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Q.
A cricket is rolled on ice with a velocity of $6 m / s$ and comes to rest after traveling $10 m$. Find the coefficient of friction. $\left( g =9.8 m / s ^{2}\right)$
Laws of Motion
Solution:
$v ^{2}- u ^{2}=2 as$
$a =\frac{\left(0^{2}\right)-(6)^{2}}{2 \times 10}=\frac{-36}{20}=-1.8$
$a =\mu g $
$\Rightarrow \mu=\frac{ a }{ g }=\frac{-1.8}{9.8}$
$ \Rightarrow \mu=-0.18$
(-ve sign indicates that friction force acts in opposite direction.)