Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
$x=2 \;\tan \;\theta$
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App