A block of mass 5,kg is placed at rest on a table of rough surface. Now, if a force of 30,N is appli

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4-2.Friction

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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, $g =10\,ms ^{-2}$)

A

$0.60$

B

$0.75$

C

$0.50$

D

$0.25$

(JEE MAIN-2023)

Solution

$S=u t+\frac{1}{2} a t^2$

$50=0+\frac{1}{2} \times a \times 100$

$a=1 m / s ^2$

$F-\mu m g=m a$

$30-\mu \times 50=5 \times 1$

$50 \mu=25$

$\mu=\frac{1}{2}$

Standard 11

Physics

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