Question

An inclined plane has upper 1/3 smooth qhile lower 2/3 rough. A block released from the top of inclined plane reaches the bottom and comes to test. Find (u) - coefficient of friction of rough part...


pls answer only with explanation
best answer- marked as brainliest.
​

Answer 1

The acceleration acting on the block when it moves over the smooth surface is $\sf{g\sin\theta}$ only. $\theta$ is the angle of inclination.

Let the length of the inclined plane be $\sf{l,}$ so that the length of the smooth surface is $\sf{\dfrac{l}{3}}$ and that of rough surface is $\sf{\dfrac{2l}{3}.}$

Let the velocity of the block when it enters the rough surface be $\sf{v.}$ It is the initial velocity of the block when it moves over the rough surface, which is the same as the final velocity of the block when it moves over the smooth surface.  

Hence by third equation of motion, since the block is released from rest,

$\longrightarrow\sf{v^2=0^2+2g\sin\theta\cdot\dfrac{l}{3}}$

$\longrightarrow\sf{v^2=\dfrac{2gl\sin\theta}{3}\quad\quad\dots(1)}$

The acceleration acting on the block when it moves over the rough surface is $\sf{g\sin\theta-\mu\,g\cos\theta.}$ Because if $\sf{m}$ is the mass of the block then the friction acting on the block is $\sf{\mu\,mg\cos\theta.}$

Since the block comes to rest, by third equation of motion,

$\longrightarrow\sf{0^2=v^2+2(g\sin\theta-\mu\,g\cos\theta)\cdot\dfrac{2l}{3}}$

$\longrightarrow\sf{v^2=\dfrac{4gl(\mu\,\cos\theta-\sin\theta)}{3}\quad\quad\dots(2)}$

From (1) and (2),

$\longrightarrow\sf{\dfrac{2gl\sin\theta}{3}=\dfrac{4gl(\mu\,\cos\theta-\sin\theta)}{3}}$

$\longrightarrow\sf{\sin\theta=2(\mu\cos\theta-\sin\theta)}$

$\longrightarrow\sf{3\sin\theta=2\mu\cos\theta}$

$\Longrightarrow\sf{\underline{\underline{\mu=1.5\tan\theta}}}$

===========================================

Shortcut

Suppose the lower $\sf{\left(\dfrac{1}{n}\right)^{th}}$ part of an inclined plane of angle of inclination $\theta$ is rough while the remaining upper part of the plane is smooth.

So if a block released from the top of the plane comes to rest at the bottom of the plane, then the coefficient of friction of the rough surface is,

$\longrightarrow\sf{\mu=n\tan\theta}$

By the question,

$\longrightarrow\sf{\dfrac{1}{n}=\dfrac{2}{3}\quad\implies\quad n=\dfrac{3}{2}=1.5}$

Hence the coefficient of friction is,

$\longrightarrow\sf{\underline{\underline{\mu=1.5\tan\theta}}}$