question_answer

The upper half of an inclined plane of inclination 0 is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between block and lower half of the plane is given by:

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 \[{{\operatorname{W}}_{all force}} =\Delta KE\] \[\frac{+mg\operatorname{x}\,sin\theta }{2} {{\operatorname{W}}_{BC}}=\frac{\left[ mg\,sin\theta -\mu \operatorname{mg} cos\theta  \right]}{2}=0\] \[\mu =2\,\tan \theta \]