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Old Nov 21st 2017, 11:47 AM   #1
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Spring compression: Force vs Energy

Hi,

I'm trying to solve a problem with two different methods, but I can't seem to get to the same answer with both methods, I don't understand where I'm making a mistake.

Here is the problem: A mass of 1 kg is resting on top of a spring in the vertical position. The spring's constant is of 20 N/m. Find the lenght of the spring's compresses.

Method 1

F spring = F gravity
KX = mg
20 * X = 1 * 9.8
X = 9.8 / 20 = 0.49 m
Spring compresses 49 cm down.

Method 2

Elastic energy = Initial potential gravity engergy
.5KX = mgX (since the height lost by the mass is the compression, X)
.5 * 20 * X = 1 * 9.8 * X
10 X - 9.8X = 0
X (10X - 9.8) = 0
X = 0 or 10X -9.8 = 0
Dismiss X=0 since it is the initial state
Hence X = .98
Spring compresses 98 cm Down.


The difference between method 1 and 2 is a factor of 2. I have also come across other situations when solving the problem with forces or energy i get the same factor of 2 difference. Can someone help me understand my mistake?

Thanks!

Last edited by Cyberrave; Nov 21st 2017 at 11:53 AM.
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Old Nov 21st 2017, 12:51 PM   #2
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Perhaps Include "Free Length."

Hi Cyberrave...

A spring has a "free length." But spring displacement can be generalized (as a dx) or written
relative to its free length. To include the "free length" makes things clearer.

These examples include the "free length."

Simple Springs | THERMO Spoken Here!
Bungee Jumper | THERMO Spoken Here!
Box Bounces Between Springs | THERMO Spoken Here!
Washer Switch | THERMO Spoken Here!

Good luck with your studies... JP
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Old Nov 21st 2017, 02:04 PM   #3
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From what I understand, when I do the equality in my second method, The points where there is no kinetic energy are at the extremes, when My spring is completely extended or compressed, so I'm in fact calculating 2x the value i'm looking for. Would you say this assumption is correct?

Thanks for the quick reply
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Old Nov 21st 2017, 02:34 PM   #4
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note there is a difference in "lowering" the mass to equilibrium from a height $x$ and "releasing" the mass from a height "x".

"Lowering" the mass to equilibrium requires an external force upward to counteract motion.

Reference the attached diagram ...
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Spring compression: Force vs Energy-spring_vertical.jpg  
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Old Nov 22nd 2017, 07:38 AM   #5
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When releasing, wont the mass end up stabilizing because the SHM is dampened by gravity?
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Old Nov 22nd 2017, 07:57 AM   #6
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The spring will be damped by the internal stresses in the spring
(heating the material up and thus removing energy from the system)
It may also be damped by aerodynamic drag
(again removing energy from the system)

However, it will not be damped by gravity.
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Old Nov 22nd 2017, 09:37 AM   #7
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Originally Posted by Woody View Post
The spring will be damped by the internal stresses in the spring
(heating the material up and thus removing energy from the system)
It may also be damped by aerodynamic drag
(again removing energy from the system)

However, it will not be damped by gravity.
agree ... gravity is a conservative force.
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Old Nov 22nd 2017, 01:16 PM   #8
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Ok, so the reason That the method 2 does not work is that before equilibrium is reached, the sring and mass ossilate up and down, where energy is lost due non conservative forces.

Last edited by Cyberrave; Nov 22nd 2017 at 01:29 PM.
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Old Nov 22nd 2017, 03:19 PM   #9
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Originally Posted by Cyberrave View Post
Ok, so the reason That the method 2 does not work is that before equilibrium is reached, the sring and mass ossilate up and down, where energy is lost due non conservative forces.
yes
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Old Nov 23rd 2017, 05:56 AM   #10
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Originally Posted by Cyberrave View Post
Ok, so the reason That the method 2 does not work is that before equilibrium is reached, the sring and mass ossilate up and down, where energy is lost due non conservative forces.
Basically, yeah.

If you tweak method 2 to be

$\displaystyle \Delta EPE = \Delta GPE + \Delta KE$

then you get

$\displaystyle \frac{1}{2} k x^2 = mgx + \frac{1}{2} mv^2$

The main important point is that the velocity is non-zero at the point of the restoring force of the spring being equal to the weight, so there is energy wrapped up as kinetic energy. That explains the discrepancy between the two methods. Because the kinetic energy is non-zero at the point, you'll get oscillations. if you try and go on to solve it, you get:

$\displaystyle 10 x^2 = 9.8 x + \frac{1}{2} \left(\frac{dx}{dt}\right)^2$

$\displaystyle \frac{dx}{dt} - \left(10x^2 - 9.8 x\right)^{1/2} = 0$

This is a horrible non-linear 1st order ODE describing the oscillation. It's solvable though... Good luck!

Last edited by benit13; Nov 23rd 2017 at 07:24 AM.
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