The coefficient of static friction between a wooden block of mass $0.5\, kg$ and a vertical rough wall is $0.2$ The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be $N$ $\left[ g =10\, ms ^{-2}\right]$
JEE MAIN 2021, Medium
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$F.B.D.$ of the block is shown in the diagram
Since block is at rest therefore
$fr - mg =0$ $........(1)$
$F- N =0$ $..........(2)$
$fr \leq \mu N$
In limiting case
$fr =\mu N =\mu F$ $........(3)$
Using eq. $(1)$ and $(3)$
$\therefore \mu F = mg$
$\Rightarrow F=\frac{0.5 \times 10}{0.2}=25 N$
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