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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
     
    Last edited: Mar 18, 2009
  2. jcsd
  3. Mar 18, 2009 #2

    lanedance

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    hi blessedcurse

    this would probably go better in the physics forum...

    But what is the force and energy stored in the spring at its natural length of 2ft?
    F = kx applies when x is defined as the distance form equilibrium position...
     
  4. Mar 18, 2009 #3
    When the spring is at it's natural position of 2 feet, x=0, so F is 0 and W is 0...

    Is that what you're asking?
     
  5. Mar 18, 2009 #4

    lanedance

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    easier to update in new posts so i can keep track of what happens...

    but yep - that looks better, first part is correct

    there is some ambiguity in the 2nd part of the question whether it means:
    ~4ft-7ft total length (2 to 5 as you have done)
    ~4ft extension from natural poistion to 7ft (would give 4 to 7)
     
  6. Mar 18, 2009 #5
    I fixed my previous post before you had posted, I thought.... sorry!

    I don't know... I read it as stretching it from 2 to 5 feet beyond its natural length...
     
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