Question

A particle moves on a rough horizontal ground with some initial velocity $v_0$. If ${\left(\frac{3}{4}\right)}^{th}$ of its kinetic energy is lost due to friction between the particle and the ground in time interval $t_0$, then find the coefficient of kinetic friction between the particle and ground. (Assume only kinetic friction is acting on the particle)

• 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 $\frac{v_0}{2g{t}_0}$

${\frac{3}{4}}^{th}$of the kinetic energy is lost. Hence final Kinetic energy is $\frac{1}{4}th$of initial.
$\frac{m}{2}{v}^2=\frac{m}{2}\frac{v_0^2}{4}$
$\therefore v=\frac{v_0}{2}$
using kinematics equation
$v=v_0-a{t}_0=v_0-{\mu }_{k}g{t}_0$
$\therefore v=\frac{v_0}{2}=v_0-{\mu }_{k}g{t}_0$
here $a={\mu }_{k}g$
${\mu }_k=\frac{v_0}{2g{t}_0}$