A cylinder of mass 12 ,kg is sliding on a pla | vedclass

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(A) Given: Mass of the cylinder, $m = 12 \,kg$. Initial velocity, $u = 20 \,m/s$. Coefficient of kinetic friction, $\mu = 0.5$. Acceleration due to gr

Vedclass · Accepted Answer

$40$

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(A) Given: Mass of the cylinder, $m = 12 \,kg$. Initial velocity, $u = 20 \,m/s$. Coefficient of kinetic friction, $\mu = 0.5$. Acceleration due to gravity, $g = 10 \,m/s^2$.The frictional force acting on the cylinder is $f = \mu N = \mu mg$.According to Newton's second law, $f = ma$, so $ma = \mu mg$, which gives the retardation $a = -\mu g$.Substituting the values, $a = -0.5 	imes 10 = -5 \,m/s^2$.Using the third equation of motion, $v^2 = u^2 + 2as$, where $v = 0$ (final velocity at rest):$0 = (20)^2 + 2(-5)s$$0 = 400 - 10s$$10s = 400$$s = 40 \,m$.Therefore, the cylinder travels a distance of $40 \,m$ before stopping.

$A$ cylinder of mass $12 \,kg$ is sliding on a plane with an initial velocity $20 \,ms^{-1}$. If the coefficient of friction between the surface and the cylinder is $0.5$, what is the distance the cylinder travels before stopping (in $\,m$)?

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