Question

A body is coming with a velocity of $72km/h$ on a rough horizontal surface with a coefficient of friction of $0.5$. If the acceleration due to gravity is $10m{s}^{-2}$, find the minimum distance, it can be stopped.

• A. $400m$
• B. $40m$
• C. $0.40m$
• D. $4m$

Answer 1

The correct option is B

$40m$

Step 1: Given data.

Initial Velocity of a body, $u=72km/h$

Coefficient of friction, $\mu$$=0.5$

Acceleration due to gravity, $g$$=10m{s}^{-2}$

Body is at rest. So, final Velocity, $v$ $=0m{s}^{-1}$

Step 2: Find the minimum distance when the body stopped.

Formula used:

$a_{friction}$ $=-\mu g$

$v^2-u^2=2as$

Now,

$$\begin{gathered} a_{friction}=-0.5\times 10 \\ =-5m{s}^{-2} \end{gathered}$$

$$\begin{gathered} u=72\times \frac{{10}^3}{3600} \\ =20m{s}^{-1} \end{gathered}$$

Step 3: According to equation of motion.

$$\begin{gathered} \Rightarrow v^2-u^2=2as \\ \Rightarrow 0-{20}^2{\left({\mathrm{ms}}^{-1}\right)}^2=2\times \left(-5\right){\mathrm{ms}}^{-2}\times s \\ \Rightarrow s=\frac{-400}{-10} \\ \Rightarrow s=40m \end{gathered}$$

Hence, Option B is correct.