Block on a rough plane Coefficient of Kinetic Friction

[boognish]

1. A 3kg block slides down a rough plane inclined at an angle of 27 degrees. If the acceleration of the block is 2.5 m/s/s find the coefficient of kinetic friction.



2. solving for components
Force Net = M x A
forces in the y direction: Fn - Fg cos 27 degrees = 0
forces in the x direction: M x G sin 27 degrees - Fk = M x A
Fk = coef-k x m x g cos 27 degrees



3. isolating Fk resolves to:
9.8 sin 27 degrees - 2.5 m/s = Fk
substituting for fk to resolve for coef of kinetic friction resolves to this:
9.8 sin 27 degrees - 2.5 m/s = mu-k x 3 kg x 9.8 cos 27 degrees
1.949 = mu-k x 26.196
mu-k = .074

 

[boognish]

is this correct..? I am wondering if all the masses should cancel out in this equation or if I have it set up correctly

 

[ap123]

Your work is section 2 is fine.

9.8 sin 27 degrees - 2.5 m/s = Fk

You've divided the left-hand side by m, but not the right-hand side.