$$
textbf{underline{textit{Solution}}}
$$

For a function $f(x)=x$ or simply $y=x$, the variable $x$ is called the independent variable and the variable $f(x)$ or $y$ is called the dependent variable.

Which physically means that the variable $x$ is the variable under control in the experiment, i.e. the variable which is changed and the variable $y$ is the measured variable whose value depends on the value of $x$.

Thus plotting time on $x$-axis and speed on $y$-axis yields the following $textbf{graph}$.

$$
textbf{underline{textit{Solution}}}
$$

One first must find the best fit line that describes the relation between the two variables with the least error, in other words the line that best describes the relation between the mass load of the spring and the length of the spring.

From the figure is best fit line, we find that when the length of the spring is 15 cm corresponds to a specific value of spring mass load, examining the figure carefully, we find that the point in $x$-axis that corresponds to $y=15$ is some where between the 15th mark and the 20th mark.

Where this point is nearly close to the midway distance between the two mark, and since one division is equal to 5 gm, hence the value of the mass that corresponds to a spring of length 15 cm is

$$
15 + dfrac{1}{2} times 5 = 17.5 ~ rm{gm}
$$

Thus, the value of the mass of the spring load that corresponds to the spring length of 15 cm is about 17.5 gm.

$textbf{underline{textit{note:}}}$ this estimation is not precise as we can’t determine exactly the actual value of the mass load, thus our estimation is accompanied by uncertainty which is about half the smallest division, thus the estimated value of the spring mass load is

$$
fbox{$17.5 pm 2.5 ~ rm{gm}$}
$$

$textbf{underline{textit{Another approach}}}$

Another approach which is more precise, to find the corresponding mass load that would cause the spring to have a length of 15 cm, is simply to find the equation of the straight line of the best fit line, knowing that the equation of the straight line is given by

$$
y=ax+b
$$

If we examined the figure carefully, we find that it is given that the $y$-intercept is $b=13.7$, it remains to find the slope of the best fit line in order to find the equation of the best fit line.

The problem in finding the slope of the best fit line, is that we want to find two precise points so we can calculate the slope accurately and by precise we mean that we can tell the exact coordination of the point, if we check the points at $x=10$ and $x=30$ we find that these 2 points are the most precise points from which we can calculate the slope of the best fit line.

At point $x=10$ the value of $y$ is 14.5 and at point $x=30$ the value if the $y$ is 16.0, thus the slope is

$$
begin{align*}
a&= dfrac{16-14.5}{30-10}\
&= 0.075\
end{align*}
$$

$textbf{underline{textit{Substituting in equation of straight line}}}$Knowing that the slope of the line is $a=0.075$ and the $y$-intercept is $b=13.7$ thus the equation of the straight line is

$$
l = (0.075) m – 13.7 tag{1}
$$

And, since we want to find the mass of the spring load that would cause the spring to have a length of 15 cm, thus we substitute in equation (1) by the value of the length and solve for the mass $m$

$$
begin{align*}
15 &= (0.075) ~ m – 13.7\
1.3 &= (0.075) m\
m &= dfrac{1.3}{0.075}\
&= 17.33 ~ rm{gm}\
&= fbox{$17.3 ~ rm{gm}$}\
end{align*}
$$

$textbf{underline{textit{note:}}}$ When calculating the slope of the line, we calculate it from any points on the best fit lines, not the plotted points as this points may have progressed some error during measurement, while the best fit line is the line describing the relation between the mass of the spring load and the length of the length of the spring with the least error, thus we calculate the slope from the points that only lies on the best fit line.