anonymous
  • anonymous
It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.
Mathematics
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SOLVED
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katieb
  • katieb
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NotTim
  • NotTim
You're looking for "x"?
anonymous
  • anonymous
Well constant is K.. I'm just confused by this whole chapter so far.
anonymous
  • anonymous
I know it uses Hooke's Law, which is F = K * X

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More answers

NotTim
  • NotTim
X is 5-2. Use the Work to find the F (there a formula? I forgot) You're looking for "K" I guess. Ignore the already given constant.
anonymous
  • anonymous
1700 = K - 3?
anonymous
  • anonymous
I mean K * 3
NotTim
  • NotTim
I don't think so. Force is not work. And yes, 3k works for the right side.
NotTim
  • NotTim
Work equals "Force times distance" right? W=f*d.
NotTim
  • NotTim
So, I guess "d" is probably your "x". Sub in 1700 for W.
anonymous
  • anonymous
F = -kx 1700/3 = -k(5-2) 1700/9 = -k k = -1700/9
anonymous
  • anonymous
1700=3F
NotTim
  • NotTim
@ Chris: 1700J isn't force...Is it?
NotTim
  • NotTim
Yup. Rearrange 1700=3f so that 1700/3=f. There's your force.
anonymous
  • anonymous
W = Fs 1700 = F(3) F = 1700/3
anonymous
  • anonymous
None of those have worked for answers.. :/
anonymous
  • anonymous
what are you trying to find
NotTim
  • NotTim
Well that's only a part of it...
NotTim
  • NotTim
He's looking for "K".
NotTim
  • NotTim
But you basically have your "f" and "X". You already tried it with f as 566.666... and x as 3?
anonymous
  • anonymous
yeah I tried +/ 1700/9 and 1700/3
anonymous
  • anonymous
try this \[F = \Delta K\] 1700 = (1/2)k(3)^2 - 0 (2*1700)/(9) = k
NotTim
  • NotTim
Did you get 188.9?
anonymous
  • anonymous
@Chris, that was correct!
anonymous
  • anonymous
:D
anonymous
  • anonymous
so how did you get that equation?
NotTim
  • NotTim
Aww. I fail at physics.
anonymous
  • anonymous
Yeah this is calculus 2, but for engineers.. Even though I'm a computer science major.
anonymous
  • anonymous
F = -kx <===== this is for when x's natural position is at 0 You have to integrate it if the natural position is elsewhere \[\int\limits_{0}^{x} F dx = \int\limits_{0}^{x}kx dx = (1/2) kx^2 |0 \to x = (1/2)kx^2 - 0\]

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