Acceleration of Forces Challenge Problem

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tdub96
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So I was given a challenge problem in physics class that reads...

Mass m1=40kg, is sitting on a frictionless floor while m2=10kg sits on top of it. Between each mass, the coefficient of static friction is .6, and the kinetic coefficient is .4. M2 is pulled by a force of 100N to the left. What are the accelerations of each mass?

I solved it and got answers that do make sense, I'm just looking for a second opinion as to whether I did it right or not. Here's my work...

F(kinetic)=.4(10kg)(9.8)=39.2N
F(static)=.6(10kg)(9.8)=58.8N
The force pulling mass 2 is of 100N, which is >58.8N of friction, so the mass will move.

m1=ƩF=ma
F(friction m2)=ma
-39.2N=40kg(a)
a=-.98m/s^2

m2=ƩF=ma
100N-39.2N=10kg(a)
A=-6.08m/s^2.

Since m1 is moving along with m2, I added the -.98 to the -6.08 to get the total acceleration of mass 2 to be -7.06m/s^2. Yet I feel like I must use static friction for something other than proving that the mass will move. However, it could be in there just to throw us students off. Did I do this correctly, or is there another route to take to a correct answer?

All help is appreciated!
-TW
 
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This should be posted in the Homework section as per PF rules. That way you can get the best help possible!
 
Your analysis is not quite right, even though the values for the two accelerations are OK.
The question here is not if m2 will move or not. The bottom surface being frictionless, any non-zero force will move the system of the two objects. The question is if they move together (same acceleration) or they have different accelerations. Here is where you need the static friction. If the analysis shows that they don't move together, then you use kinetic friction.

And regarding your last step, in what reference frame did you calculate a1?
 
I'm not quite sure I know what you mean by reference frame...

Mass 1 is moving because of the friction between it and m2, which is above it. The force of 100N is only applied to m2.