Spring constant 54 N/m, mass extends it to 35 cm

[audreylynn]

Problem)
A coil spring has a spring constant of 54 N/m. If the full length of the spring is 35 cm when a 1.0-kg mass is hung from it, what is the equilibrium length of the spring when the 1.0-kg mass is removed?

I have absolutely no idea how to even start this problem; and I know I can look in the back of the book to get the answer, but I'd reallylike to understand the problem. Any help would be greatly appreciated. :]

 

[tiny-tim]

Welcome to PF!

Hi audreylynn! Welcome to PF!

Hooke's law says that the extension of a spring (the stretched length minus the equilibrium length) is proportional to the force.

Hint: call the equilibrium length x, then the extension is 35 - x.

 

[Stonebridge]

The spring constant value (54N/m) tells you that for every 54 N of force you apply to the spring it will (if it is physically able to) stretch one meter.
Your one kilo mass has a weight of 9.8 Newtons, which it can apply to the spring if hung on the end.
If 54 N stretches it 1 meter, how far could 9.8 N stretch it?
When you have done this, you know how much the string has stretched (with mass attached) beyond its original length (no mass attached).
You are told how long the spring is with the weight on, so ...

 

[audreylynn]

Thank you so much! :]

So I would use the equation F=kx with m•g=F.
Then I could just plug in the numbers to the formula as...
9.81=54(x)
9.81=54(.35-x)

Like that's how I'd do it?

 

[tiny-tim]

9.81=54(.35-x)

That's it!

(but never use the same letter for two different things! )

 

[audreylynn]

Thank you. :]