A box is given an initial velocity of 5 m/s. It slides across a rough surface coming to rest in 2 m.?

Question

The coefficient of kinetic friction is

Answer 1

a=(v2-v1)/t
v1=5m/s
v2=0
a=-v1/t

S=v1 t + 0.5at^2
S=v1 t - 0.5( v1/t)t^2
S=v1 t - 0.5 v1 t^
t= s/(v1 - .5 v1)=2s/v1=
t=2 x 2/ 5 = 0.8 sec

from above
a= 5/(.8)=6.25 m/sec^2

The force of motion is being canceled by force of friction.
F=ma=uN
N=mg
then
ma=umg
u=a/g

u=6.25/9.81=0.64