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Orientation on Calculus Work Problem. Hooke's Law.

  1. Aug 5, 2013 #1
    A force of 16,000lb compresses a string from its natural length of 13 inch to 8 inch. Find the work done to compress it to the first inch




    W=$$\int F dx$$





    F=kx
    16000=K(5)
    3200=K


    W=$$\int F dx$$

    W=$$\int_1^{13}\!\ 3200xdx$$

    [1600x^2] from 1 to 13

    w= 268800ftlb

    Am I right?
    I think the trick here is the limits.
    If they would be from 0 to 1,, aint that to little work?
    I mean, as you compress down, it's harder to do it. The work has to increase as you reach 0 inch
    Am I right?
    If not, please explain

    Or is it from 12 to 13?

    how I viewed is that they want to know the work as you compress it to the first inch.. in this case 1

    I need a little orientation

    | 16000Lb |
    /----*------------------------*----------------------/
    0inch 1inch 8inch 13inc
    Thanks!

     
    Last edited: Aug 5, 2013
  2. jcsd
  3. Aug 5, 2013 #2

    haruspex

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    Wrong units for a force.
    Extra 0 crept in. What units do you want the work in? What units is the 5 in?
    What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches?
     
  4. Aug 5, 2013 #3
    Sorry about that
    force is 16000lb
    using hooke's law I got the K
    16000lb=K(5) 5 as the inches that the Force compressed

    The lenght of the string is 13 inches
    It was compressed 5 inches by a 16000lb force
    I need to find the work of the first inch

    Thanks
     
  5. Aug 5, 2013 #4

    haruspex

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    A spring, presumably, not a string.
    So answer my question:
    "What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches? "​
     
  6. Aug 5, 2013 #5

    1 and 13
    the spring is 13 inch long. from 0 to 13 there's no work, now, whether it compresses or streches, a work can be calculated
     
  7. Aug 5, 2013 #6

    haruspex

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    No.
    The x in your integral is the extent of compression. The spring starts at its relaxed length of 13 inches. That's x = 0, no compression. It is compressed from 13 inches to 8 inches. That's a compression of 5 inches, x = 5. What is the value of x when the spring has been compressed by only the first of those five inches of compression?
     
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