Question

find the distance traveled by a body before coming to the rest ,if it be moved on the ground with the velocity of 50.4km/h given co-efficent to friction between the body and the ground is 0.2 and g=9.8m/s​

Answer 1

Answer:

Explanation:

It may help u.......

Answer 2

Answer:

given v=0,u=50.4km/h=14m/s mew= 0.2,g=9.8m/S2

We know,

mew=F/R

F=mew.R

=ma=mew.mg (R=mg)

a=mew.g

=0.2×9.8

=1.96

by using Newtons 3rd law of motion,

S=v2-u2/2a

=0-(14)2/2×1 .96

=196/3.92

=50m (Ans)