A marble block of mass $2 k g$ 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

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0.02

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0.03

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0.06

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Solution

The correct option is D $0.06$
By the first equation of motion
$v = u + a t$ and as the block decelerates before stopping
and final velocity $v = 0$
$v = u - a t \Rightarrow u - μ g t = 0$
$∴ μ = \frac{u}{g t} = \frac{6}{10 \times 10} = 0.06$

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