1. The problem statement, all variables and given/known data A 15.0 kg box is released on a 30 degree incline and accelerates down the incline at .30 m/s^2. Find the firctiopn force impeding its motion. What is the coefficient of kinetic friction? 2. Relevant equations simga F = ma Force of friction = MU F_N 3. The attempt at a solution SIGMA F_x = m a_x = F_g_x - F_fr_k = mg sin THETA - mg cos THETA MU therefore mg cos THETA MU = mg sin THETA - m a_x I divided through by mass g cos THETA MU = g sin THETA - a_x I do not see what is wrong with this solution. This gave me 4.6 N. I'm suppose to get 69 N. I know that if I muliply 4.6 N by 15 (the mass that I canceled out) I get 69 N. So apparently I don't know why I can not divide by the mass here and would like to know why... Surpisingly enough when I rearanged for Mu I did this and got the right answer according to the back of the book F_fr_k = 4.6 N = g cos THETA MU therefore MU = 4.6 N/(g cos THETA) this gave me the right answer of .54 THANKS! Those are subscripts by the way F_fr_k is the force of kinetic friction and the all capital letters are one varialbe and are greek letters thansk!