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A cylinder of mass $12 \mathrm{~kg}$ is sliding on plane with an initial velocity $20 \mathrm{~ms}^{-1}$. If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping, the cylinder describes a distance of
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$40 \mathrm{~m}$
Given that, mass of cylinder, $m=12 \mathrm{~kg}$
Initial velocity, $u=20 \mathrm{~m} / \mathrm{s}$
Coefficient of friction, $\mu=0.5$
We know that, the retardation produced by
friction $=a=-\mu g=-0.5 \times 10=-5 \mathrm{~m} / \mathrm{s}^2$ [using, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
Let $S$ be the distance travelled by cylinder to stop.
[i.e. final velocity, $v=0$ ] Using equation of motion, $v^2=u^2+2 a s$ By substituting the value, we get
$0=(20)^2+2(-5) s \Rightarrow s=40 \mathrm{~m}$
Initial velocity, $u=20 \mathrm{~m} / \mathrm{s}$
Coefficient of friction, $\mu=0.5$
We know that, the retardation produced by
friction $=a=-\mu g=-0.5 \times 10=-5 \mathrm{~m} / \mathrm{s}^2$ [using, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
Let $S$ be the distance travelled by cylinder to stop.
[i.e. final velocity, $v=0$ ] Using equation of motion, $v^2=u^2+2 a s$ By substituting the value, we get
$0=(20)^2+2(-5) s \Rightarrow s=40 \mathrm{~m}$
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