Conservation of energy - spring problem

[shawli]

Homework Statement

A 0.500 kg mass resting on a frictionless surface is attached to a horizontal spring with a spring constant of 45 N/m. When you are not looking, your lab partner pulls the mass to oneside and then releases it. When it passes the equilibrium position, its speed is 3.375 m/s. How far from the equilibrium position did your lab partner pull the mass before releasing it?

Homework Equations

Ee = 0.5 * k * x^2

Ek = 0.5 * m * v^2

Energy is conserved within the system: Einitial = Efinal

The Attempt at a Solution

At the equilibrium position, only kinetic energy is present.
At the "unknown" position, there is elastic potential energy because the mass has strayed from the equilibrium ...but since we don't know if the mass has been pulled to the maximum distance, I'm not sure whether or not there is kinetic energy. I'm not going to include it in my calculation because otherwise I'll have two unknowns ("x" and "v2")

Ek = Ee
0.5*m*v^2 = 0.5 * k * x^2
x = SQRT [ (m*v^2) / k ]
x = SQRT [ (0.500*3.375^2) / 45) ]
x = 0.36m

The answer is actually 0.19 m. I think I probably set up my equation incorrectly. Could someone look it over please?

 

[pgardn]

Homework Statement

A 0.500 kg mass resting on a frictionless surface is attached to a horizontal spring with a spring constant of 45 N/m. When you are not looking, your lab partner pulls the mass to oneside and then releases it. When it passes the equilibrium position, its speed is 3.375 m/s. How far from the equilibrium position did your lab partner pull the mass before releasing it?

Homework Equations

Ee = 0.5 * k * x^2

Ek = 0.5 * m * v^2

Energy is conserved within the system: Einitial = Efinal

The Attempt at a Solution

At the equilibrium position, only kinetic energy is present.
At the "unknown" position, there is elastic potential energy because the mass has strayed from the equilibrium ...but since we don't know if the mass has been pulled to the maximum distance, I'm not sure whether or not there is kinetic energy. I'm not going to include it in my calculation because otherwise I'll have two unknowns ("x" and "v2")

Ek = Ee
0.5*m*v^2 = 0.5 * k * x^2
x = SQRT [ (m*v^2) / k ]
x = SQRT [ (0.500*3.375^2) / 45) ]
x = 0.36m

The answer is actually 0.19 m. I think I probably set up my equation incorrectly. Could someone look it over please?

Well no matter how far your lab partner displaces the mass, this is the max amount of energy in the system for that displacement. So the total kinetic energy has to be equal to total elastic potential energy at any moment that there is no elastic pot energy (all ke) or at any moment when there is only elastic potential energy (thus no kinetic energy so no motion). And the red above proves you know when the system only has kinetic energy.

So if you use the equation you already have above, you only have one unknown, x.

 

[shawli]

Okay I think I understand what you're saying, that at the equilibrium position there is only kinetic energy and that this magnitude of kinetic energy should be equal to the energy at any other point in the spring ...For example a point where there is all elastic potential energy only (in other words, the maximum amplitude of the spring).

But if the spring is pulled to any point between the equilibrium, is there only elastic potential energy or is there a mixture of elastic potential energy and kinetic potential energy?

Because with my equation of Ek = Ee ...my answer is wrong, and it's really confusing me :|.

 

[shawli]

In this case Ek SHOULD equal to Ee even if the spring isn't pulled all the way to the max positions since when my partner pulled the spring, it still didnt have Ek (like the spring wasn't already going)?

 

[pgardn]

Okay I think I understand what you're saying, that at the equilibrium position there is only kinetic energy and that this magnitude of kinetic energy should be equal to the energy at any other point in the spring ...For example a point where there is all elastic potential energy only (in other words, the maximum amplitude of the spring).

But if the spring is pulled to any point between the equilibrium, is there only elastic potential energy or is there a mixture of elastic potential energy and kinetic potential energy?

Because with my equation of Ek = Ee ...my answer is wrong, and it's really confusing me :|.

If the mass is pulled a little bit away from the equilibrium position, you put a small amount of total mechanical energy into the system. If the mass is pulled far away from the equilibrium position you put a larger amount of energy into the system and this is calculated as 1/2*k*x^2. If you pull the mass away from the equlibrium position and just hold it there does it have elastic potential energy, Kinetic energy, or both?

When you let the mass go and it is moving on its way to the equilibrium position does it have elastic potential energy, Kinetic energy, or both?

When you let the mass go and it is moving and at is located right at the equilibrium position does it have elastic potential energy, Kinetic energy, or both?

 

[shawli]

Just kinetic energy?

And depending on how far you pull the mass away, the speed at the equilibrium will vary.

So Ek = Ee ...and the back of the textbook must be wrong ?

 

[pgardn]

Just kinetic energy?

And depending on how far you pull the mass away, the speed at the equilibrium will vary.

So Ek = Ee ...and the back of the textbook must be wrong ?

Sorry, I totally just got kicked off the internet. I get 0.356 meters. And I did not even see you had attempted the problem before, sorry.

 

[shawli]

Don't worry about it, your explanations helped me understand what I was doing. And I'm sure that IS the correct answer and not 0.19m, like the book says. Thanks for your help!