Finding spring length when mass is removed using Hooke's Law

[catch.yossarian]

A spring with k=45 N/m is .35m when pulled
down with a 1.0 kg mass, what is the length of the
spring when the mass is taken off? The answer in the
back of the book is .13m (13 cm), I can't for the life
of me figure out how they get that.

I know that Hooke's Law is: F = -kx

What I've done so far:

F = (45)(.35) = 15.75 ---> Force it takes to pull down .35 meters, apparently using 1.0kg. Thus, if I divide 15.75 by 9.81, I should get 1.0kg, yes?

15.75 / 9.81 = 1.60 --->

I must be doing something wrong, let's start backwards.

F = (45)(.13) = 5.85 --> Force it takes to pull down .13 meters (this x is the answer, but right now I'm trying to find a proper F). 5.85 / 9.81 = .56, again, not 1.0kg. :(

What am I doing wrong? Can anybody point me in the right direction?

 

[dextercioby]

Consider the spring without the mass.It has the length $x_{0}$.When acted by a force,its length is $x$.Hooke's law says that the magnitude of the force is proportional to the the length the spring is stretched under the influence of the force:
$$F_{el}=k\Delta x=k(x-x_{0})$$

Compute the force and $\Delta x$,then use the expression for the latter to solve your problem.

Daniel.

 

[catch.yossarian]

Got it, thanks!

Man, such a small thing, too.
*sigh of relief*