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=F1-F(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=F1-F(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
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.
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