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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.

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  1. NotTim
    • 4 years ago
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    You're looking for "x"?

  2. anonymous
    • 4 years ago
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    Well constant is K.. I'm just confused by this whole chapter so far.

  3. anonymous
    • 4 years ago
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    I know it uses Hooke's Law, which is F = K * X

  4. NotTim
    • 4 years ago
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    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.

  5. anonymous
    • 4 years ago
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    1700 = K - 3?

  6. anonymous
    • 4 years ago
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    I mean K * 3

  7. NotTim
    • 4 years ago
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    I don't think so. Force is not work. And yes, 3k works for the right side.

  8. NotTim
    • 4 years ago
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    Work equals "Force times distance" right? W=f*d.

  9. NotTim
    • 4 years ago
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    So, I guess "d" is probably your "x". Sub in 1700 for W.

  10. anonymous
    • 4 years ago
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    F = -kx 1700/3 = -k(5-2) 1700/9 = -k k = -1700/9

  11. anonymous
    • 4 years ago
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    1700=3F

  12. NotTim
    • 4 years ago
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    @ Chris: 1700J isn't force...Is it?

  13. NotTim
    • 4 years ago
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    Yup. Rearrange 1700=3f so that 1700/3=f. There's your force.

  14. anonymous
    • 4 years ago
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    W = Fs 1700 = F(3) F = 1700/3

  15. anonymous
    • 4 years ago
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    None of those have worked for answers.. :/

  16. anonymous
    • 4 years ago
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    what are you trying to find

  17. NotTim
    • 4 years ago
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    Well that's only a part of it...

  18. NotTim
    • 4 years ago
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    He's looking for "K".

  19. NotTim
    • 4 years ago
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    But you basically have your "f" and "X". You already tried it with f as 566.666... and x as 3?

  20. anonymous
    • 4 years ago
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    yeah I tried +/ 1700/9 and 1700/3

  21. anonymous
    • 4 years ago
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    try this \[F = \Delta K\] 1700 = (1/2)k(3)^2 - 0 (2*1700)/(9) = k

  22. NotTim
    • 4 years ago
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    Did you get 188.9?

  23. anonymous
    • 4 years ago
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    @Chris, that was correct!

  24. anonymous
    • 4 years ago
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    :D

  25. anonymous
    • 4 years ago
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    so how did you get that equation?

  26. NotTim
    • 4 years ago
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    Aww. I fail at physics.

  27. anonymous
    • 4 years ago
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    Yeah this is calculus 2, but for engineers.. Even though I'm a computer science major.

  28. anonymous
    • 4 years ago
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    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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