The coefficient of static friction between a wooden block of mass $0.5 \mathrm{~kg}$ 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 $\mathrm{N}\left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$


Given : $\mu_{\mathrm{s}}=0.2$

$m=0.5 \mathrm{~kg}$

$\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$

we know that

$f_{s}=\mu N$ and $\ldots(1)$

To keep the block adhere to the wall

here $\mathrm{N}=\mathrm{F} \quad \ldots(2)$

$f_{s}=m g \ldots(3)$

from equation $(1),(2)$, and $(3)$, we get

$\Rightarrow m g=\mu F$

$\Rightarrow F=\frac{m g}{\mu} \Rightarrow F=\frac{0.5 \times 10}{0.2}$

$F=25 \mathrm{~N}$

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