One end of a slack wire (Young's modulus Y,le | vedclass

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(A) When the wire becomes taut,the kinetic energy of the block is converted into the elastic potential energy stored in the stretched wire.The elastic

Vedclass · Accepted Answer

$v \sqrt{\frac{m L}{A Y}}$

Vedclass · Answer

(A) When the wire becomes taut,the kinetic energy of the block is converted into the elastic potential energy stored in the stretched wire.The elastic potential energy stored in a wire is given by $U = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain} 	imes 	ext{Volume}$.Using Hooke's Law,$	ext{Stress} = Y 	imes 	ext{Strain}$,where $	ext{Strain} = \frac{\Delta l}{L}$.Therefore,$U = \frac{1}{2} 	imes Y 	imes \left(\frac{\Delta l}{L}ight)^2 	imes (A 	imes L) = \frac{1}{2} \frac{Y A}{L} (\Delta l)^2$.Equating the initial kinetic energy of the block to the potential energy stored in the wire:$\frac{1}{2} m v^2 = \frac{1}{2} \frac{Y A}{L} (\Delta l)^2$.Solving for $\Delta l$:$(\Delta l)^2 = \frac{m v^2 L}{A Y}$.$\Delta l = v \sqrt{\frac{m L}{A Y}}$.Thus,the maximum distance the block travels after the wire becomes taut is $v \sqrt{\frac{m L}{A Y}}$.

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