A block $A$ of mass $4\, kg$ is placed on another block $B$ of mass $5\, kg$, and the block $B$ rests on a smooth horizontal table. If the minimum force that can be applied on $A$ so that both the blocks move together is $12\, N$, the maximum force that can be applied to $B$ for the blocks to move together will be $..... N$
JEE MAIN 2014, Difficult
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Minimum force on A
$=12 N$
So, both block to move together
$12-\mu \times 4 \times 10=40 \times \frac{12}{9}$
$\Rightarrow \mu=\frac{1}{6}$
Now, maximum force F applied on lower block so, as to move together
$F-\mu 4 \times 10=5 \times\left(\frac{\mu \times 4 \times 10}{4}\right)$
$\Rightarrow F=\mu \times 10(5+4)$
$=\frac{1}{6} \times 10 \times 9=15 N$
$=12 N$
So, both block to move together
$12-\mu \times 4 \times 10=40 \times \frac{12}{9}$
$\Rightarrow \mu=\frac{1}{6}$
Now, maximum force F applied on lower block so, as to move together
$F-\mu 4 \times 10=5 \times\left(\frac{\mu \times 4 \times 10}{4}\right)$
$\Rightarrow F=\mu \times 10(5+4)$
$=\frac{1}{6} \times 10 \times 9=15 N$
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