Question

A cylinder of mass $10 \,kg$ is rolling on a rough plane with a velocity of $10 \,m/s$. If the coefficient of friction between the surface and cylinder is $0.5$, then before stopping it will cover a distance of (Take $g = 10\, m/s^2$):

• A. 10 m
• B. 7.5 m
• C. 5 m
• D. 2.5 m

Answer 1

The Correct Option is A

Kinetic energy of cylinder is consumed in doing work against friction in rolling motion.
A body of mass $m$ moving with velocity $v$, Possess kinetic energy given by
$K=\frac{1}{2} m v^{2} \ldots(1)$
This kinetic energy is utilized in doing work against the frictional forces.
$W=\mu m g s \ldots(2)$
Where $\mu$ is coefficient of kinetic friction, $m$ is mass, $g$ is gravity and $s$ is displacement.
Equating Egs. (1) and $(2)$, we get
$\frac{1}{2} m v^{2}=\mu m g s$
Where $\frac{1}{2} m v^{2}=\frac{1}{2} \times 10 \times(10)^{2}$
$=500\, kg - m / s$
Given, $v=10 \,m / s, \mu=0.5, m=10\, kg$
$g=10\, m / s^{2}$
$\Rightarrow s=\frac{\frac{1}{2} \times m \times(10)^{2}}{\mu mg }$
$=\frac{50}{0.5 \times 10}=10 \,m$