# Grade 12 Physics-Finding the Spring Constant

1. Jan 19, 2014

### physicsjunkie1

1. The problem statement, all variables and given/known data
This was a mini-lab and the task was to fire a rubber band by stretching it a known distance, shooting it straight horizontally from a known height. Based on where the rubber band lands determine the elastic coefficient. Known data from experiment: mass=0.30g, height=0.412m, distance flung=1.362m, stretched length (Δx)=0.047m.

2. Relevant equations
d=vt
Ee=1/2k(Δx)^2
Eg=mgh
Ek=1/2m(v)^2

3. The attempt at a solution
We tried to find the time with the vertical component
d=Vit+1/2at^2, acceleration is 9.8m/s^2
we got t=0.28997s

Then we used the horizontal component to get the velocity
d=vt
v=4.697m/s (initial velocity)

then we found the final velocity using a=(Vf-Vi)/(t)
we got Vf=7.53m/s (right before the rubber band hits the ground)

Then we thought that Eg+Ee (lost)=Ek (gained)
We solved for k, but got 6.62, which is not the right value because it is too small. There must be a problem with our method. We think maybe it has something to do with the acceleration formula used to find the final velocity. We aren't sure what we did wrong but let us know if you have any corrections or any other methods to approach this question with.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 19, 2014

### voko

Good so far.

Why would you need the final velocity? The contribution of the rubber band ends when the initial velocity is obtained.

3. Jan 19, 2014

### physicsjunkie1

We thought we needed a final velocity to put into the Ek formula. Do we just use Vi for the v variable in Ek=1/2m(v)^2.

Last edited: Jan 19, 2014
4. Jan 19, 2014

### voko

That formula is generally correct. For the initial velocity, it gives you the initial kinetic energy.

5. Jan 19, 2014

### physicsjunkie1

So what would you do after finding the initial velocity in the d=vt formula?

6. Jan 19, 2014

### voko

Where does the initial kinetic energy come from?

7. Jan 19, 2014

### physicsjunkie1

The elastic potential plus the gravitational potential energy is converted into kinetic energy.

8. Jan 19, 2014

### voko

Gravitational potential energy is converted into the initial kinetic energy? How?

9. Jan 19, 2014

### physicsjunkie1

We flung the rubber band from a height of 0.412m, so it initially had Eg.

10. Jan 19, 2014

### physicsjunkie1

So at the beginning it had Eg and Ee and then when the rubber band landed 1.3m away, it had all converted to Ek, we thought.

11. Jan 19, 2014

### voko

I thought we sorted out the (ir)relevance of "landing" in posts #1 - #3.

12. Jan 19, 2014

### physicsjunkie1

So is it Ee=Ek then ?

13. Jan 19, 2014

### voko

I am not sure what those symbols mean. Is Ee the elastic potential energy? Then it is correct.

14. Jan 19, 2014

### physicsjunkie1

yes Ee is elastic potential

15. Jan 19, 2014

### physicsjunkie1

Why dont you consider the fact that there is gravitational potential and that the elastic travels a certain disance?

16. Jan 19, 2014

### voko

Again: what relevance does the gravity have on the work of the rubber band?

17. Jan 19, 2014

### physicsjunkie1

Okay, I see what you're saying now but when I use Ee=Ek and sove for the spring constant, I only get 1.9.

18. Jan 19, 2014

### voko

19. Jan 19, 2014

### physicsjunkie1

are you converting the mass correctly? Im still getting a different answer

20. Jan 19, 2014

### voko

I can offer no further help unless I see exactly what you do.