Work and Potential Elastic Energy

In summary: You will not get the same k, but a different value, which is what you should get.In summary, the work done by gravity when you lower the load onto the spring gets split into 2 parts, elastic PE and KE. But when you hold the load stationary at a stretched position, the elastic PE is the work done by gravity, while the KE is zero. This is a static situation unlike what happens when you let the load fall into the spring, when the elastic PE keeps getting converted to KE and back to PE.
  • #1
odie5533
58
0

Homework Statement


The following is the data I collected. I set a weight on a spring and measured how far the spring was stretched. The spring was not oscillating, just stretched and left at rest.
Load (kg):
0.300
0.400
0.500
0.600
0.700
0.800
0.900

Weight (Converted the above load into N):
2.94
3.92
4.90
5.88
6.86
7.85
8.83
5.39
6.37

Stretch (m):
0.059
0.076
0.095
0.115
0.134
0.154
0.174
0.105
0.125

The question is, how do I find the work from the above and how do I find the elastic potential energy.

Homework Equations


(1/2)kx^2



The Attempt at a Solution


[tex]Work = Fd[/tex]
[tex]W = (2.94)(0.059) = 0.17346[/tex]

[tex]F = kx[/tex]
[tex]k = F/x[/tex]
[tex]2.94 / 0.059 = 49.8[/tex]
[tex]U_{elastic} = \frac{1}{2}kx^2[/tex]
[tex]U_{elastic} = \frac{1}{2}(49.8)(0.059^2) = 0.08673[/tex]

I'm pretty sure I did something wrong. It also asks me to compare the work done and the elastic potential energy. I thought W = Ue in this case, but the calculations show differently. But I think the calculations are wrong. Any help would be greatly appreciated.
 
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  • #2
let's imagine a spring, when the load, F Newtons has reached a certain extension, x, such that the force exerted by the spring is equal to F, what would the energy changes be like?

The work done by the gravity would be Fx. The elastic potential would be 1/2 Fx^2. The other half of the energy lies in the kinetic energy of the load. This position is the equilibrium position of oscillation. It would also be the position whereby the load will remain stationary when damping has taken place.

At this juncture, we can conclude that the remaining half of the energy has been dissipated into tis surroundings, the kinetic energy into forms of energy like sound.
 
  • #3
Hi odie5533,

When you gently lower a mass from, say, the height of your head to the ground, where is the work done by gravity going? It is dissipated in your arms. Here, too, the same thing is happening. If you had attached a load to the spring and just let it go, and then measured the extreme stretch while it was oscillating (without friction), then work done would have been equal to elastic PE.
 
  • #4
Oerg: So the work done by gravity equals the elastic potential? I don't see where the KE takes place, since the mass is not oscillating, v = 0, K = 0.5mv^2 = 0.

Shooting star:
[tex]W = U_{elastic} = \frac{1}{2}kx^2[/tex]
[tex]W = Fd = 2.94 * 0.059 = 0.17346[/tex]
[tex]0.17346 = \frac{1}{2}kx^2[/tex]
[tex]k = 99.66[/tex]
This any better?
 
  • #5
well, the above example can be described as an oscillating mass whereby damping has occured. The kinetic energy is lost and a stationary load hung from a spring is achieved.

Or as shooting star has described, it can also be the case whereby the mass is lowered slowly via your hand, in which case, the energy is dissipated through your hand.

Your initial post was correct, the discrepency between the 2 energy values is the loss of kinetic energy to its surroundings.

your above post equated the work done with the elastic potential energy which is wrong.
 
  • #6
odie5533 said:
Shooting star:
[tex]W = U_{elastic} = \frac{1}{2}kx^2[/tex]
[tex]W = Fd = 2.94 * 0.059 = 0.17346[/tex]
[tex]0.17346 = \frac{1}{2}kx^2[/tex]
[tex]k = 99.66[/tex]
This any better?

If the mass had simply been let drop after being attached to the spring, it would have overshot the current position with max KE and then continued downward for some distance, slowing down all the while, until when again it would reverse velocity and move up to its original position from where it was let go. Without friction, this would be undamped oscillation. The energy would shift between PE due to gravity, elastic PE and KE, the total remaining constant.

In the 1st eqn, W= Uelastic is not correct. (I will comment on the 2nd later. Time is short now. Pl don’t mind.)

Please understand that you have taken energy away from the system by not allowing the load to fall naturally after you attached it to the spring. With damping, which is there in the real world, your position of the load now would be where the load ultimately comes to equilibrium after damped oscillation.

Now, I'm sure you can refine your treatment of the problem.
 
  • #7
For finding k, use F/extension from your table, as you did in your first post.
 

1. What is work?

Work is a physical quantity that measures the amount of energy transferred to or from an object due to a force acting on it. It is typically measured in joules (J).

2. What is potential elastic energy?

Potential elastic energy is the energy stored in an object when it is deformed or stretched. This energy is a result of the elastic potential energy of the object, which is the ability to return to its original shape after being deformed.

3. How is potential elastic energy calculated?

Potential elastic energy can be calculated using the formula PE = 1/2 * k * x^2, where PE is the potential elastic energy, k is the spring constant, and x is the displacement of the object from its equilibrium position.

4. What is the relationship between work and potential elastic energy?

Work and potential elastic energy are directly related. The work done on an object is equal to the change in potential elastic energy. This means that when work is done to deform an object, its potential elastic energy increases, and when work is done to return the object to its original shape, its potential elastic energy decreases.

5. What are some real-life examples of potential elastic energy?

Some common examples of potential elastic energy include stretched rubber bands, compressed springs, and bent bows. Other examples include trampolines, diving boards, and even bungee jumping cords. These objects store potential elastic energy when they are deformed and release it when they return to their original shape.

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