A copper wire of length 3 text{ m} is stretch | vedclass

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Question

(A) The elastic potential energy $U$ stored in a stretched wire is given by the formula: $U = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain}

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(A) The elastic potential energy $U$ stored in a stretched wire is given by the formula: $U = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain} 	imes 	ext{Volume}$.Since $	ext{Stress} = Y 	imes 	ext{Strain}$,we have $U = \frac{1}{2} 	imes Y 	imes (	ext{Strain})^2 	imes 	ext{Volume}$.Given:Length $L = 3 	ext{ m}$,Extension $\Delta L = 3 	ext{ mm} = 3 	imes 10^{-3} 	ext{ m}$,Volume $V = 600 	imes 10^{-6} 	ext{ m}^3$,Young's modulus $Y = 1.1 	imes 10^{11} 	ext{ N/m}^2$.Calculate Strain: $	ext{Strain} = \frac{\Delta L}{L} = \frac{3 	imes 10^{-3}}{3} = 10^{-3}$.Now,substitute the values into the formula:$U = \frac{1}{2} 	imes (1.1 	imes 10^{11}) 	imes (10^{-3})^2 	imes (600 	imes 10^{-6})$$U = 0.5 	imes 1.1 	imes 10^{11} 	imes 10^{-6} 	imes 600 	imes 10^{-6}$$U = 0.5 	imes 1.1 	imes 600 	imes 10^{-1}$$U = 0.5 	imes 1.1 	imes 60 = 33 	ext{ J}$.

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