Young's Modulus

Question

A steel wire of length L and cross-sectional area A is stretched by a force F. The work done in stretching the wire is (Y is Young's Modulus):

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1. **Work Done in Stretching**: $W = \frac{1}{2} 	imes 	ext{Force} 	imes 	ext{Extension} = \frac{1}{2} F \Delta L$.
2. **Young's Modulus Formula**: $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta L / L} = \frac{FL}{A \Delta L}$.
3. **Substitute for $\Delta L$**: $\Delta L = \frac{FL}{AY}$.
4. **Final Calculation**: 
   $$W = \frac{1}{2} F \left(\frac{FL}{AY}ight) = \frac{F^2 L}{2AY}$$

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