Question

The upper portion of an inclined plane of inclination $\alpha$ is smooth and the lower portion is rough. A particle slides down from rest from the top and just comes to rest at the foot. If the ratio of smooth length to rough length is m : n, find the coefficient of friction.

• A. $\mu =\left(\frac{m-n}{n}\right)\tan \alpha$
• B. $\mu =\left(\frac{m+n}{m}\right)\tan \alpha$
• C. $\mu =\left(\frac{m-n}{m}\right)\tan \alpha$
• D. $\mu =\left(\frac{m+n}{n}\right)\tan \alpha$

Answer 1

Answer: (D)

$\mu =\left(\frac{m+n}{n}\right)\tan \alpha$

Using kinematics equation for distance travelled in smooth region
$v^2=0^2+2a_1{S}_1{a}_1=gsin\alpha ;S_1=m\Rightarrow v^2=2gsin\alpha m$
The ball comes to rest after traveling distance in rough region, applying kinematics equation for distance travelled in rough region
$0^2=v^2+2a_2{S}_{2}v=\sqrt{2gsin\alpha m}{a}_2=\mu gcos\alpha -gsin\alpha S_2=n$
$\Rightarrow gsin\alpha m=2\times \mu gcos\alpha -gsin\alpha \times n$
dividing equation by $gcos\alpha$
$tan\alpha m=(\mu -tan\alpha )n$
$\mu =\frac{m+n}{n}tan\alpha$