Posted by Jake on .
thanks for answering the question
I got another one...
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?
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!

Physics 
Jake,
those are subscripts by the way
F_fr_k
is the force of kinetic friction
and the all capital letters are greek leters