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Energy of moving bricks

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data

    N bricks, each having height h and mass m, are lying on the floor in a configuration of minimal potential energy. What is the minimum energy required to put all bricks one on top of another?

    2. Relevant equations

    U=mgy

    3. The attempt at a solution

    As (I assume) they are all flat on the floor with minimum potential energy, I think this problem is mainly about the most efficient way to stack the bricks. Is the energy used to stack the bricks the same as the difference in potential energy from the starting position (presumably 0 J) and the ending position (mgNh)? And how does that translate into a MINIMUM ENERGY requirement?
     
  2. jcsd
  3. Nov 30, 2009 #2

    mgb_phys

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    The minimum energy just means you lift each brick directly to it's final position.
    (It's not really an important part of the question.)
    It's also not about the most efficient stacking - you are told that they simply go on top of each other.

    Draw a stack of bricks on top of each other and think about the average potential energy,
     
  4. Nov 30, 2009 #3
    Ok, I understand what you're saying. So then the potential energy for any brick N would just be mgNh, correct? And am I correct in saying that the potential energy (mgNh) is also the "minimum energy required to put all bricking one on top of another" as the question asks?
     
  5. Nov 30, 2009 #4

    mgb_phys

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    Yes the PE for a brick is just mgH, where H is the number of bricks below it * height of a brick
    Now think about the total energy for all bricks
     
  6. Nov 30, 2009 #5
    Ah, I didn't think of the total energy of all the bricks. So would it be mghN-mgh (or of course mgh(N-1))?
     
  7. Nov 30, 2009 #6

    mgb_phys

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    I would think of in terms of the average height of a brick in a stack of N bricks
     
  8. Nov 30, 2009 #7
    I'm afraid I don't understand what you're getting at
     
  9. Nov 30, 2009 #8

    mgb_phys

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    The first (not counting the one on the floor) brick moves through height h gaining pe of mgh
    The next brick moves through 2h = mg2h
    The next = mg3h
    and the Nth mgNh

    Whats the total?
    ie what is the sum of 1+2+3+4+5......N
     
  10. Nov 30, 2009 #9
    Ok, so it's the sum of mghN from N=1 to N, but I'm not sure what the sum of 1+2+3+. . . +N is. Before I was just using an integral of mghN from 1 to N, but I guess that's not how you do it.
     
  11. Nov 30, 2009 #10

    mgb_phys

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    Hint - Karl Gauss figured it out when he was 5
     
  12. Nov 30, 2009 #11
    So. . . I believe the energy would be mgh((n^2+n)/2)?
     
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