What is the coefficient of kinetic friction on a 30 degree ramp?

[drummerjason]

Homework Statement

A box slides down a 30.0 degree ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box and the ramp.

Homework Equations

The Attempt at a Solution

Ok so what i did was different from what my teacher has, but I can't explain it.
(the answer is supposed to be .436, which is what i got)

1) F=m(9.81)(sin30.0) ...divide by m... a=4.9m/s^2 that would be the ramp with no friction right?

2) so then... F_k=(a_withoutfriction-a_with friction)m --> F_k (4.9-1.2)m= (3.7m/s^2)(m) and that would be the force of kinetic friction (I think).

3) $$\mu$$=F_k/F_n --> (3.7)(m)/(9.81)(cos30)(m)= .436

Ok so i got the right answer but I can't explain it. Just for starters, why in step 1, did it have to be sin30? would it be cos30 to find the accleration in the x direction?

 

[fizzynoob]

why in step 1, did it have to be sin30?

If you translate your coordinate system, where the X axis is on the incline you will get mgsin(t) for the forces of gravity in the X direction.

[PLAIN]http://img524.imageshack.us/img524/4726/gradianthw.png

 

[drummerjason]

Ok me a two of my freinds got together tonight and figured out most the theorizing and problems with the problem. And yeah, that diagram was basically what we came up with.

But I really appreciate you taking your time to respond to my question. Thank you!