Finding Spring Rebound Distance with Given Mass and Constant k

In summary, the conversation discusses using the law of conservation of energy to determine the distance a block will go before rebounding to the equilibrium position when dropped above a vertical spring. The participants also share different approaches and equations, ultimately arriving at the correct answer of x=2mg/k.
  • #1
amcavoy
665
0
If I held a block of mass m directly above a vertical spring with constant k and dropped it, I need to find out how far down it would go before rebounding to the equilibrium position. I don't have a specific problem; I'm just trying to learn the concept.

I know that the magnitude of the force on the block due to the spring is |F|=kx. The force on the spring due to the block is |F|=mg. I am able to find the equilibrium position by equating those two, but how can I determine how far past the equilibrium position the block will go if it is dropped? I cannot find anything about this in my book or online. Could someone give me a hint on how to begin with the information I have above? I'd appreciate it.

Thank you.
 
Physics news on Phys.org
  • #2
[tex]W = \int_A^B \mathbf{F} \cdot d\mathbf{s}[/tex]
Does this sound gibberish to you?
 
  • #3
I know what it is -- line integral -- and how to use them (in most problems), but I cannot see how it's applied here. Are you suggesting something like the following?:

[tex]\text{mgx}=\text{k}\int_0^xs\,ds[/tex]

Then I could say:

[tex]x=\frac{2\text{mg}}{\text{k}}[/tex]

Is this reasoning / answer alright?

Thanks again for your help.
 
Last edited:
  • #4
I suggest using the law of conservation of energy. but that will require the distance the block from the spring initially.
 
  • #5
Leong said:
I suggest using the law of conservation of energy. but that will require the distance the block from the spring initially.
I basically mean zero distance from the block to the spring -- as if someone was holding the block above the spring, barely making contact with it.
 
  • #6
Anyways, I partitioned the distance into x1 (the distance from the release point to the equilibrium point) and x2 (the distance from the equilibrium point to the point where the block begins to rebound back to the eq. point).

[tex]{v_1}^2=2a\Delta x_1\implies {v_1}^2=\frac{2mg^2}{k}[/tex]

For the above step, I solved for x1 and replaced it into the left equation.

[tex]{v_1}^2=2a\Delta x_2\implies\frac{2mg^2}{k}=x_2\left(\frac{k\left(x_1+x_2\right)}{m}-g\right)[/tex]

I can get x2 from this, but it's much longer than what I expected. Does anyone know a better way?
 
  • #7
Hmmm... I actually tried using something else.

[tex]F=kx-mg[/tex]

[tex]W=\int_{0}^{x}\left(ks-mg\right)\,ds=\frac{1}{2}kx^2-mgx[/tex]

But since both the initial and final velocities are zero, it is clear that [itex]\Delta K=0[/itex].

[tex]\frac{1}{2}kx^2-mgx=0\implies x=\frac{2mg}{k}[/tex]

...which is what I came up with earlier.

Is this alright?

Thank you very much.
 
  • #8
conservation of energy

Initial energy of the system = 0 (taking the initial level of the block as the reference level to calculate the gravitational potential energy)
Energy of the system when the block is at the lowest position =
1/2*kx^2+mg(-x)
Initial energy of the system = energy of the system at that position
x=2mg/k
 
  • #9
apmcavoy said:
...

[tex]\text{mgx}=\text{k}\int_0^xs\,ds[/tex]

Then I could say:

[tex]x=\frac{2\text{mg}}{\text{k}}[/tex]

Is this reasoning / answer alright?
Yup. This is correct.
The work done by its weight is equal with the opposite of the work done by the tension of the string.
apmcavoy said:
Hmmm... I actually tried using something else.

[tex]F=kx-mg[/tex]

[tex]W=\int_{0}^{x}\left(ks-mg\right)\,ds=\frac{1}{2}kx^2-mgx[/tex]

But since both the initial and final velocities are zero, it is clear that LaTeX graphic is being generated. Reload this page in a moment..

[tex]\frac{1}{2}kx^2-mgx=0\implies x=\frac{2mg}{k}[/tex]

...which is what I came up with earlier.

Is this alright?
This is also correct.
apmcavoy said:
[tex]{v_1}^2=2a\Delta x_1\implies {v_1}^2=\frac{2mg^2}{k}[/tex]
This is wrong since the acceleration is not constant. It's not 'g'.
Viet Dao,
 
Last edited:
  • #10
Yes, I understand that now. I don't know what I was thinking. Anyways, thank you all very much for your help.
 

1. How do you calculate the spring rebound distance?

The spring rebound distance can be calculated using the formula: x = (m*g)/k where x is the rebound distance, m is the mass of the object attached to the spring, g is the gravitational force (9.8 m/s^2), and k is the spring constant.

2. What is the unit of measurement for the spring constant?

The unit of measurement for the spring constant is N/m (newtons per meter).

3. How do I determine the mass to use in the calculation?

The mass used in the calculation should be the total mass of the object attached to the spring, including any additional weights or attachments.

4. Can I use this formula for any type of spring?

No, the formula is specifically for an ideal spring that follows Hooke's law, where the force exerted by the spring is directly proportional to its extension or compression.

5. How does the spring rebound distance change with a different spring constant?

The spring rebound distance will increase as the spring constant increases. This means that a stiffer spring with a higher spring constant will cause the object to rebound a greater distance compared to a softer spring with a lower spring constant.

Similar threads

  • Introductory Physics Homework Help
Replies
29
Views
823
  • Introductory Physics Homework Help
Replies
24
Views
892
  • Introductory Physics Homework Help
Replies
3
Views
830
  • Introductory Physics Homework Help
Replies
10
Views
810
  • Introductory Physics Homework Help
Replies
12
Views
673
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
31
Views
972
  • Introductory Physics Homework Help
Replies
8
Views
4K
  • Introductory Physics Homework Help
Replies
22
Views
444
  • Introductory Physics Homework Help
Replies
5
Views
2K
Back
Top