Question

A particle moves on rough horizontal ground with some initial velocity $v_0$. if 3/4 of its kinetic energy is lost in friction in time $t_0$ the coefficient of friction between the particle and ground.

• A. $\frac{v_0}{2g{t}_o}$
• B. $\frac{v_0}{4g{t}_0}$
• C. $\frac{3v_0}{4g{t}_0}$
• D. $\frac{v_0}{g{t}_0}$

Answer 1

Answer: (A)

$\frac{v_0}{2g{t}_o}$

Initial velocity=$v_0$

As 3/4 th kinetic energy lost so 1/4 kinetic energy remain after time '$t_0$'

So velocity after time '$t_0$'= $v_0/2$

retardation due to friction=$-\mu g$

apply first equation of motion

$v=u+at$

$v_0/2=v_0-\mu g{t}_0$

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