Q. A marble block of mass 2 kg lying on ice when given a velocity of $6\ {ms}^{- 1}$ is stopped by friction in 10 s. Then the coefficient of friction is
A) 0.02
B) 0.03
C) 0.06
D) 0.01
Explanation
(c) Let coefficient of friction is$\ \mu$, and then retardation will be $\mu$ g.
From equation of motion, $v = u + at$
$\Longrightarrow \ \ \ \ 0 = 6 - \mu\ g \times 10$
$\Longrightarrow \ \ \ \ \mu = \frac{6}{100} = 0.06$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.