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Difference between Work and Energy? 
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#1
Oct2613, 01:34 PM

P: 32

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? 


#2
Oct2613, 01:38 PM

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P: 11,878

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. 


#3
Oct2613, 01:49 PM

P: 32

Thanks for the clarification!



#4
Oct2613, 05:53 PM

P: 2,500

Difference between Work and Energy?
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]. 


#5
Oct2713, 09:51 AM

P: 3

In it's simplest form, energy is the ability to do work.



#6
Oct2713, 10:49 AM

P: 2,500




#7
Oct2713, 11:03 AM

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P: 29,238

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. 


#8
Oct2713, 11:09 AM

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#9
Oct2713, 11:13 AM

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#10
Oct2713, 11:20 AM

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Zz. 


#11
Oct2713, 11:30 AM

P: 2,500

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. 


#12
Oct2713, 11:39 AM

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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. 


#13
Oct2713, 12:23 PM

P: 2,500

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. 


#14
Oct2713, 01:10 PM

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P: 17,317

Thread closed due to misinformation.



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