A spring 40,mm long is stretched by the appli | vedclass

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(D) The spring constant $k$ is given by Hooke's Law: $k = \frac{F}{x}$.Given $F = 10\,N$ and $x = 1\,mm = 1 	imes 10^{-3}\,m$.So,$k = \frac{10}{1 	i

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$8$

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(D) The spring constant $k$ is given by Hooke's Law: $k = \frac{F}{x}$.Given $F = 10\,N$ and $x = 1\,mm = 1 	imes 10^{-3}\,m$.So,$k = \frac{10}{1 	imes 10^{-3}} = 10^{4}\,N/m$.The work done $W$ in stretching the spring by a displacement $x = 40\,mm = 40 	imes 10^{-3}\,m$ is given by $W = \frac{1}{2}kx^{2}$.Substituting the values: $W = \frac{1}{2} 	imes 10^{4} 	imes (40 	imes 10^{-3})^{2}$.$W = \frac{1}{2} 	imes 10^{4} 	imes (1600 	imes 10^{-6}) = \frac{1}{2} 	imes 16 = 8\,J$.

$A$ spring $40\,mm$ long is stretched by the application of a force. If $10\,N$ force is required to stretch the spring through $1\,mm$,then work done in stretching the spring through $40\,mm$ is ............. $J$.

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