Physics, asked by saumitraverma04, 6 months ago

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

Answers

Answered by rashich1219
2

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

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