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

Concept: When an external stretching force is applied to a metallic wire, work is performed against the internal interatomic restorative forces of the material. This work is stored within the molecular matrix of the wire as elastic potential energy (\(U\)). The basic mechanical relationship for energy stored during a variable spring-like stretch is given by: \[ U = \frac{1}{2} \times \text{Stretching Force} \times \text{Elongation} = \frac{1}{2} \cdot F \cdot x \] Alternatively, from the macroscopic perspective of materials science, the potential energy per unit volume (energy density, \(u\)) is related directly to mechanical stress and strain: \[ u = \frac{\text{Energy}}{\text{Volume}} = \frac{1}{2} \times \text{Stress} \times \text{Strain} \]

Step 1: Finding the required stretching force using Young's Modulus.
By definition, Young's Modulus (\(Y\)) is the ratio of longitudinal stress to longitudinal strain: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\left(\frac{F}{A}\right)}{\left(\frac{x}{L}\right)} = \frac{F \cdot L}{A \cdot x} \] Rearranging this relationship to isolate the dynamic tension force \(F\) required to sustain an elongation of \(x\): \[ F = \frac{Y \cdot A \cdot x}{L} \quad \cdots (1) \]

Step 2: Calculating the total stored elastic potential energy ($U$).
Substitute our expression for force from equation (1) into the mechanical work-energy formula: \[ U = \frac{1}{2} \cdot F \cdot x \] \[ U = \frac{1}{2} \cdot \left( \frac{YAx}{L} \right) \cdot x \] Combining the identical variable terms yields: \[ U = \frac{YAx^2}{2L} \] This directly matches the algebraic layout of Option (B).