# Homework Help: Finding a Spring Constant

1. Jul 11, 2014

### onelove8187

The question I am having difficulty with states that it requires 123 J of work to stretch a very light ideal spring from a length of 1.4m to a length of 2.9m. What is the value of the spring constant?

My thinking was that the work required would be equal to the spring force so I set up 123=.5k(1.5)^2 getting 109.33 which is wrong. I dont just want the answer I would like to understand how this problem works. Thanks!

2. Jul 12, 2014

### TumblingDice

I'm not an expert in any way in this area. I did a couple of searches out of my own curiosity...

If the 1.4 meter length is the equilibrium point, Hooke's Law should apply. However I think Joules are units of work, and need to be converted to units of force to use with F=kX, or in this case, k=F/X . Perhaps convert Joules to Newtons... (Is that right?)

EDIT: What a dummy I am. More reading and I see Joules are equal to newton-meters. <sigh> My apologies for excessive eagerness to try to help. Someone should step in soon with proper help.

Last edited: Jul 12, 2014
3. Jul 12, 2014

### Nathanael

He used the "potential spring energy equation" (I just made up the name) which is simply the integration of Hooke's law with respect to length ($=\frac{kx^2}{2}$)
(If you've learned: Energy is the integration of force with respect to distance. That is why the integration of hooke's law yields the energy used in stretching it, which is also the energy re-gained when it unstretches)

Are you given the resting length of the spring? (or any other information?)

If I'm not mistaken, the energy required to stretch the spring by an additional 1.5 meters depends on how far the spring is already stretched. So your answer would be correct if the resting length of the spring was 1.4m.

However, if the resting length is, for example, 1 meter, then the spring will have a lower constant. To calculate it you would integrate F=-kx from 0.4 to 1.9 (instead of from 0 to 1.5, like you did) giving you 71 as the spring's constant for a resting length of 1 meter (just an example)

So I think that not enough information is known.

4. Jul 12, 2014

### Delta²

Assuming that 1.4m and 2.9m do not refer to the length of the spring but to the displacements from the equilibrium position, it would be

123=0.5k(2.9)^2-0.5k(1.4)^2.

That is to find the energy needed (or the work need to be done) to strech the spring from 1.4m to 2.9m we subtract the the initial energy at 1.4m from the final energy at 2.9m. It is wrong to subtract the displacements first , to find energy we subtract or add energies we dont subtract or add displacements.

Last edited: Jul 12, 2014
5. Jul 12, 2014

### Staff: Mentor

Hi onlove8187. Welcome to Physics Forums. Your answer looks correct to me. Maybe you have a issue with significant figures. Maybe the answer should be 109 N/m. What value do they give, if any, for the "correct" answer?

Chet