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

```							is the mass of the body is given ?v2=u2+2asu=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/s2by 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
```
11 months ago
```							from v2=u2+2as 0=52+(-2a×4.9) a =2.55m/s2Now fk=ma. It is the only force acting in x direction.and normal reaction (R) =mgalso fk=u*R, u*=fk/Rputting values- u*=ma/mgu*=2.55/10coff of friction (u*) =0.25
```
11 months ago
```							Find acceleration using v2=u2+2asThen 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
```
8 months ago
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