The upper half of an inclined plane of inclination theta is perfectly smooth while lower half is rough. A block starting from rest at the

Q: The upper half of an inclined plane of inclination $\theta$ 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 the block and lower half of the plane is given by

c $\begin{array}{l} Let\,m\,be\,mass\,of\,the\,block\,and\,L\,be\,lenght\,of\\ the\,inclined\,plane.\\ According\,to\,work - energy\,theorem\\ \,\,\,\,\,\,\,\,W = \Delta K = 0\,\,(Initial\,and\,final\,speeds\,are\,zero)\\ \therefore \,Wrok\,done\,by\,friction\, + Work\,done\,by\,gravity = 0\\ \,\,\,\,\,\,\,\,\,\, - \mu mg\cos \theta \frac{L}{2} + mg\sin \theta L = 0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\mu }{2}\cos \theta = \sin \theta \\ \,\,\,\,\,\,\,\,\,\,\,\,\mu \, = \frac{{2\sin \theta }}{{\cos \theta }} = 2\tan \,\theta \end{array}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

The upper half of an inclined plane of inclination $\theta$ 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 the 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 $

AIPMT 2013, Medium

STD 11 Science
NEET
STD 11 - 4.2. friction
Physics

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