
New Member
Natural length of a spring given work
If 6 J of work are needed to stretch a spring from 10 cm to 12 cm and another 10 J are needed to stretch it from 12 cm to 14 cm, what is the natural length of the spring?
I understand that F = Kx and that work is the integration of force but I dont understand how to set this problem up and find the natural length of the spring.
Any hints are appreciated. Thanks in advance.

Elite Member
If work is the integral of force, then integrate the "force" equation you included.
Note that you have two definite integrals: from 10 to 12 and from 12 to 14. You can use this to find the constant "K", etc.
If you get stuck, please reply showing what you have tried and any progress you have made. Thank you.
Eliz.

Senior Member
let a = natural length of the spring in cm
[tex]\L \int_{.10a}^{.12a} kx dx = 6[/tex]
[tex]\L \frac{k}{2}[(.12a)^2  (.10a)^2] = 6[/tex]
[tex]\L k[(.0144  .24a + a^2)  (.01  .2a + a^2)] = 12[/tex]
[tex]\L k(.0044  .04a) = 12[/tex] (1)
[tex]\L \int_{.12a}^{.14a} kx dx = 10[/tex]
[tex]\L \frac{k}{2}[(.14a)^2  (.12a)^2] = 10[/tex]
[tex]\L k[(.0196  .28a + a^2)  (.0144  .24a + a^2)] = 20[/tex]
[tex]\L k(.0052  .04a) = 20[/tex] (2)
from the first equation ...
[tex]\L k = \frac{12}{.0044  .04a}[/tex]
substitute the above in for k in the second equation ...
[tex]\L \frac{12}{.0044  .04a}(.0052  .04a) = 20[/tex]
[tex]\L 12(.0052  .04a) = 20(.0044  .04a)[/tex]
[tex]\L .0624  .48a = .088  .8a[/tex]
[tex]\L .32a = .0256[/tex]
[tex]\L a = .08[/tex]
the natural length of the spring is 8 cm.
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