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Math Help - Using Hooke's Law to solve a problem

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    Senior Member vaironxxrd's Avatar
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    Using Hooke's Law to solve a problem

    I have an assignment with the following problems...

    The natural length of a certain spring is 16 inches and a force of 8 pounds is required to keep it stretched 8 inches. Find the work done in each case.

    (a) stretching it from a length of 18 inches to a length of 24 inches.

    First, I'll find the K constant and also convert from inches to feet to make my life easier.

    If it's being extended 8 inches it's natural length, that should be 2/3 foot its natural length.

    Here's the part where I'm slightly confused. I solved for K using two different methods and I'm getting two different answers

    k (2/3) = 8lb

    k = \frac{8}{\frac{2}{3}} = 12 ft.

    However, using this method I get the following.

    8 = \int_0^{2/3} (kx)dx = (\frac{kx^2}{2})_0^{2/3}

    8 = \frac{k(2/3)^2}{2} = 36

    I did solve this problem using 36 and obtained 15/2, but now I'm not sure if this was the correct method.

    (b) Compressing it from its natural length to a length of 12 inches

    Regarding this compression problem, I'm also a bit confused since my assignment didn't show a compressing example. I'm assuming the integral would look something like \int_0^{-1/3} = -\int_{-1/3}^0
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    Re: Using Hooke's Law to solve a problem

    I'm not sure converting to feet is going to work better for you. To get k, you just divide: F=kx, so k=F/x=8 lb/8 in = 1 lb/in.

    Work is positive if the force is in the same direction as the change in length and negative if the force is in the opposite direction as the change in length.

    So in (a), you should have the integral from 2 to 8 of kx dx. The x is the change in length from its rest position, so (18-16) to (24-16). The force is in the expanding direction and the change in length is in the expanding direction.

    In (b), you should have the integral from 0 to -4 of kx dx. Now the force is in the contracting direction and the change in length is in the contracting direction.

    In either case you are doing work on the spring, so the answer is positive.

    - Hollywood
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    Senior Member vaironxxrd's Avatar
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    Re: Using Hooke's Law to solve a problem

    Quote Originally Posted by hollywood View Post
    I'm not sure converting to feet is going to work better for you. To get k, you just divide: F=kx, so k=F/x=8 lb/8 in = 1 lb/in.

    Work is positive if the force is in the same direction as the change in length and negative if the force is in the opposite direction as the change in length.

    So in (a), you should have the integral from 2 to 8 of kx dx. The x is the change in length from its rest position, so (18-16) to (24-16). The force is in the expanding direction and the change in length is in the expanding direction.

    In (b), you should have the integral from 0 to -4 of kx dx. Now the force is in the contracting direction and the change in length is in the contracting direction.

    In either case you are doing work on the spring, so the answer is positive.

    - Hollywood
    Thanks!

    Sorry for taking a while to reply.

    Here's my solution...

    (a)

    k = 1 lb/in (as you explained)

    \int_2^8{1xdx} = \left[\frac{x^2}{2}\right]_2^8 = 32-2 = 30 inlb or 2.5 ftlb

    (b)
    k is still 1lb/in

    \int_0^{-4}{1xdx} = \left[\frac{x^2}{2}\right]_0^{-4} = 8 inlb or \frac{2}{3} ftlb

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    Re: Using Hooke's Law to solve a problem

    That looks correct.

    - Hollywood
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