Potential energy and work: Spring

Question:
A spring is stretched when 4.0-kg object is hung on it vertically. The displacement is measured as 0.020 m. Then, one stretches it farther by 0.040 m. Find the work done by this external agent.

Answer:
Work can be obtained by integrating the force $|F|=kx$ with respect to correspondent displacement.
\begin{eqnarray}
W &=& \int^x_0 F\cdot dx  \\
  &=& \int^x_0 kx dx   = \frac{1}{2}kx^2
\end{eqnarray}
In order to calculate the work, we need to find the spring constant, $k$. Consider the free-body diagram. The spring and gravitational forces act on the hanging mass. According to the equation of motion, we have the net force:
\[
\sum F = kx - mg = ma = 0
\]
This system is in equilibrium, so $a=0$. Thus,
\[
kx = mg  \\
\rightarrow k = \frac{mg}{x} = \frac{4.0\cdot 9.8}{0.02}=1960 \ \mathrm{N/m}
\]
Now, we can calculate the work done by external agent.
\begin{eqnarray}
W &=& \frac{1}{2}kx^2  \\
    &=& \frac{1}{2}1960 \cdot 0.04^2  \\
    &=& 1.6 \ \mathrm{J}
\end{eqnarray}

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