Given:
Step 1: Relationship Between Lateral and Longitudinal Strain
Poisson’s ratio is given by:
\[ \nu = \frac{\text{lateral strain}}{\text{longitudinal strain}} \]
Rearranging for longitudinal strain \( \varepsilon \):
\[ \varepsilon = \frac{\varepsilon_l}{\nu} = \frac{10^{-3}}{0.25} = 4 \times 10^{-3} \]
Step 2: Elastic Energy Density Formula
The elastic energy density \( U \) is given by:
\[ U = \frac{1}{2} Y \varepsilon^2 \]
Substituting values:
\[ U = \frac{1}{2} \times (2 \times 10^{11}) \times (4 \times 10^{-3})^2 \]
\[ U = \frac{1}{2} \times (2 \times 10^{11}) \times (16 \times 10^{-6}) \]
\[ U = \frac{32 \times 10^5}{2} = 16 \times 10^5 \text{ J/m}^3 \]
Answer: The correct option is D (16 × 105 J/m3).
The elastic energy density \( u \) is given by the formula: \[ u = \frac{1}{2} \sigma \varepsilon \] where: - \( \sigma \) is the stress, - \( \varepsilon \) is the strain. We know that stress \( \sigma \) is related to the Young's modulus \( Y \) and strain \( \varepsilon \) by: \[ \sigma = Y \cdot \varepsilon \] The lateral strain \( \varepsilon_l \) is related to the longitudinal strain \( \varepsilon \) by Poisson's ratio \( \nu \) as: \[ \varepsilon_l = - \nu \cdot \varepsilon \] Substituting the values: - \( Y = 2 \times 10^{11} \, \text{Nm}^{-2} \), - \( \nu = 0.25 \), - \( \varepsilon_l = 10^{-3} \), we can calculate the longitudinal strain \( \varepsilon \): \[ \varepsilon = \frac{\varepsilon_l}{\nu} = \frac{10^{-3}}{0.25} = 4 \times 10^{-3} \] Now, the stress \( \sigma \) is: \[ \sigma = Y \cdot \varepsilon = 2 \times 10^{11} \times 4 \times 10^{-3} = 8 \times 10^8 \, \text{N/m}^2 \] Finally, the energy density is: \[ u = \frac{1}{2} \cdot \sigma \cdot \varepsilon = \frac{1}{2} \cdot 8 \times 10^8 \cdot 4 \times 10^{-3} = 16 \times 10^5 \, \text{Jm}^{-3} \] Thus, the correct answer is (D).
A 2 $\text{kg}$ mass is attached to a spring with spring constant $ k = 200, \text{N/m} $. If the mass is displaced by $ 0.1, \text{m} $, what is the potential energy stored in the spring?