Please Help

[Fusilli_Jerry89]

does a 1.2 kg box moving 2.0 m/s along a table with an acceleration of -3.0m/s/s apply any force in the x direction? If not, would you solve for mu by simply dividing 3.6(Fnet) by 11.76(9.8x1.2kg) to get 0.3 for mu? And is the box does have a force on the x-axis, how do you find it and how would you solve for mu in that case?

[kreil]

You need to stop posting this question over and over, that's really pretty ridiculous.

I don't even know what information you're trying to present here, but use the fact that

F_f=\mu N=\mu mg \implies \mu=\frac{F_f}{mg}

and

F_{net}=F_{applied}-F_f

to solve for mu.

[Fusilli_Jerry89]

I'm sorry for posting it so many times, the first time I posted it it post 3 times for some reason. The website went down and when I went back on there were three posts. I also notice that I had spelled it wrong and didn't know how to change the title so I just made a new post. Sorry about that.