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Mechanics

Scorching goIITian

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17 Mar 2007 15:18:15 IST

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861

work energy and power

None

a uniform chain of length l and mass m is resting on a smooth table. one-third of its lenght is hanging vertically down over the table. the amount of work required to pull the hanging part back to the table is

1)mgl / 3

2)mgl / 6

3)mgl / 9

4)mgl /18

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Comments (10)

Himanshu

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Joined: 19 Feb 2007

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17 Mar 2007 15:28:44 IST

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is ans mgl/18

001sri

Scorching goIITian

Joined: 28 Dec 2006

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17 Mar 2007 15:29:20 IST

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ans is d

sneha kulkarni

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17 Mar 2007 15:31:03 IST

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yeah.....the ans is 4th option...but pls solve it........i m not getting the right ans.........

Himanshu

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17 Mar 2007 15:39:07 IST

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the effective mass which is to b lifted = M/L * L/3 (coz 1/3rd part)

so mass = M/3

work done is= (here dist taken is L/6)

so wd = M/3 * L/6 *g

= MgL/18

sneha kulkarni

Scorching goIITian

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17 Mar 2007 15:51:38 IST

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ya......thanks a lot.......

Himanshu

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17 Mar 2007 15:53:52 IST

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welcome a lot

JEETNA MERI JIDD NAHIN, MERI AADAT HAIN.........

Blazing goIITian

Joined: 22 Feb 2007

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17 Mar 2007 15:59:59 IST

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hey look...........

himanshu has done a fine job.......

but i m just adding something more.................

as U = Mgh

M= m/3.........

g= g..........

h= l/6..............( because we take effective height ..............

that is height of the centre of mass of the hanging part.........)

therefore...........

U= mgl/18.............ans

thanks...........

i know u have got the answer but just .........giving the information why himanshu has taken

l/6

hope u understands........

Himanshu

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17 Mar 2007 16:15:13 IST

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sry don ji
main information type karna bhul gaya
hume maaf kar do plzzzzzzzzzz!!!!!!!!!!!!!!!!!!

vishak p

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17 Mar 2007 18:44:41 IST

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here's an another method :
consider a length x from the bottom of the hanging chain and a dx part just above it !
now, here we can think of an integration process as the work done in lifting every element will be different as they are at different heights from the datum h= 0
now , since the dx part is too small, the work done in lifting the dx part after the x part will be actually very very close

so , dw = M/ L * dx * g * x
integrating the above and putting the limits from 0 to L/3 , we get W = mgl / 18 !!!
what i think that integration is far more better in such type of problems !!
anyways !! sorry i wnt offline so, when will u be online now ?

jibinjoyk

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26 Oct 2011 19:23:37 IST

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a body of mass m slowly hauled up a curved incline by a force,which at each point is

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