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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