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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: H.C.Verma problem
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[Post New]18 Apr 2008 17:06:07 IST
    Subject: Re:H.C.Verma problem
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sboosy (3063)

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Olaaa!! Perrrfect answer. 539  [723 rates]
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\mbox{If two masses are} \ m_1 \ \mbox{and} \ m_2 \ \mbox{then} \\ \\ \mbox{acceleration is} \ \frac{(m_1-m_2)g}{m_1+m_2} = \frac{(2m-m)g}{2m+m} = \frac{g}{3} \\ \\ \mbox{Tension is} \ \frac{2m_1m_2g}{m_1+m_2} = \frac{2(2m)(m)g}{2m+m} = \frac{4mg}{3} = 16 \\ \\ \Rightarrow mg =12 \\ \\ \mbox{Now in one second the blocks will travel} \\ \\ \frac{1}{2}*\frac{g}{3}*1 = \frac{g}{6} \\ \\ \mbox{Decrease in PE is} \ 2mg(\frac{g}{6}) - mg(\frac{g}{6}) = mg*\frac{g}{6} = 2g
 this reply: 10 points  (with 2 Olaaa!! Perrrfect answer.   in 2 votes )   [?]
 
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