Chat with tutor

Ask Questions, Get Answers

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

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} $

1 Answer

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 $x_0 = \large\frac{\tan \theta}{k}$
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.