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Q.

In the arrangement shown in figure  mA=4.0 kg and mB=1.0 kg. The system is released from rest and block B is found to have a speed 0.3 m/s after it has descended through a distance of 1m. Then the coefficient of friction between the block and the table is x. Then find 9x [Neglect friction elsewhere. (Take g = 10 m/s2 ).]

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Detailed Solution

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From constraint relations, we can see that
 vA=2vB
Therefore,  vA=2(0.3)=0.6 m/s
as vB=0.3 m/s  (given)
Applying  Wnc=ΔU+ΔK
We get
μ(4.0)(10)(2)=(1)(10)(1)+12(4)(0.6)2  +12(1)(0.3)2
Or   80μ=10+0.72+0.045
Or    80μ=9.235 ⇒    μ=0.115

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