Grade 12 Physics-Finding the Spring Constant

[physicsjunkie1]

Homework Statement

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.

Homework Equations

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

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.

 

[voko]

Homework Statement

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.

Homework Equations

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

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)

Good so far.

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)

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

 

[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.

 

[voko]

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

 

[physicsjunkie1]

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

 

[voko]

Where does the initial kinetic energy come from?

 

[physicsjunkie1]

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

 

[voko]

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

 

[physicsjunkie1]

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

 

[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.

 

[voko]

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

 

[physicsjunkie1]

So is it Ee=Ek then ?

 

[voko]

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

 

[physicsjunkie1]

yes Ee is elastic potential

 

[physicsjunkie1]

Why don't you consider the fact that there is gravitational potential and that the elastic travels a certain disance?

 

[voko]

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

 

[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.

 

[voko]

I get about 30 N/m from your data. Show your working in details.

 

[physicsjunkie1]

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

 

[voko]

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

 

[physicsjunkie1]

Ee=Ek
(1/2)kx^2= (1/2)mv^2
(1/2)k(0.047)^2=(1/2)(3.0 x 10^-4)(4.967)^2
0.0011045k= 0.00309271
k= 0.00309271/0.0011045
k= 2.8

 

[voko]

(1/2)(3.0 x 10^-4)(4.967)^2 is not 0.00309271 but a little more.

Anyway, the result is close to 3 N/m which I am getting now. I think I did make a mistake in mass conversion.

Which brings me to this: are you sure it really was 0.3 g? That is very tiny. The rubber density is about 1000 kg per cubic meter, which makes that volume 0.0000003 cubic meter, or about a cube with 7 mm side; your reported extension is about 7 times that size.

 

[physicsjunkie1]

Im almost positive that is the correct mass and I googled masses of other rubber bands and they're all less than a gram. If the mass is larger it would make sense and that's the only place I can see that is the problem.

 

[voko]

The only realistic shape I can think that would have this mass is a thin and long strip of rubber; that should indeed have low stiffness. Can you describe the shape and its dimensions, even if approximate?

 

[physicsjunkie1]

It is just a few centimetres in diameter. Its a circular rubber band, kind of thick and pretty stiff.

 

[physicsjunkie1]

4.7cm in diameter to be specific.

 

[lightgrav]

a 1 gram rubber-band is 6mm wide, almost 1mm thick, and 180mm circumference.

hmm. that 47mm diameter is not accidently the same as the stretch?
that is, Ee = ½ k s^2 , where s is the distance the rubber has been stretched.

 

[physicsjunkie1]

The elastic was originally 4.6cm long and was stretched to 9.3cm long, for a difference of 4.7cm.

 

[voko]

No way that cylinder of rubber can be just 0.3 g. It is far too large for that. Either the dimension or the mass is way off.