Kinetic friction problem

A 1.34-kg block slides across a rough surface such that it slows down with an acceleration of 1.19 m/s2. What is the coefficient of kinetic friction between the block and the surface?

I used F= M x a and kept getting the wrong answer so, help?

F = ma gets net force, not friction.
Fric = (coefficient)x(normal) is the equation you're looking for.

Force(vertical) = 0 as it does not move up or down, so normal = weight, so you to find the normal, you have to find the weight, which is
weight = massxgravity = 1.34 kg x 9.8 m/s2 = 13.132 = normal

Now that just leaves Fric, and for that you use F = ma, which in this case is Fric because the net force goes to the left, slowing the block down, which is 1.5946 (you already know).

So Fric / normal = coefficient = (1.5946) / (13.132) = 0.121 (3 sig figs).

Now, I'm just an AP Physics student, doing this stuff as well (and doing badly on the test so far), so chances are I didn't do it correctly. Is this what you got?

(1.34 x 9.8) = weight of 13.132N.
Force = (ma), = 1.34 x 1.19, = 1.5946N. friction.
Coefficient = (1.5946/13.132) = 0.121.

If you needed a friction force, you multiply the weight by the coefficient. So to find the coefficient, divide the friction by the weight.