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
U is coefficient of friction
V
()
02
2ug ug
02
ug ug
U
22
()
2M 9' ug
None of the above​

Answer 1

Answer:

Force of friction=μN=μmg

Therefore retardation = μmg/m=μg

From,

$v^2 =u^2 +2as$

or,

$S=\frac{v^2}{2g}$

from v=u+at

or, t=v/μg