Springs Equilibrium Length Question

[grenier11]

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

 

[cimon9999]

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