Answer
See the detailed answer below.
Work Step by Step
a) We will work with the assumption the author set. And we can assume that the glider's energy is conserved.
Thus,
$$E_i=E_f$$
$$K_1+U_i=K_f+U_f$$
we know that the glider starts from rest which means that its initial kinetic energy is zero.
$$0+U_i=K_f+U_f$$
Plugging $U=\dfrac{c}{x}$
$$\dfrac{c}{x_i}=\frac{1}{2}mv^2+\dfrac{c}{x_f}$$
Solving for $v^2$;
$$\dfrac{c}{x_i}-\dfrac{c}{x_f}=\frac{1}{2}mv^2$$
$$v^2= \dfrac{\dfrac{2c}{x_i}-\dfrac{2c}{x_f}}{m} $$
$$v^2= \dfrac{2c}{x_i m}-\dfrac{2c}{x_fm} $$
$$v^2= \dfrac{2c}{x_i m}-\dfrac{2c}{x_fm} $$
The final position is at $x=1$, so
$$\boxed{v^2= \dfrac{2c}{m}x_i^{-1}-\dfrac{2c}{ m} } $$
This equation is similar to the straight line equation $y=mx+b$ where $m$ is the slope of the straight line, and $b$ is the interception point with $y$-axis.
Now we need to draw $v^2$ versus $\dfrac{1}{x_i}$, by plugging the data in the table below.
\begin{array}{|c|c| }
\hline
\dfrac{1}{x}\;{\rm (m^{-1})}& v^2\;{\rm (m^2/s^2)} \\
\hline
\dfrac{1}{0.02}=50& 1.4^2=1.96 \\
\hline
\dfrac{1}{0.04}=25 & 0.98^2=0.9604 \\
\hline
\dfrac{1}{0.06}=16.667& 0.79^2=0.6241 \\
\hline
\dfrac{1}{0.08}=12.5& 0.68^2=0.4624\\
\hline
\hline
\end{array}
The figure below shows that the dots are perfectly making a straight line, which means that the author's hypothesis is right.
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b) Now we need to find the constant $c$.
From the boxed formula above, the slope is given by
$${\rm Slope}=\dfrac{2c}{m}$$
Thus,
$$c=\dfrac{m\;{\rm Slope}}{2}$$
Plug the slope from the figure below and the mass from the given information.
$$c=\dfrac{50\times 10^{-3}\times 0.04}{2}=1\times 10^{-3}\;\rm kg.m^3/s^2$$
Therefore,
$$c=\color{red}{\bf 1\times 10^{-3}}\;\rm J.m$$