A marble block of mass 2 kg lying on ice when given a velocity of 6ms^-1 is stopped by friction in 10s. Then the coefficient of friction is $\left(\mathrm{g}=10{\text{ms}}^{-2}\right)$

Solution: Given that
Mass of marble block m = 2kg
Initial velocity u = 6m/s
Final velocity v = 0
$\begin{array}{l}\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{0-6}{10}=-0.6\mathrm{m}/{\mathrm{s}}^2\\ \mathrm{f}=\mathrm{F}=\mathrm{ma}=2\times 0.6=12\mathrm{N}\\ \text{ But }\mathrm{f}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{N}\\ {\mathrm{\mu }}_{\mathrm{k}}=\frac{\mathrm{f}}{\mathrm{N}}=\frac{1.2}{2\times 10}=\frac{0.6}{10}=0.06\end{array}$