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Q. A marble block of mass $2 \,kg$ lying on ice when given a velocity of $6 \,m/s$ is stopped by friction in $10 \,s$. Then the coefficient of friction is

AIIMSAIIMS 2012

Solution:

$v=u-a t$
$0=u-\mu g t$
$\mu=\frac{u}{g t}=\frac{6}{10 \times 10}=0.06$