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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A body is moving down a long inclined plane of angle of inclination $\theta$ . The coefficient of friction between the body and the plane varies as $\mu =0.5 \;x$ Where x is the distance moved down the plane . The body will have the maximum velocity when it has travelled a distance x given by

$\begin{array}{1 1}x=2 \;\tan \;\theta \\ x= \large\frac{2}{\tan \theta}\\x= \sqrt 2 \cot \theta \\x= \large\frac{\sqrt 2}{\cot \theta}\end{array} $

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1 Answer

$x=2 \;\tan \;\theta$
The acceleration of the body down the plane is $g \sin \theta - \mu g \cos \theta -g (\sin \theta -\mu \cos \theta )= g(\sin \theta -0.5 x \cos \theta ).$
Therefore , the body will first accelerate up to $ x <2 \tan \theta $ .
The velocity will be maximum at $ x= 2 \tan \theta$, because for $ x > 2\;\tan \theta $, the body starts decelerating .
answered Aug 29, 2014 by meena.p
 

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