A block of mass $5\,kg$ is placed at rest on a table of rough surface. Now, if a force of $30\,N$ is applied in the direction parallel to surface of the table, the block slides through a distance of $50\,m$ in an interval of time $10\,s$. Coefficient of kinetic friction is (given, $g =10\,ms ^{-2}$)
JEE MAIN 2023, Medium
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$S=u t+\frac{1}{2} a t^2$
$50=0+\frac{1}{2} \times a \times 100$
$a=1 m / s ^2$
$F-\mu m g=m a$
$30-\mu \times 50=5 \times 1$
$50 \mu=25$
$\mu=\frac{1}{2}$
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