How Do You Calculate the Spring Constant of a Hanging Mass?

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Homework Help Overview

The discussion revolves around calculating the spring constant of a spring when a mass is hung from it. The mass is specified as 0.200 kg, and the spring stretches by 0.086 m. Participants are exploring the relationship between gravitational force and spring force, as well as the application of energy conservation principles in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using energy equations and Hooke's law to find the spring constant. There is an attempt to break down the energy equation and questions about the assumptions made regarding the forces acting on the mass. Some participants suggest drawing free body diagrams to clarify the forces involved.

Discussion Status

The discussion is ongoing, with participants providing guidance on using Newton's laws and Hooke's law. There is a recognition of the need for clearer definitions of variables and the importance of distinguishing between the two parts of the problem. Multiple interpretations of the approach to finding the spring constant are being explored.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can provide or the methods they can use. There is an emphasis on understanding the physical principles involved rather than simply arriving at a numerical answer.

Timebomb3750
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Finding the spring constant of a spring...

Homework Statement


A mass is hung from a vertical spring. The mass is .200kg, and the spring stretches .086m. Find the spring constant.


Homework Equations


ΔE=ΔKE+ΔUg+ΔUs



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?
 
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Please do not bumb.

Did you consider using Hook's law, F=-kx ?
 
Integral said:
Please do not bumb.

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

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?
 
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.
 
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.
 
Integral said:
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.

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.
 
Timebomb3750 said:
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.

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?
 
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.
 

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