A 42.0 kg block of ice slide down a plane with an incline of 34 degree assuming friction is negligible, what is the acceleration of the block down the incline?
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 by Refujoi
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:
$\displaystyle \sum F_x = w sin \theta = ma$
where $\displaystyle \theta$ is the angle of incline. (Note that it is not cosine!)
Since w = mg we may solve this equation for a and we get
$\displaystyle a = g sin( \theta )$.
(Note that the acceleration is less than g as it should be.)
-Dan