Question

A particle moves on a rough horizontal ground with some initial velocity $v_{0} .$ If $\frac{3}{4}$ th of its kinetic energy is lost due to friction in time $t_{0}$, the coefficient of friction between the particle and the ground is

• A. $\frac{v_{0}}{2\, g t_{0}}$
• B. $\frac{v_{0}}{4\, g t_{0}}$
• C. $\frac{3 v_{0}}{4\, g t_{0}}$
• D. $\frac{v_{0}}{g t_{0}}$

Answer 1

Answer: (A)

$KE$ lost is $\frac{3}{4}$ th, therefore, $KE$ left is $\frac{1}{4}$ th.
Hence, velocity of particle reduces from $v_{0}$ to $\frac{v_{0}}{2}=v_{0}-\mu g\, t_{0}$
or $\mu=\frac{v_{0}}{2 g t_{0}}$