Conservation of Energy vs Force Analysis for Springs

In summary: Yes, the equilbrium point (where the sum of forces = 0 ) is the same (at x below the unstretched length) in either case. The spring is not in equilibrium at '2x' if it were dropped, because even though it has temporarily come to a stop, it is still decelerating, net force is not 0 at that point.
  • #1
fee6
2
0

Homework Statement


A stone of mass [tex]m[/tex] is at rest on a vertical spring which is compressed a distance [tex]x[/tex]. Find its spring constant [tex]k[/tex].

All variables are given in the problem.


Homework Equations



I solved this problem realizing that because the mass is at equilibrium, the sum of the vertical forces on it is 0. Thus, the force of gravity downwards is equal to the force of the spring upwards, and so [tex]mg = kx[/tex].

The Attempt at a Solution



Therefore, [tex]k = \frac{mg}{x}[/tex], from the previous equation. The problem, now, is that after solving it like this, a friend asked me why we could not use the conservation of energy:

[tex]E_i = E_f[/tex]
[tex]mgx = \frac{1}{2}kx^2[/tex]
[tex]k = \frac{2mg}{x}[/tex]

which differs from the previous solution. I am inclined to believe that there is something wrong with the energy method, but I cannot put my finger on it. I would greatly appreciate any help with this. Thanks!
 
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  • #2
fee6 said:

Homework Statement


A stone of mass [tex]m[/tex] is at rest on a vertical spring which is compressed a distance [tex]x[/tex]. Find its spring constant [tex]k[/tex].

All variables are given in the problem.


Homework Equations



I solved this problem realizing that because the mass is at equilibrium, the sum of the vertical forces on it is 0. Thus, the force of gravity downwards is equal to the force of the spring upwards, and so [tex]mg = kx[/tex].

The Attempt at a Solution



Therefore, [tex]k = \frac{mg}{x}[/tex], from the previous equation. The problem, now, is that after solving it like this, a friend asked me why we could not use the conservation of energy:

[tex]E_i = E_f[/tex]
[tex]mgx = \frac{1}{2}kx^2[/tex]
[tex]k = \frac{2mg}{x}[/tex]

which differs from the previous solution. I am inclined to believe that there is something wrong with the energy method, but I cannot put my finger on it. I would greatly appreciate any help with this. Thanks!
Your first method is correct. The energy method assumes that the block is released from rest at the unstretched position of the spring, in which case the mass comes to a stop at a distance twice the 'x' distance (it will oscillate back and forth and ultimately damp out and settle in its equlibrium position). The problem did not assume this; the mass is gently lowered by your hand until it reaches it's equilibrium position.
 
  • #3
PhanthomJay said:
Your first method is correct. The energy method assumes that the block is released from rest at the unstretched position of the spring, in which case the mass comes to a stop at a distance twice the 'x' distance (it will oscillate back and forth and ultimately damp out and settle in its equlibrium position). The problem did not assume this; the mass is gently lowered by your hand until it reaches it's equilibrium position.

Ah, I see. If the mass were just dropped, which would cause it to come to a stop at twice the 'x' distance, would the equilibrium point then be at 'x' below the original position, and the amplitude of the resulting oscillations 'x' as well?
 
  • #4
fee6 said:
Ah, I see. If the mass were just dropped, which would cause it to come to a stop at twice the 'x' distance, would the equilibrium point then be at 'x' below the original position, and the amplitude of the resulting oscillations 'x' as well?
Yes, the equilbrium point (where the sum of forces = 0 ) is the same (at x below the unstretched length) in either case. The spring is not in equilibrium at '2x' if it were dropped, because even though it has temporarily come to a stop, it is still decelerating, net force is not 0 at that point. Good observation.
 

Related to Conservation of Energy vs Force Analysis for Springs

1. What is the difference between conservation of energy and force analysis for springs?

Conservation of energy and force analysis for springs are two different approaches to understanding the behavior of springs. Conservation of energy focuses on the total energy of a system, while force analysis looks at the individual forces acting on a spring.

2. How do conservation of energy and force analysis apply to springs?

Conservation of energy can be used to predict the maximum displacement of a spring, as it states that energy cannot be created or destroyed. Force analysis, on the other hand, can be used to determine the specific forces acting on a spring and how they contribute to its overall behavior.

3. Which approach is more accurate for analyzing springs?

Both conservation of energy and force analysis have their own strengths and limitations. Conservation of energy is a general principle that applies to all systems, while force analysis is more specific to the forces acting on a spring. Therefore, the most accurate analysis will depend on the specific situation and the level of detail needed.

4. Can conservation of energy and force analysis be used together?

Yes, conservation of energy and force analysis can complement each other when analyzing springs. By using both approaches, a more comprehensive understanding of the behavior of the spring can be achieved.

5. How do conservation of energy and force analysis relate to each other in the context of springs?

Conservation of energy and force analysis are interconnected in the context of springs. The forces acting on a spring, as determined by force analysis, contribute to the total energy of the system, which is conserved according to the principle of conservation of energy.

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