Question

The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is______ mm² . (Given, y = 2.0 x 10¹¹ Nm^-2)

Answer 1

Correct Answer: 40

To find the cross-sectional area of the steel wire, we use the formula for elastic potential energy:

\[ U = \frac{1}{2} \times \frac{F}{A} \times \Delta L \times L \]
Where:
- \( U \) is the elastic potential energy, 80 J
- \( F \) is the force applied
- \( A \) is the cross-sectional area
- \( \Delta L \) is the change in length, 2 cm = 0.02 m
- \( L \) is the original length, 20 m

Also, from Hooke's Law, the relation for Young's modulus \( y \) is:
\[ y = \frac{F \cdot L}{A \cdot \Delta L} \]
Given \( y = 2.0 \times 10^{11} \) N/m², we can rearrange to find \( F \):
\[ F = \frac{y \cdot A \cdot \Delta L}{L} \]
Substituting \( F \) in the energy equation:
\[ U = \frac{1}{2} \cdot \frac{y \cdot A \cdot \Delta L}{L} \cdot \frac{\Delta L \cdot L}{A} \]
\[ U = \frac{1}{2} \cdot y \cdot (\Delta L)^2 \]
Solving for \( A \):
\[ A = \frac{U \cdot 2L}{y \cdot (\Delta L)^2} \]
\[ A = \frac{80 \cdot 2 \cdot 20}{2.0 \times 10^{11} \cdot (0.02)^2} \]
\[ A = \frac{3200}{2.0 \times 10^{11} \cdot 0.0004} \]
\[ A = \frac{3200}{8 \times 10^7} \]
\[ A = 40 \times 10^{-5} \, \text{m}^2 \]
Converting to mm²:
\[ A = 40 \, \text{mm}^2 \]

The cross-sectional area of the wire is 40 mm², and it falls within the expected range [40, 40].