Calculating Force Constant of a Spring

[aishax]

A spring is suspended from a celing and a 0.256-kg mass is attatched to it and pulled down to stretch the spring by 0.182 m. The mass is released and travels through the equilibrium position with a speed of 0.746 m/s. Calculate the force constant on the spring.

I'm not really sure how to tackle this question.
What I ended up trying was looking at the two points being addressed (the point where the spring is stretched 0.182 meters downwards, and the equilibrium).

Using the conservation of energy, I put the equation:

E total1 = E total2

(Where E total1 is at the point where the spring is stretched, and E total2 is the point at which the spring passes the equilibrium)

I then put:

Ep + Ek + Ee = Ep + Ek + Ee

I canceled out Ep ( because h = 0), and Ek (because v = 0) on the left side of the equation. I canceled out Ee on the right side of the equation because x = 0. The equation became:

Ee = Ep +Ek


So far, am I on the right track, or is what I did completely wrong? Could this question be done a simplier way using the formula: Fs = -kx ?

Help is very much appreciated :)

 

[PhanthomJay]

A spring is suspended from a celing and a 0.256-kg mass is attatched to it and pulled down to stretch the spring by 0.182 m. The mass is released and travels through the equilibrium position with a speed of 0.746 m/s. Calculate the force constant on the spring.

I'm not really sure how to tackle this question.
What I ended up trying was looking at the two points being addressed (the point where the spring is stretched 0.182 meters downwards, and the equilibrium).

Using the conservation of energy, I put the equation:

E total1 = E total2

(Where E total1 is at the point where the spring is stretched, and E total2 is the point at which the spring passes the equilibrium)

I then put:

Ep + Ek + Ee = Ep + Ek + Ee

I canceled out Ep ( because h = 0), and Ek (because v = 0) on the left side of the equation. I canceled out Ee on the left
you mean right[/color]side of the equation because x = 0. The equation became:

Ee = Ep +Ek


So far, am I on the right track, or is what I did completely wrong?

it is completely right

Could this question be done a simplier way using the formula: Fs = -kx ?

This would be a much harder way to solve it, with errors guaranteed!

 

[Kreamer]

At the point where it is stretched fully all the energy is in the form of potential energy. At the equilibrium point all energy in Kinetic.

Let's start at the equilibrium. Energy is all Kinetic so..
E=(1/2)mv^2
E=(1/2)(0.256)0.746^2
E=.0712

Now since that energy is conserved the energy at the equilibrium point should be the same as at the beginning when it is stretched however it is in the form of potential energy. So...

E=(1/2)Ks*S^2
.0712=(1/2)Ks*(.182)^2
Ks=.0712/[(1/2)(.182)^2]
Ks=4.301

Hope I did that all right. If not hopefully it at least helps you out some.

 

[aishax]

Phew, good to know I'm getting somewhere. Thanks PhanthomJay!
And thank you for your input Kreamer! You just forgot to include elastic potential energy :)

 

[Kreamer]

I had a feeling I was forgetting something. We just learned about all that recently as well so I don't claim to be an expert ;) glad I could help though!