A marble block of mass 2, kg lying on ice when given a velocity of 6, m/s is stopped by frictio

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4-2.Friction

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

A$0.02$

B$0.03$

C$0.06$

D$0.01$

Solution

$u=6 m / s, v=0, t=10 s$
$a=-\frac{f}{m}=\frac{-\mu m g}{m}=-\mu g=-10 \mu$
$v=u+a t$
$0=6-10 \mu \times 10$
$\therefore \mu=0.06$

Standard 11

Physics

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