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Difference between Work and Energy?

by jacket
Tags: difference, energy, power, work
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jacket
#1
Oct26-13, 01:34 PM
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If I say 'a system has 5 Joule energy' then will it be equivalent of saying 'a system can do 5 Joule of work' ? And also will it be equivalent to say 'a system can give 5 watt of power for 1 second' ?
If so, then which is the basic property of a system? Work, Power or Energy?
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Drakkith
#2
Oct26-13, 01:38 PM
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All of them. Energy is just easier to use because you don't have to state "The system has the ability to perform X amount of work" or something. It's already implied by using a single word. Energy.

Edit: To clarify, energy and work are interchangeable in many ways. Power is not, however. As it requires an extra variable. Time.
jacket
#3
Oct26-13, 01:49 PM
P: 32
Thanks for the clarification!

SW VandeCarr
#4
Oct26-13, 05:53 PM
P: 2,499
Difference between Work and Energy?

Quote Quote by Drakkith View Post
All of them. Energy is just easier to use because you don't have to state "The system has the ability to perform X amount of work" or something. It's already implied by using a single word. Energy.

Edit: To clarify, energy and work are interchangeable in many ways. Power is not, however. As it requires an extra variable. Time.
While this is essentially correct, work is the absolute difference between the initial and final kinetic energy of a massive body with the application of a force: [itex] W=|1/2 mv^2 _{final} - 1/2 mv^2_ {initial})|[/itex].

Potential energy (PE) is the capacity to do work and is defined as [itex]PE=mv^2[/itex]. Note for the PE of a body's position in a force field such as a gravitational field PE=mgh where g is the acceleration of the field and h is the distance over which the object can (potentially) move. Since g can change over the path of motion, the mean acceleration over the path can be used as an approximation if h is large. The equation is dimensionally equivalent to [itex]PE=mv^2[/itex].
sepa0202
#5
Oct27-13, 09:51 AM
P: 3
In it's simplest form, energy is the ability to do work.
SW VandeCarr
#6
Oct27-13, 10:49 AM
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Quote Quote by sepa0202 View Post
In it's simplest form, energy is the ability to do work.
I know that "simple" definition is found in some sources, but I think it's a bit too simple. The first law of thermodynamics states energy can neither be created nor destroyed (matter represents potential energy). The second law states that the entropy of a system always increases. This means that for any system, unavailable energy is created as work is done, increasing the entropy of the system. By definition, unavailable energy is still energy, but it is not available to do work.
ZapperZ
#7
Oct27-13, 11:03 AM
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Quote Quote by SW VandeCarr View Post
While this is essentially correct, work is the absolute difference between the initial and final kinetic energy of a massive body with the application of a force: [itex] W=|1/2 mv^2 _{final} - 1/2 mv^2_ {initial})|[/itex].

Potential energy (PE) is the capacity to do work and is defined as [itex]PE=mv^2[/itex]. Note for the PE of a body's position in a force field such as a gravitational field PE=mgh where g is the acceleration of the field and h is the distance over which the object can (potentially) move. Since g can change over the path of motion, the mean acceleration over the path can be used as an approximation if h is large. The equation is dimensionally equivalent to [itex]PE=mv^2[/itex].
This is puzzling.

I have an object sitting at height h1. I then move it to a height h2 and let it sit there. In both cases, the initial and final speed of the object is zero. By your definition of work, I've done ZERO work.

Do you think this is correct?

And btw, where exactly did you get your definition of work? Please cite your source.

Zz.
Doc Al
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Oct27-13, 11:09 AM
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Quote Quote by SW VandeCarr View Post
While this is essentially correct, work is the absolute difference between the initial and final kinetic energy of a massive body with the application of a force: [itex] W=|1/2 mv^2 _{final} - 1/2 mv^2_ {initial})|[/itex].
I don't think this is a very useful general definition of work. Instead, it's a conclusion based on applying the "work"-KE theorem to a particle.
Doc Al
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Oct27-13, 11:13 AM
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Quote Quote by ZapperZ View Post
I have an object sitting at height h1. I then move it to a height h2 and let it sit there. In both cases, the initial and final speed of the object is zero. By your definition of work, I've done ZERO work.
The work-KE theorem applied to this situation will give you only the net work done on the body due to all forces, which of course is zero. Not very useful! (Which, I suspect, is your point.)
ZapperZ
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Oct27-13, 11:20 AM
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Quote Quote by Doc Al View Post
The work-KE theorem applied to this situation will give you only the net work done on the body due to all forces, which of course is zero. Not very useful! (Which, I suspect, is your point.)
It is too specific and not general. Instead, one should look at the change in the potential energy, which would not only result in the change in the KE if the object is moving, but also allows one to arrive at the work done even if the object isn't moving at the final and initial positions.

Zz.
SW VandeCarr
#11
Oct27-13, 11:30 AM
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Quote Quote by ZapperZ View Post
This is puzzling.

I have an object sitting at height h1. I then move it to a height h2 and let it sit there. In both cases, the initial and final speed of the object is zero. By your definition of work, I've done ZERO work.

Do you think this is correct?

And btw, where exactly did you get your definition of work? Please cite your source.

Zz.
http://hyperphysics.phy-astr.gsu.edu/hbase/work.html

It wasn't intended as a definition but as an example. In your example, to move an object, you need to apply a force (F) through a distance and F*d=mad=KE expended. I talk about the change in PE in the second paragraph of the post.

I apologize if I gave the impression that I was giving a general definition of work.
ZapperZ
#12
Oct27-13, 11:39 AM
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Quote Quote by SW VandeCarr View Post
http://hyperphysics.phy-astr.gsu.edu/hbase/work.html

It wasn't intended as a definition but as an example. In your example, to move an object, you need to apply a force (F) through a distance and F*d=mad=KE. I talk about the change in PE in the second paragraph of the post.
I apologize if I gave the impression that I was defining work.
But why would you use a rather specific and narrow situation like this to present a way to arrive at work done? If I don't know any better, I would think that work is somehow defined by a change in KE only, and that something must HAVE to be moving for me to know what work has been done. This is patently false.

The situation in this case is conservative. It doesn't care HOW one gets from one location to another. As long as there is a change in the potential energy between the initial and final, then there's work done equivalent to that change. Period. There is no need to resort to such a narrow example where it will work only for some specific situation.

Zz.
SW VandeCarr
#13
Oct27-13, 12:23 PM
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Quote Quote by ZapperZ View Post
But why would you use a rather specific and narrow situation like this to present a way to arrive at work done? If I don't know any better, I would think that work is somehow defined by a change in KE only, and that something must HAVE to be moving for me to know what work has been done. This is patently false.

The situation in this case is conservative. It doesn't care HOW one gets from one location to another. As long as there is a change in the potential energy between the initial and final, then there's work done equivalent to that change. Period. There is no need to resort to such a narrow example where it will work only for some specific situation.

Zz.
Well, the fact is that potential energy is also defined in terms of (potential) motion: [itex]PE=mv^2[/itex]. In the second paragraph of the post in question I do use the dimensionally equivalent formula PE=mgh which more directly addresses your example for 1 g gravity.

In hindsight, I might have started with a formal definition of work [itex]W=cos\theta Fd [/itex] where theta is the angle at which force is applied, the 0 degree angle being in line with the direction of motion. But I think it's less intuitive. I do agree that with your example, the example in the first paragraph of the post in question doesn't work.
DaleSpam
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Oct27-13, 01:10 PM
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