JEE MAIN - Physics (2005 - No. 64)
The upper half of an inclined plane with inclination $$\phi $$ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by
$$2\,\cos \,\,\phi $$
$$2\,sin\,\,\phi $$
$$\,\tan \,\,\phi $$
$$2\,\tan \,\,\phi $$
Explanation
Let the length of the inclined plane is = $$l$$. So only $${l \over 2}$$ part will have friction.
According to work-energy theorem, $$W = \Delta k = 0$$ (Since initial and final speeds are zero)
$$\therefore$$ Work done by friction + Work done by gravity $$=0$$
i.e., $$ - \left( {\mu \,mg\,\cos \,\phi } \right){\ell \over 2} + mg\ell \,\sin \,\phi = 0$$
or $${\mu \over 2}\cos \,\phi = \sin \phi $$
or $$\mu = 2\,\tan \,\phi $$
According to work-energy theorem, $$W = \Delta k = 0$$ (Since initial and final speeds are zero)
$$\therefore$$ Work done by friction + Work done by gravity $$=0$$
i.e., $$ - \left( {\mu \,mg\,\cos \,\phi } \right){\ell \over 2} + mg\ell \,\sin \,\phi = 0$$
or $${\mu \over 2}\cos \,\phi = \sin \phi $$
or $$\mu = 2\,\tan \,\phi $$
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