$\begin{array}{1 1}20Joules\\25Joules\\40Joules\\34Joules\end{array} $

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Answer : 20 Joules

We must first find the spring constant $k$.

By Hooke's law,$F(x) =kx$

Thus,$F(2) =k(2)=5$ and so we can calculate k as follows

$k=\large\frac{F(2)}{2}$

$\;\;\;=\large\frac{5N}{2m}$

$\;\;\;=2.5N/m$

Therefore,$F(x)=2.5x$ and so the work done by the force is

Work done = $\int\limits_0^4 2.5xdx$

$\qquad\qquad\;=\big[\large\frac{5}{4}$$x^2\big]_0^4$

$\qquad\qquad\;=20$ Joules

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