Question

A block is projected along a rough inclined plane making an angle of 30 0 with the
horizontal. Obtain coefficient of friction between the block and the plane if the time
of ascent is half the time of descent.
Please answer with proper explanation and do not spam

Answer 1

Given:

A block is projected along a rough inclined plane making an angle of 30degree with the  horizontal.

To Find:

Obtain coefficient of friction between the block and the plane if the time

of ascent is half the time of descent?

Solution:

here, according to question; a block is projected along a rough inclined plane,

therefore,

$mgsin\theta+f_{1}=ma_{A}\\a_{A}=-g(sin\theta+\mu cos\theta)$

from equation of motion;

$v=u+at \ , \ v^2=u^2+2as$

$0=u-a_{A}t_{A} \ , \ 0=u^2-2a_{A}s\\implies , \ t_{A}=\sqrt{\dfrac{2s}{a_{A}}}$

when block slide down then,

$t_{D}=\sqrt{\dfrac{2s}{a_{D}} }$

Since, it is given that- time  of ascent is half the time of descent.

therefore,

$t_{A}=\dfrac{1}{2}t_{D}\\\\\sqrt{\dfrac{a_{D}}{a_{A}} }=\dfrac{1}{2} \\\\\dfrac{a_{D}}{a_{A}}=\dfrac{1}{4}\\\\\dfrac{sin\theta-\mu cos\theta}{sin\theta+\mu cos\theta}=\dfrac{1}{4}$

$\mu =\dfrac{3}{5}tan\theta\\=\dfrac{3}{5}tan(\dfrac{\pi}{6})\\\\=\dfrac{3}{5}\times\dfrac{1}{\sqrt{3} }\\\\=\dfrac{\sqrt{3} }{5}\\\\=0.346$

Hence, coefficient of friction is 0.346