Posted by robert on Sunday, October 14, 2012 at 3:03am.
In Case A the mass of each block is 4.8 kg. In Case B the mass of block 1 (the block behind) is 9.6 kg, and the mass of block 2 is 4.8 kg. No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 9.4 N does act on block 2 in both cases and opposes the motion. For both Case A and Case B determine (a) the magnitude of the forces with which the blocks push against each other and (b) the magnitude of the acceleration of the blocks.

Physics  Elena, Sunday, October 14, 2012 at 11:38am
Assume that the blocks accelerate to the right with acceleration ‘a’
F1 is the force acting on the block 2 by block 1, and the same force (according to the 3 Newton's Law) acts on the block 1 by the block 2.
(A) m1=m2=4.8 kg
m1•a=F1
m2•a=F1F(fr).
Adding two equations
a• (m1+m2)= F(fr)
a=F(fr)/(m1+m2) = 9.4/9.6 =0.98 m/s²
Net force acting on the block 1
F1=  m1•a=  4.8•0.98=  4.7 N
The net forces acting on block 2
F1 F(fr) = m2•a = 4.8•0.98=4.7 N
(B)
m1•a=F1
m2•a=F1F(fr)
Adding two equations
a• (m1+m2)= F(fr)
a=F(fr)/(m1+m2) = 9.4/9.6 =0.65 m/s²
Net force acting on the block 1
F1=m1•a=4.8•0.6.5=3.12 N
The net forces acting on block 2
F1 F(fr) = m2•a = 9.6•0. 65=6.24 N