To solve the problem, we need to find Young's modulus of the material of the wire and the energy stored in the wire. Let's break down the solution step by step.
### Step 1: Calculate the Force
The force (F) applied to the wire is given by the weight of the load. The weight can be calculated using the formula:
\[ F = m \cdot g \]
where:
- \( m = 10 \, \text{kg} \) (mass of the load)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
Calculating the force:
\[ F = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \]
### Step 2: Calculate the Cross-Sectional Area of the Wire
The diameter of the wire is given as 1 mm. First, we need to convert this to meters:
\[ \text{Diameter} = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \]
The radius (r) is half of the diameter:
\[ r = \frac{1 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \, \text{m} \]
Now, we can calculate the cross-sectional area (A) using the formula for the area of a circle:
\[ A = \pi r^2 \]
Calculating the area:
\[ A = \pi (0.5 \times 10^{-3})^2 = \pi (0.25 \times 10^{-6}) \approx 7.85 \times 10^{-7} \, \text{m}^2 \]
### Step 3: Calculate the Stress
Stress (σ) is defined as the force applied per unit area:
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{100 \, \text{N}}{7.85 \times 10^{-7} \, \text{m}^2} \approx 1.27 \times 10^{8} \, \text{N/m}^2 \]
### Step 4: Calculate the Strain
Strain (ε) is defined as the change in length divided by the original length:
\[ \epsilon = \frac{\Delta L}{L_0} \]
where:
- \( \Delta L = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} \)
- \( L_0 = 5 \, \text{m} \)
Calculating the strain:
\[ \epsilon = \frac{0.3 \times 10^{-3} \, \text{m}}{5 \, \text{m}} = 0.00006 \]
### Step 5: Calculate Young's Modulus
Young's modulus (E) is defined as the ratio of stress to strain:
\[ E = \frac{\sigma}{\epsilon} \]
Substituting the values:
\[ E = \frac{1.27 \times 10^{8} \, \text{N/m}^2}{0.00006} \approx 2.12 \times 10^{12} \, \text{N/m}^2 \]
### Step 6: Calculate the Energy Stored in the Wire
The energy (U) stored in the wire can be calculated using the formula:
\[ U = \frac{1}{2} \sigma \epsilon A L_0 \]
Substituting the values:
\[ U = \frac{1}{2} \times (1.27 \times 10^{8}) \times (0.00006) \times (7.85 \times 10^{-7}) \times (5) \]
Calculating the energy:
\[ U \approx 0.0174 \, \text{J} \]
### Final Answers
- Young's Modulus \( E \approx 2.12 \times 10^{12} \, \text{N/m}^2 \)
- Energy stored \( U \approx 0.0174 \, \text{J} \)
