about spring constant and Hooke's law


by KFC
Tags: constant, hooke, spring
KFC
KFC is offline
#1
Feb25-13, 05:15 PM
P: 369
Hi all,
I am reading an online material on elastic force and Hooke's law on spring. The definition of the Hooke's law reads that the restoring force is linear proportional to the displacement of the spring with constant k. The restoring force is defined as the force bringing the object back to the equilibrium position and k characterize the system's (spring's) nature. So if we have a spring with equilibrium position sitting in the origin of the coordinate so the restoring force F satisfies
$$F = -k x$$
Now, if there is a way to add a second force ($F'$) which is exactly the same as $F$ (same direction and magnitude), since $F'$ is along $F$ all the time, so $F'$ always pointing toward the equilibrium position too. In this sense, should I conclude that
$$F+F' = 2F = -k' x$$
if so, can we say that by adding an additional force $F'$, the spring constant k changed to k'=k/2 ? But I think the spring constant should be given by the spring only, so how does the paradox come from?
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rock.freak667
rock.freak667 is offline
#2
Feb25-13, 05:18 PM
HW Helper
P: 6,214
The spring constant k will be constant for a spring of given dimensions.

If you have a force F applied on the spring such that F = kx and you apply a second identical force then the extension 'x' will just double as well.
KFC
KFC is offline
#3
Feb25-13, 05:24 PM
P: 369
Quote Quote by rock.freak667 View Post
The spring constant k will be constant for a spring of given dimensions.

If you have a force F applied on the spring such that F = kx and you apply a second identical force then the extension 'x' will just double as well.
Thanks rock. You are right, I totally forget that the displacement should be changed too if I double the force.


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