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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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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} $

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1 Answer

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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

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