Question

A block accelerates at 3.3 m/s^2 down a plane inclined at an angle of 26 degrees.Find micro(k) between the block and the inclined plane. The acceleration of gravity is 9.81 m/s^2.

This is my work but i am not sure if it is correct:

sum of forces on x axis= -kinetic friction + mass(gravity) sin theta= mass ( acceleration)x

kinetic friction=fk

sum of forces on y axis=normal force - mg cos theta=ma y

ma y=0

-fk=coefficient of kinetic friction (normal force)

-coefficient(n) + mg sin theta =ma x

-l(-coefficientkf)=(ma x -mg sin theta/n)-1

coeffictientkf=-ma x + mg sin theta/-n

n= mg cos theta

coeffictientkf = -ma x + mgsin theta/-(mg cos theta)

-a x + g sin theta/-g cos theta <-- the masses canceled out

-3.3+9.81 sin 26/ -9.81 cos 26

= -3.694 <-- that cannot be a negative number right?

Answer 1

It looks like you have a good idea of how to solve this; unfortunately, your notation makes it very difficult to follow.
Once you find the normal force (n), you will be working only along the x-axis, so just drop any x-subscripts, so the acceleration is simply a.
Let's let u = the coefficient of kinetic friction, and let T = the angle theta.
I get the equation,
m g sin T - u m g cos T = m a
Simplifying:
u g cos T = g sin T - a
Solve for u.
It looks like you somehow made a "g" -9.81 instead of +9.81.
Remember, g -- as an acceleration -- is a vector; it has direction as well as magnitude. I suspect you introduced a minus sign in an early equation (thus indicating the direction of the vector) and then later used a negative value for g, but magnitudes are always positive.