What is the minimum coefficient of static friction μs for which the 1kg block will not slip relative to the lower block if the lower block is displaced by 50 mm to the right from equilibrium position and released ? Take g=10m/s2.

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0.346

0.628

0.525

0.428

answer is D.

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Detailed Solution

amax=ω2A and ω2=kM+m w.r. to 6 kg μs mg≥mω2A⇒μsmg≥mkM+mA⇒μs≥kAM+mg ∴μsmin=600×5×10−27×10=0.428

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