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A block A of mass $4 k g$ is placed on another block of mass $5 k g$ and the block B rests on a smooth horizontal table. For sliding block A on B, a horizontal force of $12 N$ is required to be applied on it. How much maximum force can be applied on B so that both A and B move together ? Also find out acceleration produced by this force.

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$24 N$

$27 N$

$28 N$

$30 N$

answer is A.

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Detailed Solution

As $12 N$ force is needed to slide the block A over B, so limiting friction on block A $= 12 N$ .

The acceleration provided to block A by this frictional force,

$a = \frac{12}{4} = 3 m / s^{2}$

Now force required to move block together with block A ,

$F = (4 + 5) \times a = 9 \times 3 = 27 N .$

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