Q 11-05-035NEETJEE MainEasy
A $2$ kg block moving at $6$ m/s on a rough horizontal floor comes to rest after sliding $4.5$ m. The coefficient of kinetic friction is ($g = 10\ \text{m/s}^2$)
Answer: (B) $0.4$
Work-energy theorem: friction removes all the kinetic energy.
$$\mu mg\,d = \frac{1}{2}mv^2 \;\Rightarrow\; \mu = \frac{v^2}{2gd} = \frac{36}{90} = 0.4$$
Solution by Sreeraj P, M.Sc Physics