Question

A ring of mass $m$ sliding on a smooth surface with velocity $v_0$ enters rough surface with coefficient of kinetic energy ${\mu }_k$, then

• A. The linear distance moved by centre of mass before the riong starts pure rolling is $\frac{3v_0^2}{8{\mu }_{k}g}$
• B. The gain in kinetic energy is $\frac{m{v}_0^2}{8}$
• C. The loss in kinetic energy is $\frac{m{v}_0^2}{4}$
• D. The work done by friction force is $-\frac{3{mv}_0^2}{8}$

Answer 1

Answer: (C)

The loss in kinetic energy is $\frac{m{v}_0^2}{4}$

$\begin{array}{c}Whenpurerollingstarts\\ m{v}_{0}r=mvr+m{r}^2\frac{v}{r}\\ \Rightarrow m{v}_{0}r=mvr+mvr\\ \Rightarrow v=\frac{v_0}{2}\\ K.E_i=\frac{m{v_0}^2}{2}\\ K.E_f=\frac{m{v_0}^2}{4}\\ \therefore loss=\frac{m{v_0}^2}{4}\end{array}$

Hence,

option $\left(C\right)$ is correct answer.