Interpreting result of a toboggan problem (special cases) 
Assuming there is friction contrary to the question above,
I derived the equation as in my textbook:
$$a_x=g(\sin\alpha-\mu_{k}\cos\alpha)$$
Testing special cases, if $\alpha=90°$,$a_x=g$, just as we would expect in a free fall. However, letting $\alpha=0°$, we get
$$a_x=-\mu_{k}g$$
Shouldn't $a_x=0$ as on a horizontal surface?
 A: At critical angle :
$$ \alpha_{_{th}} = 2 \arctan \left( {\sqrt{\frac {1}{\mu_{k}^2} + 1} - \frac 1\mu_{k}} \right) $$
and below, toboggan will stop moving due to kinetic friction force will overcome gravity speedup. This will make $a_x = 0$, so afterwards involving kinetic friction term will be meaningless, because body will be at rest.
A: The friction force $F_nμ_k$ only applies while the body is moving, and this force's direction is opposite to the movement direction.
In your formula $a_x=g(sin\alpha-\mu_{k}cos\alpha)$, you effectively subtract the friction force from the gravitational acceleration force, assuming that there is movement in positive x direction.
So, $a_x=-\mu_{k}g$ is perfectly valid in case you started with some positive speed. Then the friction will continuously slow down the toboggan to a halt.
After that, the friction force reduces itself to the amount necessary to compensate the gravitational acceleration force.
A: Yes but in the in x-direction the value for $\hat {g}$ disappears because g is also a vector with no x-component making $a_x =0$.
