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
Answered by Dhrubatosh Das
Good
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