Conservation of Mechanical energy

  • Thread starter ssm11s
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  • #1
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1.A chain on top of a table is stretched out in a straight line perpendicular to the table's edge with one fourth of its length L hanging over the side. How much work is required to pull the hanging section of the chain back into the table. (hint: divide the hanging part into segments of length dy)



2. W = Ui - Uf ; Total energy = U + K ; F(x) = -dU/dx



3. I can't figure out how to set up the integral..
 

Answers and Replies

  • #2
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An integral is essentially like doing a lot of addition. Think for example that each chain link is 0.05 m long, and weighs 0.1 kg. Assume five links are hanging off the edge.

Think about each chain link is being pulled up discretely. The first link must be moved 0.05m, the second must be moved 0.1m, the third must be moved 0.15m, etc. How much potential energy does each individual link hold?

The integral would essentially represent having links that are infinitely small, but cover that same distance. Does that held at all?
 
  • #3
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In other words, what you need is an expression for the potential energy of a chain segment in terms of the displacement of that chain segment (a segment hanging 5 inches down would have displacement of 5 inches).

Now, integrate that over the entire portion of the chain which is displaced.
 
  • #4
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So the equation would be int( mgh dh) from zero to h ?
 
  • #5
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where h= L/4
 

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