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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: H.C.Verma problem
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[Post New]18 Apr 2008 10:53:10 IST
    Subject: H.C.Verma problem
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Xavier4 (4)

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The heavier block in an atwood machine has a mass twice that of the lighter one.The tension in the string is 16 N when the system is set into motion.Find the decrease in the potential energy during the first second after the sysem is realeased from rest?
    
[Post New]18 Apr 2008 17:06:07 IST
    Subject: Re:H.C.Verma problem
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sboosy (3011)

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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
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[Post New]18 Apr 2008 17:09:41 IST
    Subject: H.C.Verma problem
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varun.tinkle (1084)

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2MG-T=2MA
T-MG=MA
MG=3MA
A=G/3
T=2M(G-a)
16=2M(2G/3)
16=4MG/3
16*3/4=MG
12=MG
M=12/G
CONSIDER THE MOTION OF THE BIGGER BLOCK
S=1/2AT^2
=G/6
POTENTIOL ENERGY DECREASED=24/G*G/6*G=4G
CONSIDER THE MOTION OF THE SMALLER BLOCK
DISTANCE WILL BE SAME
POTENTIOL ENERGY INCREASED
=12/G*G/6*G=2G
THEREFORE POTENTIOL ENERGY DECREASED=2G=19.6
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