Spring compression: Force vs Energy

Nov 2017
5
0
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:
Jun 2010
422
33
NC
Nov 2017
5
0
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
 
Aug 2008
113
35
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 ...
 

Attachments

Nov 2017
5
0
When releasing, wont the mass end up stabilizing because the SHM is dampened by gravity?
 
Jun 2016
1,247
592
England
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.
 
Aug 2008
113
35
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 2017
5
0
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:
Aug 2008
113
35
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
 
Oct 2017
610
312
Glasgow
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!
 
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