Question 145 of 210

A block of mass $5kg$ is placed at rest on a table of rough surface. Now, if a force of $30N$ is applied in the direction parallel to surface of the table, the block slides through a distance of $50m$ in an interval of time $10s$. Coefficient of kinetic friction is (given, $g=10m{s}^{–2}$ ):

A$0.60$B$0.75$C$0.50$D$0.25$

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Answer: (C) $0.50$

Forces acting on block is shown below

From the second equation of motion,
$s=\dfrac{1}{2}a{t}^{2} \Rightarrow 50=\dfrac{1}{2}a\times {10}^{2} \Rightarrow a=1m{s}^{-2}$
Applying Newton's second law in vertical direction,
$R-mg=0 \Rightarrow R=mg$
(where $R$ is the magnitude of normal reaction on the block due to the surface)
If $\mu$ is the coefficient of kinetic friction between the block and the surface, applying Newton's second law in horizontal direction,
$30-\mu R=ma \Rightarrow 30-\mu mg=ma \Rightarrow 30-50\mu =5 \Rightarrow \mu =\dfrac{1}{2}=0.5$