Newtonian mechanics, incline, acceleration FUN

[neesh]

"Two boxes, m1=1.0 kg witha coefficient of kinetic friction of 0.10 amd m2=2.0 kg with a coefficient of 0.20, are placed on a plane inclined at 30 degrees. (a) What acceleration does each box experience? (b) If a taut string is connecting the boxes with m2 farther down the slope, what is the acceleration of each box?"

So, I have part (a) (I believe!):
x direction: m1: m1gsin30 -Ffr=m1a1 (same for m2, but sub m2 for m1)
y direction: Fn = m1gsin30

For m1, migsin30=(muk)Fn = m1a1, you place in the appropriate values from the question and solve for a1= 4.1 m/s2
You go through the similar steps for m2 and solve for a2= 3.2 m/s2

But I'm a little confused on how exactly to explain part (b):
I think that the accelerations would be the same, but we were taught that whenever boxes are connected by a taut string, the acceleration for the boxes is the same, and it is treated as a system. But, since m1 is in back and has a faster acceleration, wouldn't it go faster so that the cord is no longer taut and the boxes are essentially not connected? Or am I thinking about it wrong and it should be treated as a system?
thanks in advance!

 

[andrevdh]

If you view the two masses as a system the tension in the string cancels out since they are the same magnitude but in opposite directions and therefore constitute an internal force.