Conservation of energy and spring and pendulum bob

Question:
A pendulum bob is released from the height of $h$ to hit a spring that creates the force $F=-kx-bx^3$ in terms of the displacement. If the pendulum has mass $m$, find the compression displacement of the spring.

Answer:
Find the potential energy of spring. Since it is a conservative force, we integrate it in terms of displacement.
\[
U = -\int F dx = \int kx+bx^3 dx = \frac{1}{2}kx^2+\frac{1}{4}bx^4
\]
The potential energy of the bob is $mgh$. This can be transferred into the spring energy, so
\[
mgh =  \frac{1}{2}kx^2+\frac{1}{4}bx^4
\]
Rearrange it to solve for $x$:
\begin{eqnarray}
& & x^4 +\frac{2k}{b}x^2 = \frac{4mgh}{b}  \\
& & \left(x^2+\frac{k}{b}\right)^2 - \frac{k^2}{b^2} = \frac{4mgh}{b}  \\
& & x^2 + \frac{k}{b} = \sqrt{\frac{4mgh}{b}+\frac{k^2}{b^2}}  \\
& & x  = \sqrt{\sqrt{\frac{4mgh}{b}+\frac{k^2}{b^2}} -\frac{k}{b}}
\end{eqnarray}

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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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