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anonymous
 4 years ago
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.
anonymous
 4 years ago
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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well constant is K.. I'm just confused by this whole chapter so far.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know it uses Hooke's Law, which is F = K * X

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0X is 52. Use the Work to find the F (there a formula? I forgot) You're looking for "K" I guess. Ignore the already given constant.

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0I don't think so. Force is not work. And yes, 3k works for the right side.

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0Work equals "Force times distance" right? W=f*d.

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0So, I guess "d" is probably your "x". Sub in 1700 for W.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0F = kx 1700/3 = k(52) 1700/9 = k k = 1700/9

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0@ Chris: 1700J isn't force...Is it?

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0Yup. Rearrange 1700=3f so that 1700/3=f. There's your force.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0W = Fs 1700 = F(3) F = 1700/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0None of those have worked for answers.. :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what are you trying to find

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0Well that's only a part of it...

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0But you basically have your "f" and "X". You already tried it with f as 566.666... and x as 3?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah I tried +/ 1700/9 and 1700/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0try this \[F = \Delta K\] 1700 = (1/2)k(3)^2  0 (2*1700)/(9) = k

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Chris, that was correct!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how did you get that equation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah this is calculus 2, but for engineers.. Even though I'm a computer science major.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0F = 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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