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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Jan 2008 12:14:25 IST
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An object attached to vertical spring is slowly lowered to its equillibium position,thus stretching the b an amount d. now the same vertical spring is allowed to fall. this spring will stretch by
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Glitter Graphics
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26 Jan 2008 12:26:06 IST
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in case the spring is attached to a fixed surface..... actually the spring will compress....b coz when an object is undergoing a free fall it will experience weightlessness..... so ht e force applied on the downward direction will decrease...... in case of free fall..... the block is moving with the spring or vice versa(all the same) it can be treated as a single entity...... therefore we can conclude spring will not stretch....... anyway the amount the spring strech will depend on the spring constant....
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26 Jan 2008 12:36:15 IST
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hey amit
am i right......
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26 Jan 2008 12:42:33 IST
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Hey dude! The extension will be equal to d,i.e.the same. Initially,the relative acceleration of the block w.r.t. the spring was 0. Finally,when both fall freely,even then the relative acceleration is 0,because both move with accn=g,downwards. This is a clear explanation of your problem. Plz do RATE me!!
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26 Jan 2008 12:50:36 IST
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i think AJ is right i was thinking in terms of a spring balance....
in case of a spring balance i am right.......i am cent percent sure in your case don no
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26 Jan 2008 12:51:51 IST
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Hey crack! if you think i was right,why dont't you give me a rating.
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26 Jan 2008 13:01:26 IST
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Come on rate me!!!!!!!!!!!!!!!!!!!!!!!!
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26 Jan 2008 13:21:46 IST
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let him rate you first....its his question and if he rates you
i will rate you too......
dude i am still not sure.....abt yours and my ans.......lets see
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26 Jan 2008 13:23:01 IST
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you owe me a rate....... dude lol
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26 Jan 2008 15:46:26 IST
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The answer is 2d Kd = mg (equilibrium condition) d = mg/k -1 When the block is released it will have maximum elongation, when it has zero kinetic energy. That is at maximum elongation increase in spring's potential energy = decrease in gravitational potential energy Hence mgx = 1/2 Kx2 mg = 1/2 Kx x = 2mg/k x = 2d (from 1)
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