Question

A block of mass $M$ is moving with a velocity $v$ on straight surface. What is the shortest distance and shortest time in which the block can be stopped if $\mu$ is coefficient of friction?

• A. $\frac{v^2}{2\mu g},\frac{v}{\mu g}$
• B. $\frac{v^2}{\mu g},\frac{v}{\mu g}$
• C. $\frac{v^2}{2Mg},\frac{v}{\mu g}$
• D. none of the above

Answer 1

The correct option is A $\frac{v^2}{2\mu g},\frac{v}{\mu g}$

Force of friction opposes the motion.
Force of friction $=\mu N=\mu mg$
Therefore retardation = $\mu mg/m=\mu g$
From $v^2=u^2+2as$
or
$s=\frac{v^2}{2\mu g}$
from $v=u+at$
or $t=\frac{v}{\mu g}$