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 BYou need to login to perform this action.
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