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Question: A body takes time \(t\) to reach the bottom of an inclined plane of angle θ will be horizontal. if t...

A body takes time tt to reach the bottom of an inclined plane of angle θ will be horizontal. if the plane is rough, the time it takes now is 2t2t. The coefficient of friction of the rough surface is ?

Explanation

Solution

In the question, a condition is already given and we can easily calculate the length of the incline and eventually after having two equations, we can find out the coefficient of friction of the rough surface.

Complete step by step answer:
As per the question, we can see the body takes time tt to reach the bottom of an inclined plate of angle θ\theta with the horizontal. Therefore,
Initial velocity, u=0u=0
And acceleration down the incline, a=gsinθa=g\sin \theta
As the plane is initially smooth, the length of incline will be
s=12gsinθt2s=\dfrac{1}{2}g\sin \theta {{t}^{2}} ---(1)(1)
Now when there is a presence of friction in the second case, let the coefficient of friction be μ\mu
Therefore, the frictional force, f=μN=μmgcosθf=\mu N=\mu mg\cos \theta
Here, mm is the mass of the body

Now, by equating the force equation , we get acceleration as
a=(sinθμcosθ)ga=(\sin \theta -\mu \cos \theta )g
Also, it is given that the time taken in the 2nd case is 2t2t
Therefore, the distance equation becomes
s=12×(sinθμcosθ)g×(2t)2s=\dfrac{1}{2}\times (\sin \theta -\mu \cos \theta )g\times {{(2t)}^{2}} ---(2)(2)
After solving equation one and two, we get
sinθ=(sinθμcosθ)4 μ=34tanθ \sin \theta =(\sin \theta -\mu \cos \theta )4 \\\ \therefore \mu =\dfrac{3}{4}\tan \theta \\\
Therefore the final answer to the solution is 34tanθ\dfrac{3}{4}\tan \theta , so, the coefficient of friction in the second case is 34tanθ\dfrac{3}{4}\tan \theta .

Note: If the initial situation was changed and there was no smooth surface then such problems are just put to trick students as the coefficient of friction of a body is constant and will be the same in all rough surfaces if we keep the rule of exception aside.