Expert: Paul Klarreich - 3/7/2007

Question
Hi again, Mr. Klarreich!  I really appreciate you helping me in the previous question I had. I'm in calculus 2 and I was wondering if you can help me on the following:

If 6 J of work is needed to stretch a spring from 10 cm to 12 cm and another
10 J is needed to stretch it from 12 cm to 14 cm, what is the natural length of the spring?

So far what I know I need to do is integrate the gorce needed to stretch a spring from x1 to x2, which is W= (k/2)[(x2-x0)^2-(x1-x0)^2] where x0 is the natural length of the spring. Thne from this I get two equations for the work needed to stretch the string over the 2 distances.

Could I go something like:
Definite Integral from (10-L) to (12-L) of kxdx=6
and then Definite Integral from (12-L) to 14-L) of kxdx?

Thank you for showing me how to do this!

Much appreciation,
Ali :)

Answer
Questioner:   Ali
Category:  Calculus
 
Subject:  Length of Spring and Calculus
Question:  Hi again, Mr. Klarreich!  I really appreciate you helping me in the previous question I had. I'm in calculus 2 and I was wondering if you can help me on the following:

If 6 J of work is needed to stretch a spring from 10 cm to 12 cm and another
10 J is needed to stretch it from 12 cm to 14 cm, what is the natural length of the spring?

So far what I know I need to do is integrate the force needed to stretch a spring from x1 to x2, which is W= (k/2)[(x2-x0)^2-(x1-x0)^2] where x0 is the natural length of the spring. Then from this I get two equations for the work needed to stretch the string over the 2 distances.

Could I go something like:
Definite Integral from (10-L) to (12-L) of kxdx=6
and then Definite Integral from (12-L) to 14-L) of kxdx?

Thank you for showing me how to do this!

Much appreciation,
Ali :)
...........................................................
Hi, Ali,

I think you are on the right track.  I would do the integration a little differently, but you are still OK.

As I recall, the restoring force exerted on/by a spring is  kx, where x is the 'displacement' from the equilibrium or 'rest' position, x0.

This is really an example of solving simultaneous equations, because you have two unknowns:  x0 and k, and two sets of facts.

W[10-12] = 6
W[12-14] = 10

W[10-12] = 6 =

{12
|   k(x - x0) dx =
}10

k(x^2/2 - x0 x)  from x = 10 to x = 12.

= k[(72 - 12x0) - (50 - 10x0)]
= k[72 - 12x0 - 50 + 10x0]
= k[22 - 2x0]

...................

W[12-14] = 10 =

{14
|   k(x - x0) dx =
}12

k(x^2/2 - x0 x)  from x = 12 to x = 14.

= k[(98 - 14x0) - (72 - 12x0)]
= k[26 - 2x0]

So the two equations are:
k[22 - 2x0] = 6
k[26 - 2x0] = 10

22k - 2kx0 = 6
26k - 2kx0 = 10
------------------
4k = 4,  so k = 1.

22 - 2x0 = 6
2x0 = 16,  so  x0 = 8