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

Forces acting on block is shown below

From the second equation of motion,

$$\begin{gathered} s=\frac{1}{2}a{t}^2 \\ \Rightarrow 50=\frac{1}{2}a\times {10}^2 \\ \Rightarrow a=1 \mathrm{m} {\mathrm{s}}^{-2} \end{gathered}$$

Applying Newton's second law in vertical direction,

$$\begin{gathered} R-mg=0 \\ \Rightarrow R=mg \end{gathered}$$

(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,

$$\begin{gathered} 30-\mu R=ma \\ \Rightarrow 30-\mu mg=ma \\ \Rightarrow 30-50\mu =5 \\ \Rightarrow \mu =\frac{1}{2}=0.5 \end{gathered}$$