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Math Help - Another work and force problem

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    Another work and force problem

    Suppose that 3 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 42 cm.
    (a) How much work W in joules, is needed to stretch the spring from 36 cm to 44 cm?

    (b) How far beyond its natural length will a force of 40 N keep the spring stretched? (Round the answer to the nearest tenth.)

    Ok so work is force times distance, so I can assume that 3=16d and that the force is 3/16

    So then I try 8 times 3/16 which gives me 1.5,b ut that's incorrect. So what am I doing wrong?

    And then for b, I have no idea how to even solve it backwards like that. Could someone point me in the right direction?
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    Quote Originally Posted by fattydq View Post
    Suppose that 3 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 42 cm.
    (a) How much work W in joules, is needed to stretch the spring from 36 cm to 44 cm?

    (b) How far beyond its natural length will a force of 40 N keep the spring stretched? (Round the answer to the nearest tenth.)

    Ok so work is force times distance, so I can assume that 3=16d and that the force is 3/16

    So then I try 8 times 3/16 which gives me 1.5,b ut that's incorrect. So what am I doing wrong?

    And then for b, I have no idea how to even solve it backwards like that. Could someone point me in the right direction?

    What are the units of J. (Hint they do not have any cm in them)

    You need to convert first.
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    Quote Originally Posted by TheEmptySet View Post
    What are the units of J. (Hint they do not have any cm in them)

    You need to convert first.
    Yeah but I get the same answer. Because 3=.16x 3 divided by .16 is 18.75, 18.75 times .08 is 1.5. See what I mean?
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    Quote Originally Posted by fattydq View Post
    Yeah but I get the same answer. Because 3=.16x 3 divided by .16 is 18.75, 18.75 times .08 is 1.5. See what I mean?

    Remember with a spring that your force is not constant!!

    F=k\cdot x Hooks law

    and

    W=\int F dx

    I hope this helps
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    Quote Originally Posted by TheEmptySet View Post
    Remember with a spring that your force is not constant!!

    F=k\cdot x Hooks law

    and

    W=\int F dx

    I hope this helps
    ok so how do I figure out what is x and what is k? Wouldn't x just be .26? So I'd have 3=.26x, 3 divided by .26 is 11.53846154 and that times .08 is another long decimal, which is an incorrect answer. What am I doing wrong!?!!?
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    Ok so 3=.16k and F would equal 18.75x. 18.75x^2/2 evaluated at 1.8, minus it evaluated at .1 gives me .21, still a wrong answer.
    Last edited by fattydq; March 5th 2009 at 12:35 PM.
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    OK I just realized something, they give you the WORK needed to hold a spring at those dimensions, not force, so now I'm REALLY lost! How does that come into play?
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    Quote Originally Posted by fattydq View Post
    OK I just realized something, they give you the WORK needed to hold a spring at those dimensions, not force, so now I'm REALLY lost! How does that come into play?
    3=\int_{0}^{0.16}kx dx

    3=\int_{0}^{.16} kx dx=\frac{k}{2}x^2 \bigg|_{0}^{.16}=\frac{k(.16)^2}{2}<br />

    \frac{6}{(.16)^2}=k


    Now we know that F=\frac{3}{(0.16)^2}x \iff \frac{F}{\frac{3}{(0.016)^2}}=x

    Now we just plug in 40 for F and we are done.
    Last edited by Jameson; March 5th 2009 at 06:57 PM. Reason: corrections made by request
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    Quote Originally Posted by TheEmptySet View Post
    3=\int_{0}^{0.16}kx dx

    3=kx^2\bigg|^{0.16}_{0} \iff 3 =k(0.16)^2 \iff k=\frac{3}{(0.16)^2}

    Now we know that F=\frac{3}{(0.16)^2}x \iff \frac{F}{\frac{3}{(0.016)^2}}=x

    Now we just plug in 40 for F and we are done.
    But then my answer would just be .00, and since the problem asks me to round to the nearest hundredth....00 is an incorrect answer.
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    Quote Originally Posted by TheEmptySet View Post
    3=\int_{0}^{0.16}kx dx

    3=kx^2\bigg|^{0.16}_{0} \iff 3 =k(0.16)^2 \iff k=\frac{3}{(0.16)^2}

    Now we know that F=\frac{3}{(0.16)^2}x \iff \frac{F}{\frac{3}{(0.016)^2}}=x

    Now we just plug in 40 for F and we are done.
    I thought F=kx not kx^2??
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