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
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
$0.01$
$0.02$
$0.03$
$0.06$
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
$v = u - at \Rightarrow u - \mu gt = 0$
$\therefore \mu = \frac{u}{{gt}} = \frac{6}{{10 \times 10}} = 0.06$
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