Answer :

Mass of the block on the table, m1 = 4 kg


Mass of the block hanging from the table, m2 = 1 kg


Acceleration of the system = a


Initially when the system was at rest, speed u = 0 m/s


Speed at a later time, v = 0.3 m/s


Distance travelled, s = 1 m


From equation of motion, 2as = v2-u2


2×a×1 = 0.3×0.3 - 0


a = 0.045 m/s2


For the 1 kg block hanging from the table, Let the tension in the string be 2T


m2a = m2g – 2T


2T = 1×(0.045 + 9.8)


2T = 9.845


T = 4.9225 N


For the 4kg block on the table, Let the force of friction be F and tension in the string will be T


T – F = 2m1a


If coefficient of friction is μ, F = μm1g


T – μm1g = 2m1a


4×μ×9.8 = 4.9225 - 2×4×0.045


39.2μ = 4.5625


μ = 0.116 ≈ 0.12


Coefficient of friction between the block and the table = 0.12


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