A 3 kg block A is placed on the top of a 4 kg block B as shown. To make the block A slip on B, assuming frictionless table, a horizontal force of 9 N is to be applied to the top block. Find the minimum horizontal force (in N) that can be applied to lower block so that A slips on B.
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answer is 12.
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Detailed Solution
fmax=3μgWhen 3 kg and 4 kg blocks move together, common acceleration, a=94+3=97m/s2 For 4kg block, fmax=4a⇒μ3g=4×97⇒ μ=127×10=635If force F is applied common acceleration, a=F7m/s2 For 3kg block fmax=3μg=3×a⇒ 3635×10=3F7⇒F=12N