Work energy theorem clarification.

In summary: Hi V0ODO0CH1LD! :smile:First the formula for kinetic energy ( \frac{1}{2} mv^2 ). I realize that in the processes of deriving the formula they always assume that the force doing work on the system is constant. Is that because the kinetic energy is only dependent on the final speed that the object reaches? Kind of like it doesn't care how it got to that state of kinetic energy, so it might as well assume it was via a constant force?Yes, that is correct. Kinetic energy is the amount of work, done by a constant force, needed to accelerate a body of a
  • #1
V0ODO0CH1LD
278
0
I just wanted to start a thread about the basic concepts of work and energy to clarify some things I have been wondering about.

Firstly, work and kinetic energy are concepts independent of force fields; right? Am I right to think that the relationship of force and kinetic energy with respect to distance is analogous to the relationship of force and momentum with respect to time? Because if I apply a constant force for a certain time period I change my momentum from an initial momentum to a final one ([itex] F \Delta t = \Delta P [/itex]). Similarly; if I apply a constant force for a certain distance I change my kinetic energy ([itex] F \Delta x = \Delta E_k [/itex]). Work is like if I gave a name to the quantity [itex] F \Delta t = \Delta P [/itex], which as far I know has no name and is only know as change in momentum or F times distance. Is all of that too far off?

Anyway.. potential energy only exists in the presence of a force field; correct? So in the absence of a force field, work done on a system will always equal the kinetic energy of that system. Like if there was no force fields in the universe there wouldn't even be a concept called potential energy. So is it that in the presence of a force field [itex] W=E_p + E_k [/itex]?

I guess the only thing that is confusing me at this point is the fact that the expression for potential energy depends on what force field you're dealing with. Is that correct to assume or is there a general expression for potential energy? Is it the integral of force with respect to position?

EDIT: Also, what would be the analogue to potential energy in my momentum analogy? Maybe some special case since force fields are not necessarily dependent on time?
 
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  • #2
Hi V0ODO0CH1LD! :smile:
V0ODO0CH1LD said:
Firstly, work and kinetic energy are concepts independent of force fields; right?

right!
Am I right to think that the relationship of force and kinetic energy with respect to distance is analogous to the relationship of force and momentum with respect to time? Because if I apply a constant force for a certain time period I change my momentum from an initial momentum to a final one ([itex] F \Delta t = \Delta P [/itex]). Similarly; if I apply a constant force for a certain distance I change my kinetic energy ([itex] F \Delta x = \Delta E_k [/itex]). Work is like if I gave a name to the quantity [itex] F \Delta t = \Delta P [/itex], which as far I know has no name and is only know as change in momentum or F times distance. Is all of that too far off?

i see what you're getting at … ∫ F dt = momentum, and ∫ F.dx = work
Anyway.. potential energy only exists in the presence of a force field; correct?

yes, potential energy is just another name for minus the work done by a conservative force

(so there has to be a force everywhere, ie a field)
So in the absence of a force field, work done on a system will always equal the kinetic energy of that system. Like if there was no force fields in the universe there wouldn't even be a concept called potential energy. So is it that in the presence of a force field [itex] W=E_p + E_k [/itex]?

yes (assuming we say that heat and radiation are also forms of kinetic energy) :smile:
…*is there a general expression for potential energy? Is it the integral of force with respect to position?

yes, it's -∫ F.dx, the work done :wink:
EDIT: Also, what would be the analogue to potential energy in my momentum analogy?

not following you :confused:
 
  • #3
tiny-tim said:
not following you :confused:

What is confusing me is:

First the formula for kinetic energy ([itex] \frac{1}{2} mv^2 [/itex]). I realize that in the processes of deriving the formula they always assume that the force doing work on the system is constant. Is that because the kinetic energy is only dependent on the final speed that the object reaches? Kind of like it doesn't care how it got to that state of kinetic energy, so it might as well assume it was via a constant force?

Could I specify that kinetic energy is the amount of work, done by a constant force, needed to accelerate a body of a given mass from rest to its stated velocity? And I don't mean that work can only be done by a constant force. But if a system has a certain kinetic energy and it got to that state though some variant force acting on it for a certain distance, the theoretical force that acted on it, by definition, is whatever constant force had to act on it for that specified distance to get it to that state of kinetic energy? Is that correct? Is it because by the end of the day, one force is the actual variant force times distance and the other is the average force times distance and the quantity "work" is the same for both?

The only interaction with potential energy, in the other hand, is with movement in the direction of the force field. So if I apply force that keeps a point mass in a certain potential but moves it perpendicular to the force field, even though I have to apply a certain force to keep my point mass in that potential, the only component of the force who's changing the energy of the particle is the one perpendicular to the force field which is increasing the kinetic energy; right?

But when I lift a point mass up in the air with some constant velocity I am not changing the kinetic energy of that point mass, because acceleration is zero therefore force is zero so work has to be zero. But I am giving that point mass some potential; what is happening?

EDIT: Also, I just though of something else: When I say (potential energy) + (kinetic energy) = (constant); is the kinetic energy I am talking about also due to the force field? Like (kinetic energy due to force field) + (potential energy due to force field) = constant? Or am I talking about any exterior forces that may be acting on the system inside the force field as well as the actual forces of the force field?

