anonymous
  • anonymous
Please answer this, because I am suffering horrible calculus-related feelings of self doubt, shattered dreams, and existential dread. Suppose that 2J of work is needed to stretch a spring from its natural length of 30cm to a length of 42cm. (a) I've already solved this, and every time someone else has posted both a and b, they only answer a and for some reason ignore that b exists. (b) How far beyond its natural length will a force of 30N keep the spring stretched? Hooke's Law: F = kx k = 138.9 The answer is 10.8cm. Explain step by step because I feel more stupid the longer I spend on these godforsaken problems. Explain in integral form. Please.
Mathematics
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anonymous
  • anonymous
Please answer this, because I am suffering horrible calculus-related feelings of self doubt, shattered dreams, and existential dread. Suppose that 2J of work is needed to stretch a spring from its natural length of 30cm to a length of 42cm. (a) I've already solved this, and every time someone else has posted both a and b, they only answer a and for some reason ignore that b exists. (b) How far beyond its natural length will a force of 30N keep the spring stretched? Hooke's Law: F = kx k = 138.9 The answer is 10.8cm. Explain step by step because I feel more stupid the longer I spend on these godforsaken problems. Explain in integral form. Please.
Mathematics
schrodinger
  • schrodinger
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Mimi_x3
  • Mimi_x3
whats question a?
anonymous
  • anonymous
(a) How much work is needed to stretch the spring from 35cm to 40cm? I am only posting this for you in the hopes it will give you useful information. I have already solved this problem, so if you try to tell me its answer, there isn't a point and it will only depress me further.
anonymous
  • anonymous
isnt it just \(F=kx\) for part b ?

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Mimi_x3
  • Mimi_x3
you use hookes law for b
anonymous
  • anonymous
Can you work that out for me, because I've tried working that out, checking it against W = Fd, etc. etc. and I keep getting the wrong answer. Oh yeah, and the answer to part b is 10.8cm. Doing what you said gives me .22
anonymous
  • anonymous
F = kx; 30 = 138.9x; x = .2159(...) So I thought hey, maybe I plug it into b for the integral from a to b. \[W = \int\limits_{a}^{b} f(x)dx\] to get\[\int\limits_{0}^{.22} 138.9(x)dx\] but it came out as 3.239740821 when I integrated it. I also tried finding W = Fd in comparison to F = kx. W/d = 30; .2159(...) = x. I can't compare them at all. It's useless.
anonymous
  • anonymous
I also tried checking the equation against previous equations in the problem. Nothing works. I can hardly find any examples of this problem on the internet, and the ones I can find don't answer b, as I mentioned. What do I do? what is there to do?
Mimi_x3
  • Mimi_x3
are you sure that the answer you got for a is correct?
anonymous
  • anonymous
1.04J. If I'm wrong, tell me why. The book says it's 1.04J. So I don't know.
anonymous
  • anonymous
a doesn't matter, like I said. It's b that's giving me an existential crisis.
Mimi_x3
  • Mimi_x3
because a is related to b; so it does matter
anonymous
  • anonymous
They're separate problems. a solves for a variable depending on two limits in a definite integral, so it completely changes the result. It isn't connected to b in any way other than that it uses the spring constant. I'm not aware of a ratio law or anything. Please correct me if I'm wrong. I would literally love to hear why.
anonymous
  • anonymous
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Mimi_x3
  • Mimi_x3
lol i found that on google as well
anonymous
  • anonymous
lol
Mimi_x3
  • Mimi_x3
and i got a retarded answer; that is not the same as the answer given by the asker..
Mimi_x3
  • Mimi_x3
http://answers.yahoo.com/question/index?qid=20081015165946AAcZCdL this looks exactly the same; but the answer is different to stonemask answer
anonymous
  • anonymous
\[2=\int\limits\limits_{0}^{0.12}F(x)dx=\int\limits_{0}^{0.12}kxdx=\frac{ k(0.12)^{2} }{ 2 }\] \[k=277.78\] 30=277.78x x=.108m or 10.8cm You had k wrong.
anonymous
  • anonymous
How did I get the right answer for a with a wrong k? :( Thank you, also, @okaywhynot. This was a dark time in my life.

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