Recent content by physicsjunkie1

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

4.7cm in diameter to be specific.

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

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.

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

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

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.

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

yes Ee is elastic potential

So is it Ee=Ek then ?

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.

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

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

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

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.