A load of 100 N acts on a bar of length 10 cm and area of cross section 1 cm. Find the energy stored in it. `(Y= 10^(11) N//m^(2))`

Answer

To solve the problem of finding the energy stored in a bar subjected to a load, we can follow these steps:

### Step 1: Identify the given values
- Load (F) = 100 N
- Length of the bar (L) = 10 cm = 0.1 m (conversion from cm to m)
- Area of cross-section (A) = 1 cm² = 1 × 10⁻⁴ m² (conversion from cm² to m²)
- Young's modulus (Y) = 10¹¹ N/m²

### Step 2: Calculate the stress (σ)
Stress is defined as the force applied per unit area. The formula for stress is:
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{100 \, \text{N}}{1 \times 10^{-4} \, \text{m}^2} = 1 \times 10^{6} \, \text{N/m}^2 \]

### Step 3: Calculate the strain (ε)
Strain is defined as the change in length per unit original length. The relationship between stress and strain is given by Young's modulus:
\[ Y = \frac{\sigma}{\epsilon} \]
From this, we can express strain as:
\[ \epsilon = \frac{\sigma}{Y} \]
Substituting the values:
\[ \epsilon = \frac{1 \times 10^{6} \, \text{N/m}^2}{10^{11} \, \text{N/m}^2} = 1 \times 10^{-5} \]

### Step 4: Calculate the energy stored (U)
The energy stored in the material due to deformation can be calculated using the formula:
\[ U = \frac{1}{2} \sigma \epsilon A L \]
Substituting the values:
\[ U = \frac{1}{2} \times (1 \times 10^{6} \, \text{N/m}^2) \times (1 \times 10^{-5}) \times (1 \times 10^{-4} \, \text{m}^2) \times (0.1 \, \text{m}) \]
Calculating this step by step:
1. Calculate \( \sigma \epsilon \):
\[ \sigma \epsilon = (1 \times 10^{6}) \times (1 \times 10^{-5}) = 10 \]
2. Now multiply by \( A \) and \( L \):
\[ U = \frac{1}{2} \times 10 \times (1 \times 10^{-4}) \times (0.1) = \frac{1}{2} \times 10 \times 1 \times 10^{-5} = 5 \times 10^{-5} \, \text{J} \]

### Final Answer
The energy stored in the bar is:
\[ U = 5 \times 10^{-5} \, \text{J} \]

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