A) \[\mu =\frac{1}{\tan \,\theta }\]
B) \[\mu =\frac{2}{\tan \,\theta }\]
C) \[\mu =\,2\,\tan \,\theta \]
D) \[\mu =\,\,\tan \,\theta \]
Correct Answer: C
Solution :
\[\operatorname{AB}=BC=\frac{x}{2}(assume)\] \[\operatorname{AB} journey {{F}_{net}}= mg sin\theta \mu mg cos\theta \] \[{{\operatorname{W}}_{AB}}=mg sin\theta \frac{\operatorname{x}}{2} cos\] \[{{\operatorname{W}}_{AB}}=\frac{+mg\operatorname{x}\,sin\theta }{2} \] \[\therefore \vec{F}and\,\vec{S}\,are\,same\ in\,dirction\,\] \[{{\operatorname{W}}_{BC}}=+\left( mg\,sin\theta -\mu \operatorname{mg} cos\theta \right)\times \frac{1}{2}\] Apply work energy theorem \[{{\operatorname{W}}_{all force}} =\Delta KE\] \[\frac{+mg\operatorname{x}\,sin\theta }{2} {{\operatorname{W}}_{BC}}=\frac{\left[ mg\,sin\theta -\mu \operatorname{mg} cos\theta \right]}{2}=0\] \[\mu =2\,\tan \theta \]You need to login to perform this action.
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