Set up your Free-Body Diagram. I am going to assume the incline is down and to the right. For a coordinate system I am going to say that +x is down the incline and +y is perpendicular to and coming out of the plane of the incline.Originally Posted byRefujoi

There are two forces in the FBD: the weight, w, acting straight downward, and the normal force, N, acting in the +y direction.

The only force having a component in the x direction is the weight, so:

where is the angle of incline. (Note that it isnotcosine!)

Since w = mg we may solve this equation for a and we get

.

(Note that the acceleration is less than g as it should be.)

-Dan