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A marble block of mass $2 \mathrm{~kg}$ lying on ice when given a velocity of $6 \mathrm{~m} / \mathrm{s}$ is stopped by friction in $10 \mathrm{~s}$. Then the coefficient of friction is (Take $\mathrm{g}=$ $10 \mathrm{~ms}^{-2}$ )
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The correct answer is:
$0.06$
$\mathrm{a}=\mu \mathrm{g}=\frac{6}{10} \quad[$ using $\mathrm{v}=\mathrm{u}+\mathrm{at}]$
$\Rightarrow \mu=\frac{6}{10 \times \mathrm{g}}=\frac{6}{10 \times 10}=0.06$
$\Rightarrow \mu=\frac{6}{10 \times \mathrm{g}}=\frac{6}{10 \times 10}=0.06$
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