First slide

Friction on inclined plane of angle more than angle of repose

Question

The upper half of an inclined plane of inclination $θ$ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

Moderate

A

$μ = 2 \tan θ$
B

$μ = \tan θ$
C

$μ = \frac{1}{\tan θ}$
D

$μ = \frac{2}{\tan θ}$

Solution

Let m be mass of the block and L be length of the inclined plane.
            According to work-energy theorem
$W = Δ K = 0$
( $∵$ Initial and final speeds are zero)
Work done by friction+ Work done by gravity= 0
$- μ m g \cos θ \frac{L}{2} + m g \sin θ L = 0 \frac{μ}{2} \cos θ = \sin θ μ = \frac{2 \sin θ}{\cos θ} = 2 \tan θ$
  Alternate solution
For upper half smooth plane Acceleration of the block, $a = g \sin θ$
Here, u = 0 ( $∵$ block starts from rest)
$a = \sin θ, s = \frac{L}{2}$
Using, $v^{2} - u^{2} = 2 a s,$ we have
$v^{2} - 0 = 2 \times g \sin θ \times \frac{L}{2} v = \sqrt{g L \sin θ} - - - (i)$
For lower half rough plane
Acceleration of the block, $a^{'} = g \sin θ - μ g \cos θ$
where $μ$ is the coefficient of friction between the block and lower half of the plane
Here, $u = v = \sqrt{g L \sin θ},$
$v = 0 (∵ b l o c k c o m e s t o r e s t) a = a^{'} = g \sin θ - μ g \cos θ, s = \frac{L}{2} A g a i n, u \sin g v^{2} - u^{2} = 2 a s, w e h a v e 0 - {(\sqrt{g L \sin θ})}^{2} = 2 \times (g \sin θ - μ g \cos θ) \times \frac{L}{2} - g L \sin θ = (g \sin θ - μ g \cos θ) L - \sin θ = \sin θ - μ \cos θ μ \cos θ = 2 \sin θ μ = 2 \tan θ$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App