energy = work?


by Hypo
Tags: energy, work
Hypo
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#1
May1-12, 02:19 AM
P: 59
Hallo everyone!

energy in a physical system is the ability to do work.

E=W

Where Work = F x D

Now if the force increase obviously "work" will increase! does that mean the energy also increases?

Now if I had a system and the force in its start is 20 Newtons x 2 meters = 40J = The amount of energy in the system right? If the force increase to lets say 50 Newtons x 2 meters= 100J so its goes on and on and on and the energy is going to increase much much more right?

So if force in a system's start at 20N then more forces is added the energy increase right?

Hope I'm making sense! In a way im trying to say "energy" in a system can increase if "force" or "distance" increase right?
Where both force and distance are the main valuables of the system.
Thanks!
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haruspex
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#2
May1-12, 02:31 AM
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Yes, but from the way you pose the question I've a funny feeling you're going to misapply the answer ;-) Do you have a particular set-up in mind?
jay.yoon314
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#3
May1-12, 03:11 AM
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Yeah, that sounds about right. (Remember that W = F * D is only strictly valid as written when F is constant throughout the entire displacement D that the force F "carried the object through.")

For example, if you have two sleds of the same mass, and you pull the second one twice as hard as the first one (with twice the force), and keep that force "steady" each time through the same distance, then the second sled will have twice the kinetic energy as the first upon "release."

Hypo
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#4
May1-12, 01:12 PM
P: 59

energy = work?


Im breaking down "Energy" to the fundamental structure.
You see... I was asking my self for a while about F and Energy confused always about the two. E depends on F so in a way I wanted to imagine and understand it.

Now I'll apply an example to what I'm trying to say.
If I had a system that is able to do work"Kinetic energy". It started for example with 1000J'S over 10 Meters the forces will = 100N's

So if I simply increased the "Force" alone in the system to lets say 500N'S x 10Meters = 5000J of kinetic energy!

Now the things is "FORCE" can be created while the distance is constant and I'd like to increase the energy of the system.

Just studying this system more and more I'm kinda a fan of "ENERGY" and its mysterious ways
Hypo
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#5
May1-12, 01:14 PM
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Quote Quote by haruspex View Post
Yes, but from the way you pose the question I've a funny feeling you're going to misapply the answer ;-) Do you have a particular set-up in mind?
Actually I do not.

Although I plan to do something if! I understood energy more.

But what did you think I was going to do? Or imagined me doing or whatever ?
Ken G
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#6
May1-12, 03:57 PM
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Quote Quote by Hypo View Post
If I had a system that is able to do work"Kinetic energy". It started for example with 1000J'S over 10 Meters the forces will = 100N's

So if I simply increased the "Force" alone in the system to lets say 500N'S x 10Meters = 5000J of kinetic energy!

Now the things is "FORCE" can be created while the distance is constant and I'd like to increase the energy of the system.
You are applying the formula correctly, but your interpretation sounds a bit off. Instead of thinking of work as energy, think of it is a change in energy, in particular, in kinetic energy (this is the "work-energy theorem", it is the crux of the usefulness of both the work and kinetic energy concepts). Now, since we like to imagine that total energy does not change, if kinetic energy does change, we have to dream up some other form of energy that can account for that change. The work-energy theorem tells us just how to do that-- invent "potential energy" and equate it to -work, and poof, the change in kinetic energy shows up in "potential energy", and the work done tells you how much energy was transferred from kinetic energy to potential energy (or from kinetic energy to heat, which is just more kinetic energy somewhere else).

In your example, therefore, if you consider how much force F and distance D is needed to stop a moving object, then F*D is a measure of how much kinetic energy (which is also mv2/2) the object must have had if it stops from force F over distance D. If you want to independently vary F and keep D constant, then it means you are not necessarily stopping the object, you are merely taking F*D kinetic energy away from it (note I am imagining that F points opposite to the object's motion-- you can also add to the object's kinetic energy of course by using a positive F instead of a negative one). If the object starts out at rest, so has 0 kinetic energy originally, then you can give it kinetic energy F*D using a force F over distance D (which perhaps is what you were talking about). By the way this all follows directly from F = ma.
thebrainstorm
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#7
May1-12, 04:04 PM
P: 6
Yes.
Hypo
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#8
May1-12, 04:39 PM
P: 59
See the thing is I'm studying force & energy all together.

I understand I due follow the laws of thermo + conservation all together in the system so I understand how things are going.However,Im making example and playing with their key valuables in the system such as "F" AND "D".

