Question

A particle moves on a rough horizontal ground with some initial velocity say $v_0$. If $\frac{3}{4}$ of its kinetic energy is lost due to 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 K.E

$K.E_i=\frac{m{v}_0^2}{2}$

Energy loss in friction is the work done by the friction

Workdone by friction

$W=\left(\frac{3}{4}\right)\frac{m{v}_0^2}{2}=\frac{3m{v}_0^2}{8}$

Speed be $v$

$\frac{m{v}^2}{2}=\frac{m{v}_0^2}{8}$

$v=\frac{v_0}{2}$

Let the coefficient be $\mu$

$a=-\mu g$

$-\mu g=\frac{\frac{v_0}{2}-v_0}{t_0}$

$\mu g=\frac{\frac{v_0}{2}}{t_0}$

$\mu =\frac{v_0}{2g{t}_0}$