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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 ....... $N$
$30$
$25$
$15$
$48$
Solution
$\begin{array}{l}
{\rm{Minimum}}\,{\rm{force}}\,{\rm{on}}\,{\rm{A}}\\
{\rm{ = 12}}\,{\rm{N}}\\
{\rm{So,}}\,{\rm{both}}\,{\rm{block}}\,{\rm{to}}\,{\rm{move}}\,{\rm{together}}\\
12 – \mu \times 4 \times 10 = 40 \times \frac{{12}}{9}\\
\Rightarrow \mu \, = \frac{1}{6}
\end{array}$
$\begin{array}{l}
Now,\,{\rm{maximum}}\,force\,F\,applied\,on\\
lower\,block\,so,\,as\,to\,move\,together\\
F – \mu \,4 \times 10 = 5 \times \left( {\frac{{\mu \times 4 \times 10}}{4}} \right)\\
\Rightarrow \,F = \mu \times 10\left( {5 + 4} \right)\\
= \frac{1}{6} \times 10 \times 9 = 15\,N
\end{array}$