So all in all in a system that has "KE" of 1000J by increasing ONLY its Force the system could get more "KE" transfered from a source.

Thus resulting a win/win situation for both laws force is added and energy is transfered HAH!"just go that eureka moment" Really really interesting they are!

Thanks everyone any more inputs I'd be grateful for them!

Actually I might come back with another thing related to this question and post it.
Pengwuino
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May1-12, 04:59 PM
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Quote Quote by Hypo View Post
So all in all in a system that has "KE" of 1000J by increasing ONLY its Force the system could get more "KE" transfered from a source.

Thus resulting a win/win situation for both laws force is added and energy is transfered HAH!"just go that eureka moment" Really really interesting they are!
Yes but there's nothing mysterious or interesting about it as you are implying. Take a baseball players, for example. If a baseball player throw a ball with a certain force, he would impart a certain kinetic energy during his throw. If he threw it with twice the force, he would impart twice the kinetic energy during his throw (assuming the rotation of his arm and subsequent release covers the same distance). However, it's a lot more strenuous to apply a larger force when throwing a ball so it's not exactly win-win for the entire system. The same idea applies for a car engine moving a car uphill vs. flat land. It's a lot more strenuous and the engine must do more work to move a vehicle uphill (since it's giving it a potential energy as well).
Hypo
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#10
May1-12, 11:00 PM
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Quote Quote by Pengwuino View Post
Yes but there's nothing mysterious or interesting about it as you are implying. Take a baseball players, for example. If a baseball player throw a ball with a certain force, he would impart a certain kinetic energy during his throw. If he threw it with twice the force, he would impart twice the kinetic energy during his throw (assuming the rotation of his arm and subsequent release covers the same distance). However, it's a lot more strenuous to apply a larger force when throwing a ball so it's not exactly win-win for the entire system. The same idea applies for a car engine moving a car uphill vs. flat land. It's a lot more strenuous and the engine must do more work to move a vehicle uphill (since it's giving it a potential energy as well).


For me I've been struggling to relate force and energy in a system. I looked at them in a different way but now I can see the picture with perfect colors.

I understand its all depended on certain conditions areas and many factors. Physics is the breakdown of EVERYTHING physical in a system so I do keep account everything surrounding the system.

Thank you for you're input!

Now if force is that "KICK" or "PUSH" in a system is energy that power to supply the "PUSH" or "KICK"?
haruspex
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#11
May1-12, 11:42 PM
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Quote Quote by Hypo View Post
But what did you think I was going to do?
E.g. if you were thinking of pushing an object a certain distance across a frictional floor. You might have thought that if you pushed twice as hard you would use twice the energy to get it the same distance.
Hypo
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#12
May2-12, 12:03 AM
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Quote Quote by haruspex View Post
E.g. if you were thinking of pushing an object a certain distance across a frictional floor. You might have thought that if you pushed twice as hard you would use twice the energy to get it the same distance.
hahaha the beauty of simplicity thanks for that example gives me another view of my question thanks though!
sweet springs
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#13
May2-12, 01:24 AM
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Hi.

Bank account = salary paid? Yes, in a sense. No, in another sense.

Regards.
Hypo
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#14
Jun2-12, 06:51 AM
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Quote Quote by sweet springs View Post
Hi.

Bank account = salary paid? Yes, in a sense. No, in another sense.

Regards.
hallo... What?!
HallsofIvy
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#15
Jun2-12, 07:06 AM
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He is pointing out that, as you said initially, change in energy= work done on or by a system. It's a bit simplistice to just say "energy= work".
jtbell
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#16
Jun2-12, 07:43 AM
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I would say that work is a method of transferring energy from one object or system to another one. In thermodynamics, heat is another method. In fact, the first law of thermodynamics says those are the only two methods for changing the internal energy of an system: ΔU = Q + W, or ΔU = Q - W, depending on the sign convention that you use for work.
Hypo
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#17
Jun2-12, 09:11 AM
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Quote Quote by HallsofIvy View Post
He is pointing out that, as you said initially, change in energy= work done on or by a system. It's a bit simplistice to just say "energy= work".
Ah, thanks HallsofIvy

Quote Quote by jtbell View Post
I would say that work is a method of transferring energy from one object or system to another one. In thermodynamics, heat is another method. In fact, the first law of thermodynamics says those are the only two methods for changing the internal energy of an system: ΔU = Q + W, or ΔU = Q - W, depending on the sign convention that you use for work.
Interesting.


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