Question

A body is projected along a rough horizontal surface with a velocity of $6m/s$. If the body comes to rest after traveling a distance of $9m$, the coefficient of sliding friction is ($g=10m{s}^2$)

Answer 1

Step 1: Given data

A body is projected along a rough horizontal surface with a velocity of 6 m/s. i.e., $u=6m/s$

The displacement of the body is $9m$. i.e., $s=9m$

Step 2: Formula used

According to third equation of motion ,$v^2=u^2+2as[wherev=finalvelocity,u=initialvelocity,a=acceleration,s=displacement]$

Step 3: Calculating acceleration

Due to motion along rough horizontal surfaces, the body comes to rest. Let us assume its final velocity as $v$. So, $v=0m/s$

$$\begin{gathered} v^2=u^2+2as \\ {0}^2=6^2+2a\left(9\right) \\ 0=36+18a \\ a=2m/s^2 \end{gathered}$$

Negative acceleration means velocity is decreasing and friction is responsible for this decrement.

Step 4: Coefficient of sliding friction
$$\begin{gathered} \mu =\frac{F}{N}[where\mu =coefficientoffriction,F=frictionalforce,N=normalforce] \\ F=\mu \times N \end{gathered}$$

Due to friction, the body comes to rest.
$Appliedforce=frictionforce$

$$\begin{gathered} m\times a=\mu \times N[AsF=m\times a] \\ m\times a=\mu \times mg[AsN=mg] \\ a=\mu \times g \\ \mu =\frac{a}{g} \\ \mu =\frac{2}{10} \\ \mu =0.2 \end{gathered}$$

Hence, the coefficient of sliding friction is $0.2$.