Is it as if the conservation of energy theorem exists with respect to a force field only? But, in the case with gravity, if my hand is laying on the floor with a rock in it, there is no potential/kinetic energy in the hand/rock system. But I can lift it, and give it some potential energy. Where did that come from? Did my hand have potential energy on the floor because I could lift the rock if I wanted to?
 
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  • #4
Hi V0ODO0CH1LD! :smile:
V0ODO0CH1LD said:
First the formula for kinetic energy ([itex] \frac{1}{2} mv^2 [/itex]). I realize that in the processes of deriving the formula they always assume that the force doing work on the system is constant. Is that because the kinetic energy is only dependent on the final speed that the object reaches? Kind of like it doesn't care how it got to that state of kinetic energy, so it might as well assume it was via a constant force?

No, they do not assume F is constant …

1/2 mv2 = ∫ mv dv = ∫ mv dv/dx dx = ∫ ma dx = ∫ F dx

kinetic energy is the integral of F wrt distance, even if F is not constant :wink:

(F includes all forces, even those with an associated potential energy, such as gravity or spring force)
The only interaction with potential energy, in the other hand, is with movement in the direction of the force field. So if I apply force that keeps a point mass in a certain potential but moves it perpendicular to the force field, even though I have to apply a certain force to keep my point mass in that potential, the only component of the force who's changing the energy of the particle is the one perpendicular to the force field which is increasing the kinetic energy; right?

sentence too long: brain hurts :redface:
But when I lift a point mass up in the air with some constant velocity I am not changing the kinetic energy of that point mass, because acceleration is zero therefore force is zero so work has to be zero. But I am giving that point mass some potential; what is happening?

what is happening is that there are two forces, the force from your hand, and the force of gravity

they are (obviously) equal and opposite, so F here is 0 :wink:
EDIT: Also, I just though of something else: When I say (potential energy) + (kinetic energy) = (constant); is the kinetic energy I am talking about also due to the force field? Like (kinetic energy due to force field) + (potential energy due to force field) = constant? Or am I talking about any exterior forces that may be acting on the system inside the force field as well as the actual forces of the force field?

kinetic energy is 1/2 mv2 … that's it! …

it comes from all the forces
Is it as if the conservation of energy theorem exists with respect to a force field only? But, in the case with gravity, if my hand is laying on the floor with a rock in it, there is no potential/kinetic energy in the hand/rock system. But I can lift it, and give it some potential energy. Where did that come from? Did my hand have potential energy on the floor because I could lift the rock if I wanted to?

it came from chemical potential energy in your muscles :smile:
 
  • #5


I would like to clarify some of the concepts you have mentioned. Firstly, you are correct in stating that work and kinetic energy are independent of force fields. Work is defined as the product of force and displacement, while kinetic energy is the energy an object possesses due to its motion. Both of these concepts are applicable in any situation, regardless of the presence of a force field.

In regards to your analogy between force and kinetic energy with respect to distance, and force and momentum with respect to time, it is a valid comparison. In both cases, a change in the respective quantity is caused by the application of a constant force over a certain period of time or distance.

You are also correct in stating that potential energy only exists in the presence of a force field. This is because potential energy is associated with the position of an object in a force field. In the absence of a force field, there is no potential energy. However, I would like to clarify that work done on a system does not always equal the kinetic energy of that system. This is because work can also change the potential energy of a system.

The expression W = E_p + E_k is correct, but it is important to note that this is the total energy of a system, not just the work done. As you mentioned, potential energy is dependent on the force field, and therefore the expression for potential energy will vary depending on the type of force field present. There is no general expression for potential energy, but it can be calculated using the integral of force with respect to position.

In your momentum analogy, there is no direct analogue to potential energy. However, one could argue that the initial momentum of a system could be considered as a type of "potential momentum", as it has the potential to change with the application of a force.

I hope this clarifies some of your questions about work and energy. If you have any further inquiries, please do not hesitate to ask. As scientists, it is important to have a clear understanding of these fundamental concepts in order to accurately analyze and interpret data in our research.
 

Related to Work energy theorem clarification.

1. What is the work-energy theorem?

The work-energy theorem states that the work done on an object by a net force is equal to the change in kinetic energy of that object. In other words, the work done on an object will either increase or decrease its kinetic energy, depending on the direction of the force.

2. How is work defined in the work-energy theorem?

In the work-energy theorem, work is defined as the force applied to an object multiplied by the distance the object moves in the direction of the force. This is mathematically represented as W = F*d, where W is work, F is force, and d is distance.

3. What is the relationship between work and energy in the work-energy theorem?

The work-energy theorem shows that work and energy are directly related. The amount of work done on an object will result in a change in its energy. This can be seen in the equation W = ΔKE, where W is work, Δ is the symbol for change, and KE is kinetic energy.

4. Can the work-energy theorem be applied to non-conservative forces?

Yes, the work-energy theorem can be applied to both conservative and non-conservative forces. Conservative forces, such as gravity, follow the law of conservation of energy, while non-conservative forces, such as friction, do not. However, the work done by non-conservative forces can still result in a change in the object's energy.

5. How is the work-energy theorem used in real-world applications?

The work-energy theorem is used in many real-world applications, including engineering, physics, and sports. It is used to calculate the amount of work and energy needed for various tasks, such as lifting objects, accelerating vehicles, and launching projectiles. It is also used in the design of machines and structures to ensure they have enough energy to perform their intended functions.

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