Question

A particle moves on a rough horizontal ground with initial velocity $v_0$. If half of its velocity is decreased 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 $\frac{v_0}{2g{t}_0}$

Given: a particle moves on a rough horizontal ground with initial velocity $v_0$. If half of its velocity is decreased due to friction in time $t_0$

To find the coefficient of friction between the particle and the ground

Solution:

From the equation

$v=u+at$

We can calculate acceleration

$\frac{v_0}{2}=v_0+a{t}_0⟹a=-\frac{v_0}{2t_0}$

Friction force will be

$F=ma⟹F=m\times (-\frac{v_0}{2t_0})⟹\mu mg=\frac{m{v}_0}{2t_0}⟹\mu =\frac{v_0}{2t_{0}g}$

is the coefficient of friction between the particle and the ground