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Grade 12th passGeneral Physics

A body pushed giving initial velocity of 5m/s.and come to rest travelling 4.9m.find coefficient of fraction

Profile image of Irfan khan
7 Years agoGrade 12th pass
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3 Answers

Profile image of Soumya
7 Years ago
is the mass of the body is given ?
v2=u2+2as
u=5 m/s and v=0 as the body comes at rest.
so 0=52+(-2a×4.9)
from here we get a =2.55m/s2
by newton’s 2nd law, in this case -
fk=m×a. you can get coff. of friction if you know the mass by putting a=2.55m/s2
 
Profile image of Soumya
7 Years ago
from v2=u2+2as
 0=52+(-2a×4.9)
 a =2.55m/s2
Now fk=ma. It is the only force acting in x direction.
and normal reaction (R) =mg
also fk=u*R, u*=fk/R
putting values- u*=ma/mg
u*=2.55/10
coff of friction (u*) =0.25
 
 
 
 
Profile image of Azim ahmed
7 Years ago
Find acceleration using v2=u2+2as
Then we know value of a and value of g (9.8)
Coefficient  pf friction = F/N
                                       =ma/mg
                                       =a/g=2.5/9.8= approx 0.25