Components of gravity in an inclined plane

The reason the block accelerates down the inclined plane is that the resultant of $N$ and $mg$ acts down the incline. However, why is the force causing the block to accelerate down the incline often cited as the component of gravity $mg\sin\theta$ parallel to the plane. I'm not quite sure how this works.

Since acceleration is a vector you can decompose it in the coordinate system you find convenient. If you define a cartesian coordinate system whose axis are along the normal to the plane and the plane itself you see there is a component of the acceleration $g\sin\theta$ along the plane. This is why the block accelerate in this direction. Notice that along the normal axis, $N$ cancels $mg\cos\theta$ and the block does not leave the plane.

The normal force is not playing a role in this case because the force is perpendicular to the moving direction of the box. By "not playing a role" I mean, during this motion, $\ F_N$ is only be used to balance out $\ mgcos\theta$ so that the object won't be able to move "into the inclined plane".

To find the force causing the block to accelerate down, we need to find the $\ F_{net}$ in the moving direction i.e. down along the plane. The $\ F_{net}$ in this case would be $\ mgsin\theta$ (if $\mu_k$ is negligible).

Hope it helps :)

There's a physical reason why the block slides down along the slope of the inclined plane. If you perform this experiment, and if frictional force is small enough, you will observe that the block slides down along the slope of the plane. It will never move in a direction perpendicular to the slope.

For it to never move in the perpendicular direction, we know from Newton's first law that the forces must be balanced, i.e. $\Sigma F_\perp = 0$

For the block to slide downwards along the plane, there must exist a parallel component of force along the slope in the direction of sliding (again from the first law), i.e. $\Sigma F_\parallel = n$

Now, the weight of the block is acting vertically downward due to gravity. Using trigonometry, we can resolve the forces into their perpendicular and parallel components as shown below:

Now, for our perpendicular condition to be satisfied, $R = W\cos\theta$ and for the second condition, $W\sin\theta - f_k = n$.

Therefore, we see that the block slides down along the slope and satisfies Newton's laws of motion.