Question

Launching a spring at an angle from an elevated surface: derive a simplified expression for "x," stretch of the spring, using conservation of energy.

Answer 1

You're launching the spring?

Answer 2

Well, you're creating a formula, so you'll work with general variables.

Also, you have to use conservation of energy because it asked you to.

Conservation of energy says this:
The energy change of everything is zero. \(\Delta E =0\)
That means the energy before is equal to the energy after.
For a system, if energy enters, then the same amount leaves.

For this, you want to know that to start the energy is potential from gravity and potential from spring energy.

That magnitude of energy will remain constant.
Now, as the spring changes, it's potential is now realized! Potential energy becomes kinetic. How much the potential changes (and so the kinetic) depends on \(x\).

Good luck! Feel free to ask questions.

Answer 3

So far we've separated the problem into two parts, the conservation of energy section at the beginning (before launching the spring) and then the kinematics section afterwards. For the first part we thought it would be just Ee = Ek where all the potential energy in the spring (being pulled back) is converted into kinetic energy. However we're not entirely sure where to go from here. Sorry for the essay ^^;

Answer 4

@KinzaN

Answer 5

For the start of the launch, before the spring is released, we were advised not to take gravitational potential energy into consideration...

Answer 6

Well, the prompt states that the surface is "elevated." Wouldn't this imply that it had height, and could potentially fall to the ground, making the ground below a good reference point for gravitational potential energy?
Isn't the least possible height good for a \(0\) gravitational potential energy?

I like working with people. If I say something that's confusing, just let me know where it caught you.

Answer 7

The elevation I though of was:
Notice that it could be any value.