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.

1. NotTim

You're looking for "x"?

2. anonymous

Well constant is K.. I'm just confused by this whole chapter so far.

3. anonymous

I know it uses Hooke's Law, which is F = K * X

4. 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.

5. anonymous

1700 = K - 3?

6. anonymous

I mean K * 3

7. NotTim

I don't think so. Force is not work. And yes, 3k works for the right side.

8. NotTim

Work equals "Force times distance" right? W=f*d.

9. NotTim

So, I guess "d" is probably your "x". Sub in 1700 for W.

10. anonymous

F = -kx 1700/3 = -k(5-2) 1700/9 = -k k = -1700/9

11. anonymous

1700=3F

12. NotTim

@ Chris: 1700J isn't force...Is it?

13. NotTim

Yup. Rearrange 1700=3f so that 1700/3=f. There's your force.

14. anonymous

W = Fs 1700 = F(3) F = 1700/3

15. anonymous

None of those have worked for answers.. :/

16. anonymous

what are you trying to find

17. NotTim

Well that's only a part of it...

18. NotTim

He's looking for "K".

19. NotTim

But you basically have your "f" and "X". You already tried it with f as 566.666... and x as 3?

20. anonymous

yeah I tried +/ 1700/9 and 1700/3

21. anonymous

try this $F = \Delta K$ 1700 = (1/2)k(3)^2 - 0 (2*1700)/(9) = k

22. NotTim

Did you get 188.9?

23. anonymous

@Chris, that was correct!

24. anonymous

:D

25. anonymous

so how did you get that equation?

26. NotTim

Aww. I fail at physics.

27. anonymous

Yeah this is calculus 2, but for engineers.. Even though I'm a computer science major.

28. 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 $\int\limits_{0}^{x} F dx = \int\limits_{0}^{x}kx dx = (1/2) kx^2 |0 \to x = (1/2)kx^2 - 0$