# Finding a spring constant

1. Oct 29, 2011

### Timebomb3750

Finding the spring constant of a spring...

1. The problem statement, all variables and given/known data
A mass is hung from a vertical spring. The mass is .200kg, and the spring stretches .086m. Find the spring constant.

2. Relevant equations
ΔE=ΔKE+ΔUg+ΔUs

3. The attempt at a solution
I first broke up the equation, to see which values would go to zero. I know ΔKE drops out, because the initial and final velocities would be zero. Eventually, I was left with this...
mgΔy=(1/2)K(x^2)

Then, I solved for K and got...
K=(2mgΔy)/(x^2)

Putting the numbers in, I get...
K=(2*.200kg*9.80m/s^2*.086m)/(.086m^2)

The final answer I got was approximately 45.6 N/M. I just want to make sure this is correct, along with one more quick thing. The next part of the problem asks, "If the mass is now lifted .086m back to the spring's unstretched position and released from rest, how far will it drop before recoiling back?"

I'm assuming that if this is a closed system, wouldn't the mass just drop down the same distance over and over again? (oscillate the same?) Or do I need to calculate the new distance?

Last edited: Oct 29, 2011
2. Oct 29, 2011

### Timebomb3750

Bump...

3. Oct 29, 2011

### Integral

Staff Emeritus

Did you consider using Hook's law, F=-kx ?

4. Oct 29, 2011

### Timebomb3750

No, didn't think of it at the time. Could I find the force with my given information, then use that to find K? Does this mean my answer is incorrect?

5. Oct 29, 2011

### QuarkCharmer

When the mass is at equilibrium you know that one force is mg right, then apply newton's third law et al. That might help you find the force you need for the equation.

6. Oct 29, 2011

### Integral

Staff Emeritus
You are given the mass, so you can compute the gravitational force.

Your result seems to big, done carefully your energy approach should work. I do not understand where your equations come from but you have lost something along the way. It would help if you gave more information. We need a definition of your variables... what is Δy?

Also a bit more history on how you arrived at your starting expression would be nice.

7. Oct 29, 2011

### Timebomb3750

You have a point. I could have explained better. Okay, ΔY is change in height, which in this case, I said was the .086m the spring stretches. The x is the spring compression/stretch.

As for the equations. ΔUg=mgh(final) - mgh(initial). The mgh(final), I said mgh(final) cancels because it's final height is zero. ΔUs= (1/2)m*x^2 (Final) - (1/2)m*x^2 (initial). I said (1/2)m*x^2 (Initial) cancels, because initially, the spring is relaxed.

But okay, I think I see what I could of done. I should of used Hooke's Law. But I'm not exactly sure how to find the F, so I can solve for the K.

8. Oct 29, 2011

### QuarkCharmer

Do a free body diagram of the mass on the end of the spring. It has a force mg pulling it down, what is stopping it from falling to the floor?

9. Oct 29, 2011

### PhanthomJay

Do not confuse the two parts of the problem.

In the first part, the mass is just hanging at rest in its equilibrium position. Identify the force the spring exerts on the mass, using Newton's First Law, and calculate the spring constant from Hooke's law.

In the 2nd part, when the mass is released from rest from the spring's unstretched position, use your energy method to calculate the max stretch of the spring.