Finding Spring Rebound Distance with Given Mass and Constant k

[amcavoy]

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.

 

[Päällikkö]

$$W = \int_A^B \mathbf{F} \cdot d\mathbf{s}$$
Does this sound gibberish to you?

 

[amcavoy]

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?:

$$\text{mgx}=\text{k}\int_0^xs\,ds$$

Then I could say:

$$x=\frac{2\text{mg}}{\text{k}}$$

Is this reasoning / answer alright?

Thanks again for your help.

 

[Leong]

I suggest using the law of conservation of energy. but that will require the distance the block from the spring initially.

 

[amcavoy]

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.

 

[amcavoy]

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).

$${v_1}^2=2a\Delta x_1\implies {v_1}^2=\frac{2mg^2}{k}$$

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

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

I can get x₂ from this, but it's much longer than what I expected. Does anyone know a better way?

 

[amcavoy]

Hmmm... I actually tried using something else.

$$F=kx-mg$$

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

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

$$\frac{1}{2}kx^2-mgx=0\implies x=\frac{2mg}{k}$$

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

Is this alright?

Thank you very much.

 

[Leong]

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

 

[VietDao29]

...

$$\text{mgx}=\text{k}\int_0^xs\,ds$$

Then I could say:

$$x=\frac{2\text{mg}}{\text{k}}$$

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.

Hmmm... I actually tried using something else.

$$F=kx-mg$$

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

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..

$$\frac{1}{2}kx^2-mgx=0\implies x=\frac{2mg}{k}$$

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

Is this alright?

This is also correct.

$${v_1}^2=2a\Delta x_1\implies {v_1}^2=\frac{2mg^2}{k}$$

This is wrong since the acceleration is not constant. It's not 'g'.
Viet Dao,

 

[amcavoy]

Yes, I understand that now. I don't know what I was thinking. Anyways, thank you all very much for your help.