Question 172: Work, Energy and Power & Circular Motion

Question

A point particle of mass $m,$ moves long the uniformly rough track $PQR$ as shown in the figure. The coefficient of friction, between the particle and the rough track equals $\mu .$ The particle is released, from rest from the point $P$ and it comes to rest at point $R.$ The energies, lost by the ball, over the parts, $PQ$ and $QR$, of the track, are equal to each other , and no energy is lost when particle changes direction from $PQ$ to $QR$.
The value of the coefficient of friction $\mu $ and the distance $x$ $(=QR),$ are, respectively close to:

Accepted Answer

Using work energy theorem for the motion of the particle,
Loss in $P.E.=$ Work done against friction from $p 	o Q$
$ + $ work done against friction from $Q 	o R$
$mgh = \mu \left( {mg\cos 	heta } ight)PQ + \mu mg\left( {QR} ight)$
$h = \mu \,\cos \,	heta  	imes PQ + \mu \left( {QR} ight)$
$2 = \mu  	imes {{\sqrt 3 } \over 2} 	imes {2 \over {\sin \,{{30}^ \circ }}} + \mu x$
$2 = 2\sqrt 3 \mu  + \mu x$  $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(i)$
[ $\sin \,\,{30^ \circ } = {2 \over {PQ}}$ ]
According to the question, work done $P 	o Q = $ work done $Q 	o R$
$ \Rightarrow $$\mu \left( {mg\cos 	heta } ight)PQ + \mu mg\left( {QR} ight)$
$	herefore$ $2\sqrt 3 \,\mu  = \mu x$
$	herefore$ $x \approx 3.5m$
From $(i)$
$2 = 2\sqrt 3 \mu  + 2\sqrt 3 \mu  = 4\sqrt 3 \mu $
$	herefore$ $\mu  = {2 \over {4\sqrt 3 }} = {1 \over {2 	imes 1.732}} = 0.29$

Answer

$0.29$ and $3.5$ $m$

Answer

$0.29$ and $6.5$ $m$

Answer

$0.2$ and $6.5$ $m$

Answer

$0.2$ and $3.5$ $m$

Work, Energy and Power & Circular Motion

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