Answer
$0.165$
Work Step by Step
First of all, we need to draw a pictorial sketch of Sam's motion, and then we need to draw a force diagram of him.
From the figure below, we can see that the net force exerted on Sam in the $y$-direction is zero.
$$\sum F_y=0$$
Hence, from the last figure below,
$$F_n-mg\cos10^\circ=0$$
$$F_n=mg\cos10^\circ\tag 1$$
The net force exerted on him in the $y$-direction is given by
$$\sum F_x=ma_x$$
Hence, from the last figure below,
$$ mg\sin10^\circ+F_{\rm Thrust}-f_k=ma_x$$
we know that $f_k=\mu_k F_n$, and from (1), $f_k=\mu_kmg\cos10^\circ$
$$ mg\sin10^\circ+F_{\rm Thrust}-\mu_kmg\cos10^\circ=ma_x$$
Solving for $\mu_k$
$$ mg\sin10^\circ+F_{\rm Thrust} -ma_x =\mu_kmg\cos10^\circ $$
$$\mu_k=\dfrac{ mg\sin10^\circ+F_{\rm Thrust} -ma_x }{mg\cos10^\circ }\tag 2$$
Now we need to find $a_x$ and to do that we need to find the distance traveled in the $x$-diection first.
From the geometry of the first figure below, we can see that
$$\sin 10^\circ=\dfrac{50}{\Delta x}$$
Thus,
$$\Delta x=\dfrac{50}{\sin 10^\circ}=\bf 288\;\rm m$$
$$v_{1x}^2=\overbrace{v_{ix}^2}^{0}+2a_x\Delta x$$
Therefore,
$$a_x=\dfrac{v_{1x}^2}{2\Delta x}=\dfrac{40^2}{2\cdot 288}=\bf 2.78\;\rm m/s^2$$
Now we need to plug the acceleration magnitude into (2) and the rest known as well.
$$\mu_k=\dfrac{ (75\cdot 9.8\cdot \sin10^\circ)+200 -(75\cdot 2.78)}{75\cdot 9.8\cos10^\circ }=\color{red}{\bf 0.165} $$
