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Math Help - Physics Spring problem. Work is shown!

  1. #1
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    Physics Spring problem. Work is shown!

    Suppose that 2 Joules of work is needed to stretch a spring from its natural length of 35 cm to a length of 54 cm. How much work is needed to stretch it from 50 cm to 55 cm?

    First change cm to m.
    W = the integral from Xi to Xf of kx dx

    so W=(1/2)k(Xf^2) - (1/2)k(Xi^2)
    So solve for k
    2=(1/2)(k(.54^2) - k(.35^2))
    4=k(.1691)
    k=23.6546

    them solve for W with your new k value but with a different distance
    W=(23.6546/2)(.55^2 - .50^2)
    W=.620975 J

    But the answer is wrong....
    Last edited by softballchick; March 19th 2011 at 01:52 PM. Reason: wrote the wrong answer, corrected it. Still wrong though
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  2. #2
    Junior Member RaisinBread's Avatar
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    Not sure if you simply made a typo, but I get W=0.6209 instead of 0.06209. Also do you know what is the right answer, or just that yours is wrong?
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  3. #3
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    Quote Originally Posted by RaisinBread View Post
    Not sure if you simply made a typo, but I get W=0.6209 instead of 0.06209. Also do you know what is the right answer, or just that yours is wrong?

    Hi, I made a typo on typing the answer. I did also get .6209 which is the answer I put and it was wrong...

    I have no idea what I did wrong. It is in Joules right? I am pretty sure it is..... so frustrating!
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  4. #4
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    W = \dfrac{1}{2} kx^2

    where x is the length of stretch or compression in meters measured from the spring's natural length.

    if the spring's natural length is 35 cm and the spring is stretched to a length of 54 cm , then ...

    x = .54 - .35 = .19

    ... use this value of x to determine the spring constant, k , then determine the work required to stretch the spring from 50 to 55 cm
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