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January 29, 2015

January 29, 2015

Posted by **Jake** on Sunday, December 20, 2009 at 2:12pm.

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**, Sunday, December 20, 2009 at 2:14pmthose are subscripts by the way

F_fr_k

is the force of kinetic friction

and the all capital letters are greek leters

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