Springs Equilibrium Length Question

AI Thread Summary
To determine the force required to hold a spring at twice its equilibrium length, Hooke's Law (F = Kx) is applied. Given a spring constant (K) of 260 N/m and an equilibrium length of 0.25 m, extending the spring to 0.50 m results in an extension of 0.25 m. The force needed for this extension is calculated as F = 260 N/m * 0.25 m, yielding a force of 65 N. Therefore, the magnitude of the force required is 65 N.
grenier11
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Homework Statement



The equilibrium length of a certain spring with a force constant of K = 260N/m is 0.25m

What is the magnitude of the force that is required to hold this spring at twice its equilibrium length?

Homework Equations



Hookes Law F=Kx

The Attempt at a Solution



260 N/m = F/.25
twice the length is .50m so 260 N/m * .50 = 130
 
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Here's a picture:

http://solomon.physics.sc.edu/~tedeschi/demo/explain/images/hooks7.jpg

The equilibirium, original, X=0 point in this case is where X=0.25m. You're looking to extend the spring to 0.5m in length - that's an extension of 0.25m. So I think the correct figure should be 260 * 0.25 = 65N
 
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