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Sping Problem



....
Much thanks to all who responded to my Spring Problem questions last
week. I'm beginning to see the light! With one more push in the right
direction, I might have this thing figured out. I'm like a vulture hovering
over this problem, ever circling, eventually diving in for the kill. However,
there is still one concept holding me back. If I can get this last point
figured out, I think I may have it.
Let's say we have a horizontal, massless, Hookean spring (k=2N/m)
attached at its left end to a rigid support. Let's say I pull with a CONSTANT
6N force to the right. The spring stretches out to position x=3m and comes to
rest, correct? Now, my question is in regard to the spring force which is
acting back toward the left as I exert my force to the right. Does it also
equal 6N in magnitude as the 3rd law would suggest, or does it vary from 0 to
-6N as F=-kx would suggest?
Any light that anyone could shed on this question will be greatly
appreciated.

If you are exerting a force of 6N to keep the spring extended 3m beyond
its equilibrium length, then the spring must be exerting a force of -6N
on you, i. e., Newton's third law certainly should apply.
(I think there may be some ambiguity in your statement "The spring
stretches out to position x = 3m." In Hooke's law the x in -kx refers to
distance from the equilibrium position of the spring's end, and it is not
measured from the rigid support to which the other end of the spring is
attached.)

A. R. Marlow E-MAIL: marlow@loyno.edu
Department of Physics, Box 124 PHONE: (504) 865 3647 (Office)
Loyola University 865 2245 (Home)
New Orleans, LA 70118 FAX: (504) 865 2453