Spring force question with a mass suspended on a spring

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Homework Help Overview

The discussion revolves around a problem involving a mass suspended from a spring, focusing on the calculation of the spring constant and the dynamics of the system as the mass falls and stretches the spring. The subject area includes concepts from mechanics, specifically spring forces and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the spring constant using different values for displacement and question the assumptions behind the setup. There is exploration of energy considerations as an alternative approach to understanding the system's behavior.

Discussion Status

Participants are actively engaging with the problem, raising questions about the role of gravitational potential energy and the total energy of the system. Some guidance has been offered regarding the use of energy considerations, and there is a recognition of the need to clarify the components of the energy involved.

Contextual Notes

There is a noted confusion regarding the displacement values used in calculations and the implications of how the mass is released versus being lowered. The discussion also touches on the distinction between considering the spring-mass system versus including gravitational effects in the analysis.

Amik
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Homework Statement
. A 50-cm-long spring is suspended from the ceiling. A 230g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 18cm before coming to rest at its lowest point. It then continues to oscillate vertically.

a. What is the spring constant? (K=)
Relevant Equations
F=kx
Fnet=0
kx-mg=0
k=mg/x=0,25*9.8/0.2=12.25
But answer is 25(And they use 0.1 instead of 0.2 Why?)
 
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I am sorry.
The problem should be
A 50-cm-long spring is suspended from the ceiling. A 250g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 20cm before coming to rest at its lowest point. It then continues to oscillate vertically.

a. What is the spring constant? (K=)
 
Fnet=0
kx-mg=0
k=mg/x=0,25*9.8/0.2=12.25
But answer is 25(And they use 0.1 instead of 0.2 Why?)
 
In this problem the mass falls lower than if you slowly lowered it until the spring and mass were at a stationary equilibrium postition at the bottom. Do you see how the spring gets stretched a bit extra because the mass falls from the spring's unstretched position?

Try using energy considerations instead. Can you say how to do that? :smile:
 
The reason why it has extra is because it is released instead of slowly lowering it?
 
Yes!

Does "using energy considerations instead" ring a bell? What is the total energy of the system? Is it constant?
 
The thing I am confusing about is is the system spring and the mass or spring mass and the earth?
 
What are the two components of the total energy of this spring+mass system (and yes, using gravity as part of it)?

Should that total energy stay constant?
 
I think we have elastic potential, gravitational potential and the kinetic energy(although it is 0 when we are using it)
 
  • #10
If it is spring+mass system(Why should we include the gravity, since the Earth exert it)
 
  • #11
Yes, gravitational potential energy is one of the two parts of the total energy of this spring+mass system. What is the other part?

Then how do you use that to help you solve this problem? :smile:
 
  • #12
If it is spring+mass system(Why should we include the gravity, since the Earth exert it)
 
  • #13
Amik said:
If it is spring+mass system(Why should we include the gravity, since the Earth exert it)
You can do it either way: consider the spring, mass Earth system, or consider the spring+mass system with gravity as a given, constant, external force.
 
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  • #14
1/2k*x*x+mghi=mghf
I put x as 0.1 and I got 49?
 
  • #15
I got it now.The equilibiluum position is when x=0.1(not x=0.2!)
 
  • #16
Note here the masss stops only temporarily before moving up. At its lowest point the total energy is the spring energy. At the equilibrium point x = 0.09 m, the net force is zero and the forces balance. Hence the acceleration is zero and the velocity is maximum. ( since as the spring approaches from the top it continues to speed up as it falls but at a slower rate as the acceleration goes to zero (at the equilibrium point acc = 0)

[Post edited by Mentor to delete solution]
 
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