Answer :


From the free body diagram, we get,
R1 + ma − mg = 0
R1 = m (g − a)
= mg − ma …(i)
Now,
F − T − μR1 = 0 and
T − μR1 = 0
F − [μ (mg − ma)] − μ(mg − ma) = 0
F − μ mg − μma − μmg + μma = 0
F = 2 μmg − 2 μma
= 2 μm (g − a)


(b) Let the acceleration of the blocks be a1.


R1 = mg − ma ….(i)
2F −T −
μR1 = ma1 …(ii)
Now,
T =
μR1 + Ma1
=
μmg − μma + Ma1


Substituting the value of F and T in equation (ii), we get,
2[2μm(g − a)] − (μmg − μma + Ma1) − μmg + μma = ma1
4μmg − 4μma − 2μmg + 2μma= ma1 + Ma1



Thus, both the blocks move with same acceleration a1 but in opposite directions.


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