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Grade: 12th pass
        
A body pushed giving initial velocity of 5m/s.and come to rest travelling 4.9m.find coefficient of fraction
19 days ago

Answers : (2)

Soumya
33 Points
							
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
 
18 days ago
Soumya
33 Points
							
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
 
 
 
 
18 days ago
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