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 ($g=10m{s}^{2}$)

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## Step 1: Given dataA 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 usedAccording to third equation of motion ,${v}^{2}={u}^{2}+2as\left[wherev=finalvelocity,u=initialvelocity,a=acceleration,s=displacement\right]$Step 3: Calculating accelerationDue to motion along rough horizontal surfaces, the body comes to rest. Let us assume its final velocity as $v$. So, $v=0m/s$${v}^{2}={u}^{2}+2as\phantom{\rule{0ex}{0ex}}{0}^{2}={6}^{2}+2a\left(9\right)\phantom{\rule{0ex}{0ex}}0=36+18a\phantom{\rule{0ex}{0ex}}a=2m/{s}^{2}$Negative acceleration means velocity is decreasing and friction is responsible for this decrement.Step 4: Coefficient of sliding friction$\mu =\frac{F}{N}\left[where\mu =coefficientoffriction,F=frictionalforce,N=normalforce\right]\phantom{\rule{0ex}{0ex}}F=\mu ×N$Due to friction, the body comes to rest. $Appliedforce=frictionforce$$m×a=\mu ×N\left[AsF=m×a\right]\phantom{\rule{0ex}{0ex}}m×a=\mu ×mg\left[AsN=mg\right]\phantom{\rule{0ex}{0ex}}a=\mu ×g\phantom{\rule{0ex}{0ex}}\mu =\frac{a}{g}\phantom{\rule{0ex}{0ex}}\mu =\frac{2}{10}\phantom{\rule{0ex}{0ex}}\mu =0.2$Hence, the coefficient of sliding friction is $0.2$.

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