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Work problem chain on table

  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data
    BXiQ2b0.png

    2. Relevant equations
    W = F d
    F = ma
    3. The attempt at a solution
    so in order to get the whole chain on the table, we need to pull the chain 0.65 meters onto the table.

    since 0.65 meters is hanging off the table, the gravity is acting on it, therefore F=ma where m is half the chain (8kg /2 = 4kg) and a = gravity(g), which is 39.2 N,since we work against this force we need to find the force required to pull 8 Kg a distance of 0.65m and then subtract 39.2 N.

    is this the correct approach?
     
  2. jcsd
  3. Mar 20, 2015 #2

    CWatters

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    As you pull it up the force required get less. I would use a different approach.

    Hint: Work = change in energy
     
  4. Mar 20, 2015 #3
    W = KEf - KEi

    KEf = 0 since v =0

    KEi = mgh = 8 kg x 9.8m/s^2 x 0.65 = 51

    W = 51 Js

    is this correct?
     
  5. Mar 20, 2015 #4

    CWatters

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    Close.
    How much mass changes height?
    Where is the centre of that mass and how much height does it gain?
     
  6. Mar 20, 2015 #5
    I didn't learn about center of masses yet , is that required to do this problem? :(
     
  7. Mar 20, 2015 #6

    CWatters

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    There is another way but I suspect its a lot harder.

    Read up on the centre of mass or centre of gravity.

    Consider a uniform ruler that's 12" long. To find the centre of mass you could balance it on a knife edge to make a seesaw (tetter-totter in the USA). If you adjust the position of the ruler until it balances the centre of mass will be found to be around the 6" position.

    The centre of mass of the part hanging over the edge would be at 0.65/2 = 0.325m below the top of the table.
     
  8. Mar 20, 2015 #7
    then wouldn't we need to find the center of mass for .325m and so forth?
     
  9. Mar 20, 2015 #8

    CWatters

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    No.

    The part of the chain that hangs over the edge can be replaced/modelled by a single 4kg mass on a rope that hangs 0.352m over the edge.
     
  10. Mar 20, 2015 #9
    so the work required to pull the lingering half of the chain up is 4kg x 9.8 x .352m?

    and the work required to pull the other half which is already on the table is : W = F d

    we just need to add (F d) + (4kg x 9.8 x 0.352m) right?
     
  11. Mar 20, 2015 #10

    CWatters

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    Correct.

    No. It actually takes no energy/work to move that bit along the table. The problem states you can ignore friction so the "F" in W = Fd is zero.
     
  12. Mar 20, 2015 #11
    ok, i got the answer to be 12.7 j
     
  13. Mar 20, 2015 #12

    CWatters

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    I agree.
     
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