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  4. 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 ):

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Q. 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 MainJEE Main 2023Laws of Motion

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}$