To find the force required to stretch a steel wire with a cross-sectional area of \(1 \, \text{cm}^2\) to increase its length by \(1\%\), we can use the formula related to Young's modulus:
\[
Y = \frac{F \cdot L_0}{A \cdot \Delta L}
\]
Where:
- \(Y\) is Young's modulus,
- \(F\) is the force applied,
- \(L_0\) is the original length of the wire,
- \(A\) is the cross-sectional area,
- \(\Delta L\) is the change in length.
### Step 1: Convert units
First, we need to convert the cross-sectional area from \( \text{cm}^2 \) to \( \text{m}^2 \):
\[
1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2
\]
### Step 2: Define the parameters
Let’s define the parameters we have:
- Young's modulus, \(Y = 2 \times 10^{11} \, \text{N/m}^2\)
- Cross-sectional area, \(A = 1 \times 10^{-4} \, \text{m}^2\)
- Change in length, \(\Delta L = 1\% \text{ of } L_0 = 0.01 L_0\)
### Step 3: Substitute into the Young's modulus formula
We can rearrange the Young's modulus formula to solve for \(F\):
\[
F = \frac{Y \cdot A \cdot \Delta L}{L_0}
\]
### Step 4: Substitute \(\Delta L\)
Substituting \(\Delta L = 0.01 L_0\) into the equation:
\[
F = \frac{Y \cdot A \cdot (0.01 L_0)}{L_0}
\]
### Step 5: Simplify the equation
The \(L_0\) cancels out:
\[
F = Y \cdot A \cdot 0.01
\]
### Step 6: Plug in the values
Now, substituting the values for \(Y\) and \(A\):
\[
F = (2 \times 10^{11} \, \text{N/m}^2) \cdot (1 \times 10^{-4} \, \text{m}^2) \cdot 0.01
\]
### Step 7: Calculate the force
Calculating the force:
\[
F = 2 \times 10^{11} \cdot 1 \times 10^{-4} \cdot 0.01 = 2 \times 10^{11} \cdot 1 \times 10^{-6} = 2 \times 10^5 \, \text{N}
\]
### Final Answer
The force required to stretch the steel wire is \(F = 2 \times 10^5 \, \text{N}\).
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