A body of mass $1 \mathrm{~kg}$ is moving with a velocity $10 \mathrm{~ms}^{-1}$ due to a constant force on…
A body of mass $1 \mathrm{~kg}$ is moving with a velocity $10 \mathrm{~ms}^{-1}$ due to a constant force on a horizontal rough surface having coefficient of kinetic friction 0.4 . If the constant force is removed, the body comes to rest in a time (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
$2.5 \mathrm{~s}$
$4 \mathrm{~s}$
$0.4 \mathrm{~s}$
$0.25 \mathrm{~s}$
Solution
Mass of body, $\mathrm{m}=1 \mathrm{~kg}$
velocity, $v=10 \mathrm{~m} / \mathrm{s}$
Coefficient of kinetic friction, $\mu=0.4$
from the equation of motion
$\begin{aligned}
& \mathrm{v}=\mathrm{u}+\mathrm{at} \\
& \mathrm{t}=\frac{\mathrm{v}}{\mathrm{a}} \quad \quad(\because \mathrm{F}=\mu \mathrm{mg}) \\
& =\frac{\mathrm{vm}}{\mathrm{F}}=\frac{\mathrm{vm}}{\mu \mathrm{mg}}=\frac{\mathrm{v}}{\mu \mathrm{g}}
\end{aligned}$
The body comes to rest in a time
$\mathrm{t}=\frac{\mathrm{v}}{\mu \mathrm{g}}=\frac{10}{0.4 \times 10}=2.5 \mathrm{~s}$