question_answer

A body of mass \[\text{5}\times \text{1}{{0}^{-3}}\text{kg}\] is launched upon a rough inclined plane making an angle of \[\text{3}{{0}^{O}}\]with the horizontal. Obtain the coefficient of friction between the body and the plane if the time of ascent is half the time of descent.                                           

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 \[\mu =\frac{3}{5}\tan \theta =\frac{3}{5}\tan {{30}^{o}}=\frac{3}{5\sqrt{3}}=0.346\]