^{-4})^2 = \pi (2.25 \times 10^{-8}) \approx 7.06858 \times 10^{-8} \, \text{m}^2
\]
3. **Calculate the stress (σ)**:
\[
\sigma = \frac{F}{A} = \frac{20 \, \text{N}}{7.06858 \times 10^{-8} \, \text{m}^2} \approx 2.83 \times 10^8 \, \text{N/m}^2
\]
### Step 2: Calculate the Strain
Strain (ε) is defined as the change in length (ΔL) divided by the original length (L).
1. **Given values**:
- Change in length (ΔL) = 2 mm = \( 2 \times 10^{-3} \, \text{m} \)
- Original length (L) = 3 m
2. **Calculate the strain (ε)**:
\[
\epsilon = \frac{\Delta L}{L} = \frac{2 \times 10^{-3}}{3} \approx 6.67 \times 10^{-4}
\]
### Step 3: Calculate Young's Modulus (E)
Young's modulus (E) is defined as the ratio of stress to strain:
\[
E = \frac{\sigma}{\epsilon}
\]
Substituting the values we calculated:
\[
E = \frac{2.83 \times 10^8}{6.67 \times 10^{-4}} \approx 4.25 \times 10^{11} \, \text{N/m}^2
\]
### Step 4: Calculate the Potential Energy Stored in the Wire
The potential energy (U) stored in the wire can be calculated using the formula:
\[
U = \frac{1}{2} \sigma \epsilon V
\]
where \( V \) is the volume of the wire, given by \( V = A \times L \).
1. **Calculate the volume (V)**:
\[
V = A \times L = (7.06858 \times 10^{-8}) \times 3 \approx 2.12 \times 10^{-7} \, \text{m}^3
\]
2. **Calculate the potential energy (U)**:
\[
U = \frac{1}{2} \sigma \epsilon V = \frac{1}{2} \times (2.83 \times 10^8) \times (6.67 \times 10^{-4}) \times (2.12 \times 10^{-7})
\]
\[
U \approx 0.92 \times 10^{-5} \, \text{J} \approx 0.92 \, \text{J}
\]
### Final Answers:
- Young's Modulus \( E \approx 4.25 \times 10^{11} \, \text{N/m}^2 \)
- Potential Energy \( U \approx 0.92 \, \text{J} \)
