question_answer

In the figure shown, a particle of mass m is released from the position A on a smooth track. When the particle reaches at B, then normal reaction on it by the track is  

A) mg                   

B)        2 mg

C) \[\frac{2}{3}\,\,mg\]      

D)        \[\frac{{{m}^{2}}g}{h}\]

Correct Answer: A

Solution :

By conservation of energy \[\operatorname{mg} (3h) = mg (2h) + \frac{1}{2}m{{v}^{2}} \left( v = velocity at B \right)\] \[\operatorname{mgh}=\frac{1}{2}m{{v}^{2}};\,\,\,\,\,v=\sqrt{2\,gh}\] From free body diagram of block at B \[\operatorname{N}+mg=\,\,\frac{m{{v}^{2}}}{h}=2mg\,\,;\,\,\,\,\,N=mg\]o