Asked by Sudhanshu
                a body rolled on a surface with speed of 10m/s comes to rest after covering distance of 50 m find cofficient of friction 
            
            
        Answers
                    Answered by
            Kunal
            
    v^2 - u^2 = 2as
0-100=2*a*50
-100/100=a
a=-1 m/s^2
Negative sign shows reduction.
let n be the coeff. of friction,
n*mg=m*a
n=a/g
n=1/9.8
n=0.102
    
0-100=2*a*50
-100/100=a
a=-1 m/s^2
Negative sign shows reduction.
let n be the coeff. of friction,
n*mg=m*a
n=a/g
n=1/9.8
n=0.102
                    Answered by
            Dhrubatosh Das
            
    Good
    
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