Answer:
Here,
\[m=\text{5}\times \text{1}{{0}^{-3}}\text{kg},\text{ }\theta
=\text{3}0{}^\circ ,\mu =?\]
If f is the force of friction between the body
and the inclined plane, then
\[f=\mu R=\mu \,mg\,\cos \theta \] ??.
(i)
Let \[{{a}_{1}}\] be the
acceleration while ascending (when factional force acts down the plane). Fig.
3(HT).8(a).
(a)
\[\therefore \] \[mg\,\sin \theta
+f=m{{a}_{1}}\] ?? (ii)
If \[{{a}_{2}}\] is
acceleration, while descending (when frictional force acts up the plane). Fig.
3(HT).8(&).
(b)
\[\therefore \] \[mg\,\sin \theta
-f=m{{a}_{2}}\] ??.(iii)
Dividing (ii) by (iii), we get
\[\frac{mg\,\sin \theta +f}{mg\,\sin \theta
-f}=\frac{{{a}_{1}}}{{{a}_{2}}}\]
Using (i), \[\frac{mg\,\sin \theta +\mu \,mg\,\cos \theta
}{mg\,\sin \theta -\mu \,mg\,\cos \theta }=\frac{{{a}_{1}}}{{{a}_{2}}}\]
\[\frac{mg\,\cos \theta (\tan \theta
+\mu )}{mg\,\cos \theta \,(\tan \theta -\mu )}=\frac{{{a}_{1}}}{{{a}_{2}}}\]
If \[{{t}_{1}}\] is time of
ascent and \[{{t}_{2}}\] is time of descent for length of plane I, then
\[l=\frac{1}{2}{{a}_{1}}t_{1}^{2}=\frac{1}{2}{{a}_{2}}t_{2}^{2}\] or
\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{t_{2}^{2}}{t_{1}^{2}}={{\left(
\frac{2}{1} \right)}^{2}}=4\]
From (iv), \[\frac{\tan
\theta +\mu }{\tan \theta -\mu }=4\], which gives \[\mu =\frac{3}{5}\tan
\theta =\frac{3}{5}\tan {{30}^{o}}=\frac{3}{5\sqrt{3}}=0.346\]
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