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Trouble with hooks law

  1. Feb 18, 2004 #1
    the question:

    A balloon is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the strecthing of the hose obeys hooke's law with a spring constant 102 N/m. if the hose is stretched by 5.30m and then released, how much work does the force from the hose do on the baloon in the puch by the time the hose reaches its relaxed lenghty

    The answer:

    this is how i tried to solve it...F=-kx according to hooke's law..so i did -(102)(5.30)= -540.6 N ..and work is W=Fd..so i mulitiped the force with distance 5.30 and got -2865.18 which is not the answer...what am i doing wrong? thanks
     
  2. jcsd
  3. Feb 18, 2004 #2
    See, force is varying in a spring.. meaning you can't use [itex]W = F \times D[/itex]. Since work is the integral of force, the equation you want to use is:
    [tex]W = \int_{0}^{x} F(x) dx[/tex]
    or
    [tex]W = \frac{1}{2} k x^2[/tex]

    (In this case we integrate from x to 0, so we should remember to add a negative sign to our answer to show negative work)
     
  4. Feb 18, 2004 #3
    so its (1/2)kx^2 - (1/2)kx^2)

    but since initial is 0 we get -(1/2)kx^2

    and k=102 N/m and distance = 5.30 m

    so -(1/2)(102)(5.30^2) = -1432.59 J

    however -1432.59 J is not the right answer..am i missing something here?
     
  5. Feb 18, 2004 #4
    what is the right answer?
     
  6. Feb 18, 2004 #5
    i have no idea..its those websites where u input the answer and if u get it wrong it tells u its wrong and when u get it right it tells u its right
     
  7. Feb 18, 2004 #6
    I see.. have you tried rounding off to a different certain number of places? I know that did it for me last year for physics problems online when I used too many sigfigs
     
  8. Feb 18, 2004 #7

    Doc Al

    User Avatar

    Staff: Mentor

    The work done by the hose is positive since the force and displacement are in the same direction.
     
  9. Feb 18, 2004 #8
    thx!!
     
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