Physics Help Forum Spring compression: Force vs Energy

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 Nov 21st 2017, 10:47 AM #1 Junior Member   Join Date: Nov 2017 Posts: 5 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 .5KX² = mgX (since the height lost by the mass is the compression, X) .5 * 20 * X² = 1 * 9.8 * X 10 X² - 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 10:53 AM.
 Nov 21st 2017, 11:51 AM #2 Senior Member   Join Date: Jun 2010 Location: NC Posts: 398 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
 Nov 21st 2017, 01:04 PM #3 Junior Member   Join Date: Nov 2017 Posts: 5 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
 Nov 21st 2017, 01:34 PM #4 Senior Member     Join Date: Aug 2008 Posts: 113 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 ... Attached Thumbnails
 Nov 22nd 2017, 06:38 AM #5 Junior Member   Join Date: Nov 2017 Posts: 5 When releasing, wont the mass end up stabilizing because the SHM is dampened by gravity?
 Nov 22nd 2017, 06:57 AM #6 Senior Member     Join Date: Jun 2016 Location: England Posts: 590 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. __________________ ~\o/~
Nov 22nd 2017, 08:37 AM   #7
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Posts: 113
 Originally Posted by Woody 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.

 Nov 22nd 2017, 12:16 PM #8 Junior Member   Join Date: Nov 2017 Posts: 5 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 12:29 PM.
Nov 22nd 2017, 02:19 PM   #9
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 Originally Posted by Cyberrave 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

Nov 23rd 2017, 04:56 AM   #10
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 Originally Posted by Cyberrave 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 06:24 AM.

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