2 blocks, a light string and kinetic friction

[Trista]

So, there are two boxes of fruit on a frictionless horizontal surface and are connected by a light string. m1=10kg, m2=20kg. A force of 50N is applied to the 20-kg box.
There are two questions to answer, I already answered a:
Determine the acceleration of each box, and the tension in the string:
a= F/m1+m2 = 50N/30kg = 1.7 m/s^2 and T=m1a ... T=10kg(1.7 m/s) = 17 N.
This is the one I can't figure out, b:
Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.10. So, mu k = .10
I figured that the equation to use is Newtons 2nd Law, sum Fx=T-fk=m1a1. But I can't get it to work. I actually know the answer, but I don't know how to get there. Any help?

 

[nevetsnosaj]

Remember that the force of friction is always hold your force. So the sum of the forces will be:

Fp - Ffriction = ma

Find the friction and plug the number in.

Then you can find our answer

 

[Trista]

Thank You

got it. I appreciate your help.