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Math Help - Integrals and work

  1. #1
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    Integrals and work

    Question: If the work required to stretch a spring 1 ft beyond its natural length is 9 ft-lb, how much work W is needed to stretch it 10 in. beyond its natural length?

    So my teacher today gave us some physics equations and said to do this homework. Im a little stuck on this, usually they give us the force required to stretch a spring, not the work. Here is what i tried:

    W=FD
    so
    9=F*1ft
    F=9

    Then I did F=KX (hooke's law) to solve for the k constant of the spring
    so
    9 = K * 1 ft
    K = 9

    So I now know that F = 9X (equation for force of a spring when you know the k constant of that spring)
    And then i said the:
    INTEGRAL of: 9x from 0-10/12 = Total work needed to stretch the spring 10inches.
    I get 7.25, but this answer is wrong.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fogel1497 View Post
    Question: If the work required to stretch a spring 1 ft beyond its natural length is 9 ft-lb, how much work W is needed to stretch it 10 in. beyond its natural length?

    So my teacher today gave us some physics equations and said to do this homework. Im a little stuck on this, usually they give us the force required to stretch a spring, not the work. Here is what i tried:

    W=FD
    so
    9=F*1ft
    F=9

    Then I did F=KX (hooke's law) to solve for the k constant of the spring
    so
    9 = K * 1 ft
    K = 9

    So I now know that F = 9X (equation for force of a spring when you know the k constant of that spring)
    And then i said the:
    INTEGRAL of: 9x from 0-10/12 = Total work needed to stretch the spring 10inches.
    I get 7.25, but this answer is wrong.
    you are right up until where you set up the integral. so something must have happened after that

    the work in ft-lb is given by:

    W = \int_0^{10/12}9x~dx = \frac 92x^2 \bigg|_0^{10/12}
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  3. #3
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    Your answer comes out to be 3.125, which my online homework tells me is an incorrect value.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fogel1497 View Post
    Question: If the work required to stretch a spring 1 ft beyond its natural length is 9 ft-lb, how much work W is needed to stretch it 10 in. beyond its natural length?

    So my teacher today gave us some physics equations and said to do this homework. Im a little stuck on this, usually they give us the force required to stretch a spring, not the work. Here is what i tried:

    W=FD
    so
    9=F*1ft
    F=9

    Then I did F=KX (hooke's law) to solve for the k constant of the spring
    so
    9 = K * 1 ft
    K = 9

    So I now know that F = 9X (equation for force of a spring when you know the k constant of that spring)
    And then i said the:
    INTEGRAL of: 9x from 0-10/12 = Total work needed to stretch the spring 10inches.
    I get 7.25, but this answer is wrong.
    Here's a different approach:

    W=\int_a^b F(x)\,dx

    We are told that the distance stretched is 1 ft, and the work done is 9 ft-lbs.

    So we see that 9=\int_0^1 kx\,dx, where k is the spring constant.

    Thus, 9=\left.\left[\frac{1}{2}kx^2\right]\right|_0^1\implies 9=\tfrac{1}{2}k\implies \color{red}\boxed{k=18}

    Thus, when its stretched 10 inches, we see that W=18\int_0^{\frac{10}{12}}x\,dx=9\left.\bigg[x^2\bigg]\right|_0^{\tfrac{10}{12}}=9\left(\frac{10}{12}\ri  ght)^2=\color{red}\boxed{6.25~ft-lb}

    That should be the desired result.

    Does this make sense?

    --Chris
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