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Work done on a spring

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    The amount of WORK to stretch a spring 4 feet beyond its natural length of 2 feet is 10 ft-lbs. Find the work required to stretch the spring from 4 feet to 7 feet.

    2. Relevant equations

    W=[tex]\int^{b}_{a}Fdx=\int^{b}_{a}kxdx=[kx^{2}/2]^{b}_{a}[/tex]

    3. The attempt at a solution

    10=[tex][kx^{2}/2]^{4}_{0}[/tex]
    10=k(16)/2-k(0)/2
    10=8k
    k=4/5

    W=[tex][kx^{2}/2]^{b}_{a}[/tex]
    W=4/5[tex][x^{2}/2]^{5}_{2}[/tex]
    W=4(25)/10-4(4)/10
    W=42/5

    W=8.4 ft*lb
     
  2. jcsd
  3. Mar 19, 2009 #2
    Could you explain your limits of integration? The spring is either being stretched from 0 to 6 ft or from 2 to 6 ft, I don't see where you got 0 to 4 from.
     
  4. Mar 19, 2009 #3
    It's being stretched 4 feet beyond its natural length of 2 feet, so the natural length is the 0 and then 4 feet beyond that is 4.

    At first I had 2 to 6, but then I realized that that meant stretching it from 2 feet beyond its natural length to 6 feet beyond it because the limits refer to change in distance... so the change in distance is 0 to 4.
     
  5. Mar 19, 2009 #4
    Oh, I see. Sorry, that is a badly worded question (or more likely I read it badly). I thought it meant it's being stretched four feet beyond it's elestic limit! My bad.

    Given that is true I think your answer is correct, at least in it's principle (Ihaven't checked the numerics).
     
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