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Question

In the arrangement shown in figure, mA=4 kg and mB=1 kg. The system is released from rest and block B is found to have a speed of 0.3 m/s after it has descended through a distance of 1 m. Find the coefficient of friction between the block and the table. Neglect friction elsewhere. (Take g=10 m/s2).


A
μ=0.187
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B
μ=0.115
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C
μ=0.272
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D
μ=0.323
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Solution

The correct option is B μ=0.115
From constraint relations, we can say that
VA=2VB
Therefore, VA=2(0.3) m/s=0.6 m/s
Also if distance moved by B, SB=1 m
SA=2SB=2 m

On applying work-energy theorem
Wnoncons=ΔU+ΔK.E
where the only non conservative force acting on the system is
friction f=μmAg
We get: μmAgSA=mBgSB+12mAV2A+12mBV2B
μ(4)(10)(2)=(1)(10)(1)+12(4)(0.6)2+12(1)(0.3)2
(80μ)=(10)+(0.72)+(0.045)
80μ=9.235
μ=0.115

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