$\begin{array}{1 1}\large\frac{27}{10}\\\large\frac{29}{10}\\27\\\large\frac{17}{10}\end{array} $

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Answer : $\large\frac{27}{10}$

From Hooke's law : $F=kx$ where $x$ is the distance the sprin is stretched from its natural length,and k is the (usually unknown) spring constant.

The spring's length is 30cm,so it is stretched 12cm from its natural length,we can use

$W=\int dW$

$\;\;\;\;=\int Fdx$

$2J=\int_0^{.12m} kxdx$

$\quad\;=\large\frac{k}{2}$$x^2\bigg|_0^{.12m}$

$\quad\;=\large\frac{k}{2}$$(0.0144m^2)$

$k =\large\frac{4Nm}{0.0144m^2}$

$\;\;\;=\large\frac{100}{9}$N/m

Once we have $k$,we can calculate the distance,since we know that $F=30N$ :

$F=kx$

$30=\large\frac{100}{9}$$x$

$x=\large\frac{27}{10}$

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