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Math Help - Science Application of Integrals

  1. #1
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    Science Application of Integrals

    A spring has a natural length of 10 in. An 800-lb force stretches the spring to 14 in.

    A. Find the force constant.
    work= force times displacement- though that applies here I do not know.

    B. How much work is done in stretching the spring from 10 in to 12 in?
    I have no idea.

    C How far beyond its natural length will a 1600-lb force stretch the spring?
    No idea.

    All I understand is the boundaries are [10,14].

    Thanks!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Truthbetold View Post
    A spring has a natural length of 10 in. An 800-lb force stretches the spring to 14 in.

    A. Find the force constant.
    work= force times displacement- though that applies here I do not know.

    B. How much work is done in stretching the spring from 10 in to 12 in?
    I have no idea.

    C How far beyond its natural length will a 1600-lb force stretch the spring?
    No idea.

    All I understand is the boundaries are [10,14].

    Thanks!
    Part (a):
    This is an application of Hooke's Law F=kx. Solving for k, we get k=\frac{F}{x}. F is the force that stretches the spring (800 lbs) and x is the displacement of the spring from rest (4 in). Thus, k=\frac{800lbs}{4in}=200\frac{lbs}{in}.


    Part (b):
    Now that we have our k value, we can substitute into the equation for force:
    F=200x.
    If we integrate force with respect to x, we get work. The limits of integration now are from rest (x=0) to the displaced distance (x=2). The integral can be written as:
    W=200\int_{0}^{2}x\,dx=200(\frac{x^2}{2})_{0}^{2}=  400.
    It takes 400 Joules of work to displace the spring 2 inches.

    Part (c)
    Again, we would use Hooke's Law. Since we're dealing with the same spring, we would still have the same k value. Since F=200x and F=1600, we see x (the displacement) is equal to \frac{1600{lbs}}{200\frac{lbs}{in}}=8in.
    Thus, a 1600 lb force can displace the spring 8 in

    Hope that answers your question!!!!
    Last edited by Chris L T521; May 7th 2008 at 09:10 PM. Reason: wrong units... :-\
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  3. #3
    Forum Admin topsquark's Avatar
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    You have the correct answer, but be warned. The unit for work here is NOT Joules! It is lb in. (The typical unit here would be lb ft.)

    -Dan
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by topsquark View Post
    You have the correct answer, but be warned. The unit for work here is NOT Joules! It is lb in. (The typical unit here would be lb ft.)

    -Dan
    Thanks for catching that!
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