Q. A particle moves on a rough horizontal ground with some initial velocity $v_{0}.$ If $\frac{3}{4}$ th of its KE is lost in friction in time $t_{0}$, the coefficient of friction between the particle and the ground is
A) $\frac{v_{0}}{2gt_{0}}$
B) $\frac{v_{0}}{4gt_{0}}$
C) $\frac{3v_{0}}{4gt_{0}}$
D) $\frac{v_{0}}{gt_{0}}$
Explanation
(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 $\mu = \frac{v_{0}}{2gt_{0}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.