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How much work

  • Thread starter forty
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135
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An infinitesimally thin rope is held on a frictionless table with one-fourth of its length hanging over the edge. If the rope has length L and total mass m (assuming uniform mass distribution), how much work is required to pull the hanging part back onto the table.

U = mgh

The length of the overhanging piece of string is (L/4) and the mass of this piece is (m/4) so do i just plug in the values, presuming I'm using the right equation... work = mLg/16 ??
 

Answers and Replies

144
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Not really. Since the length of the rope hanging of the table varies when you pull it, the mass on the part hanging of the table varies, and therefore also the force. This means that you cant just take force times distance to get the total energy, since the force is not constant.
 
Last edited:
5
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take an elementary length dl,then dm=(m/l)dl.now integrate it within proper limits.remeber work done=-w by mg.
good luck!
 
135
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I'm really stuck at how to solve this "properly" using an integral. I can solve this by treating the piece of the string over the edge as a point mass..

mass of string = m/4
distance moved = L/8

.: workdone = mgh = (m/4) * g * (L/8) = mgL/32

If anyone could shed some light on the above mentioned method it would be greatly appreciated.
 

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