Spring force question with a mass suspended on a spring

AI Thread Summary
The discussion revolves around calculating the spring constant for a mass-spring system where a 250g mass stretches a spring by 20cm after being released from rest. The initial calculations using the formula k=mg/x yield a spring constant of 12.25, but the correct answer is 25, leading to confusion over the use of 0.1m instead of 0.2m for the stretch. It is clarified that the mass falls further than it would if lowered slowly, resulting in additional stretch due to kinetic energy. Participants explore the total energy of the system, which includes gravitational potential energy and elastic potential energy, emphasizing that the equilibrium position is reached at 0.1m. The conversation concludes with the understanding that the system's dynamics involve both the spring and gravitational forces.
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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Likes berkeman
  • #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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