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

- chestercat

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- NotTim

You're looking for "x"?

- anonymous

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

- anonymous

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

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## More answers

- 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

1700 = K - 3?

- anonymous

I mean K * 3

- NotTim

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

- NotTim

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

- NotTim

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

- anonymous

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

- anonymous

1700=3F

- NotTim

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

- NotTim

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

- anonymous

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

- anonymous

None of those have worked for answers.. :/

- anonymous

what are you trying to find

- NotTim

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

- NotTim

He's looking for "K".

- NotTim

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

- anonymous

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

- anonymous

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

- NotTim

Did you get 188.9?

- anonymous

@Chris, that was correct!

- anonymous

:D

- anonymous

so how did you get that equation?

- NotTim

Aww. I fail at physics.

- anonymous

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

- 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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