Saturday

April 19, 2014

April 19, 2014

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

**Related Questions**

physics - A 5-kg box is released on a 30 degree incline and accelerates down the...

Physics - A 15.0 kg box is released on a 32 degree incline and accelerates down ...

Physics - A 13.0 kg box is released on a 33 degree incline and accelerates down ...

Physics - A 24 kg box is released on a 21 degree incline and accelerates down ...

Physics - A 13.0 -kg box is released on a 33 degree incline and accelerates down...

physics - A 23.0 kg box is released on a 39.0° incline and accelerates down the ...

Physics - A 13.0 -kg box is released on a 33 degree incline and accelerates down...

11th grade - A 16.0 kg box is released on a 39.0° incline and accelerates down ...

Accelerated Physics - An 18.0 kg box is released on a 35.0° incline and ...

AP Physics - An 18.0 kg box is released on a 35.0° incline and accelerates down ...