Question

A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$. If it is stretched by an amount $x$, the elastic potential energy stored in the wire is:

• A. \( \frac{YAx^2}{L} \)
• B. \( \frac{YAx^2}{2L} \)
• C. \( \frac{2YAx^2}{L} \)
• D. \( \frac{YAx}{L} \)

Hint:

Save time by treating a stretching wire exactly like a mechanical spring! A wire behaves as a spring with an effective stiffness constant of \(k = \frac{YA}{L}\). Since the potential energy stored inside a spring stretched by a distance \(x\) is universally given by \(\frac{1}{2}kx^2\), substituting our effective wire stiffness yields \(\frac{1}{2}\left(\frac{YA}{L}\right)x^2 = \frac{YAx^2}{2L}\) in under 5 seconds!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the elastic potential energy stored in a wire of length \(L\), cross-sectional area \(A\), and Young's modulus \(Y\) when it is stretched by a distance \(x\).
When an external stretching force is applied to a wire, work is performed against the internal interatomic restoring forces of the material.
This work is stored within the molecular matrix of the wire as elastic potential energy.
Step 2: Key Formula or Approach:
Young's modulus \(Y\) is defined as the ratio of tensile stress to tensile strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} \]
The force \(F\) required to stretch the wire by an amount \(x\) can be expressed as:
\[ F = \frac{YAx}{L} \]
The elastic potential energy \(U\) stored in the wire is given by the average work done:
\[ U = \frac{1}{2} \times \text{Stretching Force} \times \text{Elongation} = \frac{1}{2} F x \]
Step 3: Detailed Explanation:
1. We rearrange the Young's modulus equation to express the tension force \(F\) in terms of the elongation \(x\):
\[ F = \frac{YAx}{L} \]
2. Now, substitute this force expression into the work-energy formula for the stored potential energy:
\[ U = \frac{1}{2} \cdot \left( \frac{YAx}{L} \right) \cdot x \]
3. Combining the terms simplifies the expression:
\[ U = \frac{YAx^2}{2L} \]
4. This result matches Option (B).
Step 4: Final Answer:
The stored elastic potential energy in the wire is \( \frac{YAx^2}{2L} \), which corresponds to Option (B).