• # question_answer A particle moves on a rough horizontal ground with some initial velocity say ${{v}_{0}}.$ If 3/4th of its kinetic energy is lost in friction in time ${{t}_{0}}.$ Then coefficient of friction between the particle and the ground is - A) $\frac{{{v}_{0}}}{2g{{t}_{0}}}$         B) $\frac{{{v}_{0}}}{4g{{t}_{0}}}$ C) $\frac{3{{v}_{0}}}{4g{{t}_{0}}}$                   D) $\frac{{{v}_{0}}}{g{{t}_{0}}}$

[A] $\frac{3}{4}th$ energy is lost i.e.,$\frac{1}{4}th$ kinetic energy is left. Hence, its velocity becomes $\frac{{{v}_{0}}}{2}$ under a retardation of $\mu g$ in time to. $\therefore \,\,\frac{{{v}_{0}}}{2}={{v}_{0}}-\mu g\,\,{{t}_{0}}$ or $\mu g\,\,{{t}_{0}}=\frac{{{v}_{0}}}{2}$ or $\mu =\frac{{{v}_{0}}}{2g{{t}_{0}}}$