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Conservation of Energy + Hooke's Law

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1. Homework Statement
A 2.70kg mass is pushed against a horizontal spring of force constant 25.0 N/cm on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring. When the spring has been compressed enough to store 12.0 J of potential energy, the mass is suddenly released from rest. What is the greatest speed the mass reaches?

2. Homework Equations
Conservation of Energy
Hooke's Law (F = -kx)

3. The Attempt at a Solution
As soon as they put a spring in the conservation of energy problems, I become confused. First, I don't understand why it's 1/2kx^2 (which doesn't apply to this problem I don't think), and second, I have a hard time determining if it's positive or negative. Here is what I come up with for this problem, there are more parts to it and I was hoping that you guys could help me work through my spring-a-phobia.

Edit: I see a mistake already, fixing the image I posted.
 
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First, it should have been 25N/cm so that's really 2500 N/m.

Then:
[tex]12.0 = \frac{1}{2}(2.7)v^{2}[/tex]
[tex]v = 2.98[/tex]

That seems correct to me. The second part is to determine when this speed occurs. I'm not sure how to do this exactly. Intuition tells me that it happens right as the block leaves the spring, or when the spring is at equilibrium. I don't know hot to express that mathematically though. If I say that x is zero in the conservation equation I get that 0 = KE which is clearly not the case.

This is my answer for the second part:
The maximum velocity of 2.98 m/s occurs when the spring is at it's equilibrium point. After that point, the spring begins to accelerate in the opposite direction in order to comply with Hooke's Law, and the mass leaves the spring as predicted with Newton's first law.
Am I on the right track so far?
 
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Hi QuarkCharmer! :smile:

A compressed spring holds the potential to release energy.
In neutral position it obviously does not hold energy, therefore a compressed spring holds positive energy.

Why is it [itex] {1\over 2}kx^2[/itex]?
Well:
[tex]W=\int Fdx = \int kxdx = {1\over 2}k x^2[/tex]

And yes, your result for v is right.

When the (perfect massless) spring is back in its equilibrium point, it has transferred all of its energy to the mass and the mass.
(The same integral formula shows that all stored power has been returned at this point.)
Therefore the spring won't accelerate in opposite direction.
Hooke's Law is observed, because there is no mass fixed to the spring anymore, so F=ma has to be zero.
 
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Thank you ILS,

Wouldn't the spring continue to oscillate after the mass disconnects? Also, wouldn't the integral contain a negative kx, since F = -kx, such that:
[tex]W_{elastic/spring} = \int -kx dx = -\frac{1}{2}kx^{2}[/tex]

The next part want's to know what the greatest acceleration of the mass is. Here is how I worked it out:

Since I assume the acceleration of the mass stops when the mass leaves the spring, I needed to determine the length the spring was compressed. I did this via:

[tex]\frac{1}{2}kx^{2}=12[/tex]
[tex]\frac{1}{2}(2500)x^{2}=12[/tex]

I put k as 2500, since the desired units are N/m and I was given 25N/cm, so:
[tex]\frac{25N}{cm}\frac{100cm}{1m} = \frac{2500N}{m}[/tex]

This tells me that x is 0.0980 m (seems too small...)

Then I used kinematics to find the accleration:
[tex]V_{f}^{2} = V_{i}^{2}+2aΔx[/tex]
[tex]2.98^{2} = 0^{2}+2a(0.0980)[/tex]
[tex]a = 45.4 m/s^{2}[/tex]
Which is incorrect?
 

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I suddenly realize what your problem with the minus sign is.

The magnitude of elastic force is F=kx.
The vector of the elastic force is [itex]\vec F = -k \vec x[/itex].

I can imagine this can be confusing.
I usually only use the magnitude F=kx.
When relevant we need to realize of course that its direction is opposite to the deviation.


Thank you ILS,

Wouldn't the spring continue to oscillate after the mass disconnects?
When the mass disconnects, the remaining mass attached to the spring is m=0.
If the spring would get to some position x, its elastic force would be F=kx.
Using F=ma the resulting acceleration a would be infinite!

Suffice to say: no, it does not oscillate after the mass disconnects (assuming a perfect massless spring).


The next part want's to know what the greatest acceleration of the mass is. Here is how I worked it out:

Since I assume the acceleration of the mass stops when the mass leaves the spring, I needed to determine the length the spring was compressed. I did this via:

[tex]\frac{1}{2}kx^{2}=12[/tex]
[tex]\frac{1}{2}(2500)x^{2}=12[/tex]

I put k as 2500, since the desired units are N/m and I was given 25N/cm, so:
[tex]\frac{25N}{cm}\frac{100cm}{1m} = \frac{2500N}{m}[/tex]

This tells me that x is 0.0980 m (seems too small...)
Looks good.


Then I used kinematics to find the accleration:
[tex]V_{f}^{2} = V_{i}^{2}+2aΔx[/tex]
[tex]2.98^{2} = 0^{2}+2a(0.0980)[/tex]
[tex]a = 45.4 m/s^{2}[/tex]
Which is incorrect?
No, this formula is not applicable.
It only applies when the acceleration "a" is constant, which it isn't.

You should use F=kx to find the maximum acceleration.
 
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When the mass disconnects, the remaining mass attached to the spring is m=0.
If the spring would get to some position x, its elastic force would be F=kx.
Using F=ma the resulting acceleration a would be infinite!

Suffice to say: no, it does not oscillate after the mass disconnects (assuming a perfect massless spring).
Understood. So, "yes" it would still oscillate, but not in these imaginary perfect condition physics problems.


I understand the error with using kinematics there! (and the part about the negative in Hooke's Law, thank you)

[tex]F=ma[/tex]
[tex]F=kx[/tex]
[tex]kx=ma[/tex]
[tex](2500)(0.0980)=(2.7)a[/tex]
[tex]a_{max} = 90.7 m/s^{2}[/tex]

That seems outrageously high? So I assume I am applying this wrong as well?
 

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[tex]a_{max} = 90.7 m/s^{2}[/tex]

That seems outrageously high? So I assume I am applying this wrong as well?
No, you applied it correctly.
It means you can successfully shoot your mass upward (it will overcome gravity). :smile:
 
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Just seems like a large number for acceleration with a spring! That's nearly 300ft/s, or about 1/2 the speed of an average hand gun round. That's some spring they have there...


The last part of this practice question want's to know when this max acceleration occurs.

That's going to be when the spring reaches equilibrium correct?
 

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Just seems like a large number for acceleration with a spring! That's nearly 300ft/s, or about 1/2 the speed of an average hand gun round. That's some spring they have there...


The last part of this practice question want's to know when this max acceleration occurs.

That's going to be when the spring reaches equilibrium correct?
No, to the contrary.
Are you mixing acceleration and velocity?

Remember that you calculated a speed v=2.98 m/s!

Note that a=F/m.
When is F at its maximum?
 
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Is this the idea then?

15xls2x.jpg
 
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I was thinking of it's acceleration vs time, but vs distance might be more appropriate yes. Either way, that would make the max acceleration occur right when the spring starts to expand back to it's equilibrium state correct?
 

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Yep! :approve:
 
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Great! That's the section I am starting now. Trying to work ahead a few weeks in homework/subjects because I plan to be super busy with another class next week. I just finished all of the homework involving springs and such using knowledge gained in this thread. Thank you!

It's Pendulum time now!
 

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