Applying Newton's second law in the vertical and horizontal directions:

$$ \Sigma F_y = N - mg - F\sin\theta = 0 \implies N = mg + F\sin\theta $$ $$ \Sigma F_x = F\cos\theta - f_k = ma $$

The kinetic friction force is $f_k = \mu_k N$. Substituting $N$ and $f_k$ into the x-equation gives:

$$ F\cos\theta - \mu_k (mg + F\sin\theta) = ma $$

Solving for acceleration $a$ as a function of $\mu_k$:

$$ a(\mu_k) = \left(\frac{F\cos\theta}{m}\right) - \mu_k \left(g + \frac{F\sin\theta}{m}\right) $$

This is a linear equation of the form $a = A - B\mu_k$. From the graph, we identify the y-intercept and the slope. The y-intercept occurs at $\mu_k=0$, which is given as $a=a_1$.

$$ A = \frac{F\cos\theta}{m} = a_1 $$

The slope can be found using the points $(0, a_1)$ and $(\mu_{k2}, 0)$.

$$ \text{Slope} = -B = \frac{0 - a_1}{\mu_{k2} - 0} = -\frac{a_1}{\mu_{k2}} $$ $$ B = g + \frac{F\sin\theta}{m} = \frac{a_1}{\mu_{k2}} $$

We now have a system of two equations:

1. $\frac{F}{m}\cos\theta = a_1$
2. $\frac{F}{m}\sin\theta = \frac{a_1}{\mu_{k2}} - g$

Dividing equation (2) by equation (1) eliminates the term $F/m$:

$$ \frac{\frac{F}{m}\sin\theta}{\frac{F}{m}\cos\theta} = \tan\theta = \frac{\frac{a_1}{\mu_{k2}} - g}{a_1} $$ $$ \tan\theta = \frac{1}{\mu_{k2}} - \frac{g}{a_1} $$