A steel wire of diameter 0.50 mm and length 1.20 m has Young's modulus 2.0 × 10¹¹ N/m²…

Step 1. The greatest load. $F = \text{stress} \times A$ $= 2.5 \times 10^{8}\ \text{N/m}^2 \times 1.96$ $\times 10^{-7}\ \text{m}^2$ $= 49.1$ N. So $M = \dfrac{F}{g} = \dfrac{49.1\ \text{N}}{9.8\ \text{m/s}^2} = 5.0$ kg.

Step 2. The extension. $\Delta L = \dfrac{\text{stress} \times L}{Y}$ $= \dfrac{2.5 \times 10^{8}\ \text{N/m}^2 \times 1.20\ \text{m}}{2.0 \times 10^{11}\ \text{N/m}^2}$ $= 1.5 \times 10^{-3}\ \text{m} = 1.5$ mm.

Step 3. The energy. $U = \tfrac{1}{2} F\,\Delta L$ $= \tfrac{1}{2} \times 49.1\ \text{N} \times 1.5 \times 10^{-3}\ \text{m}$ $= 0.037$ J.

Why stop. Beyond the elastic limit the extension grows faster than the load, and the wire keeps a permanent stretch. The graph is no longer a straight line, so the slope does not give $Y$.