Question

# A body takes time t to reach the bottom of an inclined plane of angle θ will the horizontal. if the plane made rough, time takes now is 2t. The coefficient of friction of the rough surface is

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Solution

## Step 1: Given dataThe angle of the inclined plan is $\theta$.The initial velocity of the body is zero.Time is taken by the body to reach from top to bottom on a smooth inclined plane is t sec.Time is taken by the body to reach from top to bottom on a rough inclined plane is 2t sec.Step 2: Exerted force on a body placed on an inclined planeFor a block on an inclined plane, there are basically three types of force exerted, frictional force, normal force, and gravitational force.The frictional force is a force, that is directed opposite to the body's motion and that prevents the movement of the body at the point of contact.The frictional force is defined by the form, $F=\mu N$, where $\mu$is the coefficient of friction and n is the normal force.Step 3: DiagramStep 4: Finding the coefficient of friction for the smooth inclined planethe distance (hypotenuse) between the top point and bottom point is, ${s}_{smooth}=\frac{1}{2}g\mathrm{sin}\theta ×{t}^{2}$ ……………..(1)(applying the formulae of kinematics)where g is the acceleration due to gravity and t is the taken time.Step 5: Finding the coefficient of friction for the Rough inclined planethe normal force (N) on the body is, $N=mg\mathrm{cos}\theta$ ……………(2)Now, the frictional force of the body on the rough plane is,$F=\upsilon N=\mu mg\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}orF=\mu mg\mathrm{cos}\theta ...................\left(3\right)$And the acceleration of the body is,$a=g\mathrm{sin}\theta -\mu g\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}ora=g\left(\mathrm{sin}\theta -\mu \mathrm{cos}\theta \right)................\left(4\right)$Again, ${s}_{rough}=\frac{1}{2}\left(\mathrm{sin}\theta -\mu \mathrm{cos}\theta \right)×g×{\left(2t\right)}^{2}.\phantom{\rule{0ex}{0ex}}or{s}_{rough}=\left(\mathrm{sin}\theta -\mu \mathrm{cos}\theta \right)×g×2{t}^{2}...............\left(5\right)$Now, comparing equations (1) and (5) we get,$\frac{1}{2}g\mathrm{sin}\theta ×{t}^{2}=\left(\mathrm{sin}\theta -\mu \mathrm{cos}\theta \right)×g×2{t}^{2}.\phantom{\rule{0ex}{0ex}}or\left(\mathrm{sin}\theta -\mu \mathrm{cos}\theta \right)=\frac{\mathrm{sin}\theta }{4}\phantom{\rule{0ex}{0ex}}or\frac{\left(\mathrm{sin}\theta -\mu \mathrm{cos}\theta \right)}{\mathrm{sin}\theta }=\frac{1}{4}\phantom{\rule{0ex}{0ex}}or1-\mu cot\theta =\frac{1}{4}\phantom{\rule{0ex}{0ex}}or\mu cot\theta =\frac{3}{4}\phantom{\rule{0ex}{0ex}}or\mu =\frac{3}{4cot\theta }=\frac{3}{4}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}or\mu =\frac{3}{4}\mathrm{tan}\theta$So, the coefficient of friction is $\mu =\frac{3}{4}\mathrm{tan}\theta$.

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