A) \[2\sin \phi \]
B) \[2\cos \phi \]
C) \[2\tan \phi \]
D) \[\tan \phi \]
Correct Answer: C
Solution :
According to work-energy theorem, (initial and final speeds are zero) Work done by friction + Work done by gravity = 0 \[-\,(\mu \,mg\,cos\phi )\,\frac{l}{2}\,+mgl\,\sin \phi =0\] (friction will start at 1/2) orYou need to login to perform this action.
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