how do you find the net force when the velocity is not constance meaning that acceleration is not zero

a= .3 m/s2 m=15kg

Force is equal to mass times acceleration. F=ma. So if this is all that it gives you then the object has a force of .3*15 which is 4.5N. Your question is strange because you can only have a force when the velocity is not constant or in other words, when there is an acceleration.

15kg box is released on a 32˚ incline and accelerates down the incline at .3 m/s2 find the friction force impeding its motion. What is the coefficient of kinetic friction?

There is an acceleration! 3 m/s2

I knew there was more to it. Kinetic friction equals the normal force times the coefficient.(f=un). So first find the normal force. Since it is on a slope of 32, the force of gravity is not acting on it straight down, but at mgcos32=n=122.6. The force of friction is in the x direction. F(x)=mgsin32-friction. From the info given, the 15kg block is accelerating at .3m/s^2 which means the F(x) is equal to 4.5N. From this you have 4.5=81.1-friction. Friction=76.6. So then you go back to friction=normal(coefficient of friction), plug in the numbers, 76.6=122.6(u), u=.625(THESE CALCULATIONS WERE DONE IN RADIANS NOT DEGREES I'M SORRY FOR THE CONFUSION. THE THEORY IS CORRECT BUT THE NUMBERS ARE NOT. I POSTED THE STEPS TO SOLVE THIS TWO POSTS UP!)

I wonder you are taking the value of g as how much... I'm using 10 m/s^2 and 9.8 m/s^2 and am getting values a little different from yours... =/

Well lets look at the calculations without values. I wish I knew how to put all the correct symbols in! Fk=kinetic friction u=coefficent of kinetic friction n=normal force Fx=forces in the x direction Fy=forces in the y direction and a=angle. Fk=u(n). Fy=n=mgcos(a), Fx=mgsin(a)-Fk. It gives you the value of Fx because it tells you the objects mass and acceleration in the x direction. So solve for Fk. Fk=mgsin(a)-Fx. Then solve for u. u=Fk/n.

Try having a look at the mimetex tutorial for LaTeX for the forum :)

https://www.askmehelpdesk.com/math-s...las-50415.html

Oh thanks that's helpful! And thanks for checking my work, though it is a bit embarrassing to make such a simple mistake. I keep having to change it from radians to degrees for different classes.