Question

A particle moves on a rough horizontal ground with some initial velocity say $v_0$. If $3/4$th of its kinetic energy is lost in friction in time $t_0$, then coefficient of friction between the particles 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

Answer: (A)

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

Given $\frac{3}{4}$ of kinetic energy is lost due to friction

Thus velocity of the object will reduce to $\frac{v}{2}$ under the action of friction of $\mu$g

Now at time $t_0$ coefficient of friction $\mu$ can be written as

$\frac{v}{2}=v-\mu g{t}_0-\mu =\frac{-v}{2g{t}_0}\mu =\frac{v}{2g{t}_0}$