Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
+1 vote

A body is sliding down a rough inclined plane of angle of inclination $\theta$ for which the coefficient of friction varies with distance x as $\mu (x)=kx$, where k is a constant . Here x is the distance moved by the body down the plane . The net force on the body will be zero at a distance $x_0$ given by :

$\begin{array}{1 1}\frac{\tan \theta}{k}\\k\;tan\;\theta\\\frac{\cot \theta}{k} \\k\;cot \theta\end{array} $

Can you answer this question?

1 Answer

0 votes
Answer : $\large\frac{\tan \theta}{k}$
The net downward force on the body at a distance x is
$f(x) =mg \sin \theta - \mu mg \cos \theta$
$\qquad= mg( \sin \theta - \mu \cos \theta)$
$\qquad= mg (\sin \theta- kx \cos \theta)$
$\therefore f(x) =0$ at a value of $x =x_0$ given by
$\sin \theta -kx_0 \cos \theta=0$
Which gives $x_0 = \large\frac{\tan \theta}{k}$
answered Aug 29, 2014 by meena.p

Related questions

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