Question

A particle moves on a rough horizontal ground with some initial velocity say $v_0$. If ${\frac{3}{4}}^{th}$ of its kinetic energy is lost overcoming 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. $\frac{3v_0}{4g{t}_0}$
• D. $\frac{v_0}{g{t}_0}$

Answer 1

The correct option is A $\frac{v_0}{2g{t}_0}$

Initial kinetic energy $=\frac{1}{2}m{v}_0^2$
$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_0^2-\frac{3}{4}\times \frac{1}{2}m{v}_0^2$
$v=\frac{v_0}{2}$
${\frac{3}{4}}^{th}$ 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 $t_0$.
$\therefore \frac{v_0}{2}=v_0-\mu g{t}_0$
or $\mu g{t}_0=\frac{v_0}{2}$ or $\mu =\frac{v_0}{2g{t}_0}$