What is the work done in stretching a uniform | vedclass

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(B) The work done in stretching a wire is given by the formula for elastic potential energy stored in the wire:$W = \frac{1}{2} 	imes 	ext{Stress} \

Vedclass · Accepted Answer

$0.8$

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(B) The work done in stretching a wire is given by the formula for elastic potential energy stored in the wire:$W = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain} 	imes 	ext{Volume}$$W = \frac{1}{2} 	imes \left( Y \frac{\Delta \ell}{\ell} ight) 	imes \left( \frac{\Delta \ell}{\ell} ight) 	imes (A \ell)$$W = \frac{1}{2} \frac{Y A}{\ell} (\Delta \ell)^2$Given:$Y = 2 	imes 10^{11} \ N/m^2$$A = 10^{-6} \ m^2$$\ell = 2 \ m$$\Delta \ell = 2.004 - 2 = 0.004 \ m = 4 	imes 10^{-3} \ m$Substituting the values:$W = \frac{1}{2} 	imes \frac{2 	imes 10^{11} 	imes 10^{-6}}{2} 	imes (4 	imes 10^{-3})^2$$W = \frac{1}{2} 	imes 10^5 	imes 16 	imes 10^{-6}$$W = 0.5 	imes 1.6 = 0.8 \ J$

What is the work done in stretching a uniform metal wire of length $2 \ m$ to $2.004 \ m$ with an area of cross-section $10^{-6} \ m^2$ (in $J$)? [Young's modulus of the wire = $2 \times 10^{11} \ N/m^2$]

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