The conservation of energy, why does it work? Negative work vs positive work

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SUMMARY

The discussion centers on the conservation of energy in physics, specifically addressing the equation ΔKE + ΔPE + ΔTE + ΔU = 0 in the absence of external forces. Participants clarify that gravitational potential energy (PE) is defined as positive when an object is raised, leading to a positive work done by gravity, while the work done against gravity is negative. The conversation highlights the importance of understanding the signs in energy equations and the distinction between conservative and non-conservative forces, particularly in the context of friction and internal energy.

PREREQUISITES
  • Understanding of basic physics concepts such as kinetic energy (KE) and potential energy (PE).
  • Familiarity with the work-energy theorem and its application.
  • Knowledge of gravitational acceleration, specifically -9.81 m/s².
  • Ability to differentiate between conservative and non-conservative forces.
NEXT STEPS
  • Study the work-energy theorem in detail, focusing on its applications in various physical scenarios.
  • Explore the concept of conservative forces and their relationship with potential energy.
  • Learn about the role of friction in energy conservation and its classification as a non-conservative force.
  • Examine real-world applications of energy conservation principles, such as in mechanical systems and inclined planes.
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Students of physics, educators teaching energy conservation principles, and anyone interested in understanding the dynamics of forces and energy in mechanical systems.

  • #31
flyingpig said:
Anyways, this also brings me the attention to friction doing work, is the "d" actually displacement or distance?
When computing the work done by any force, it's always displacement that counts. (For simple cases, distance is just the magnitude of the displacement.) The relative direction of force and displacement determines the sign of the work done.
 
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  • #32
Doc Al said:
When computing the work done by any force, it's always displacement that counts. (For simple cases, distance is just the magnitude of the displacement.) The relative direction of force and displacement determines the sign of the work done.

I thought that when dealing with non conservative force we take the total distance and not only the displacement or is that wrong? i mean even if you slide a ball along a rough surface then it hit the wall and return to it's original position wouldn't the work done by friction = N times the total distance and not zero?
 
  • #33
The work done by force F along a path C is always given by

W=\int_C \mathbf{F}\cdot d\mathbf{x}

where x is the displacement. In the case of friction, because its direction is always opposite the displacement's, you have that

\mathbf{F}\cdot d\mathbf{x} = -|\mathbf{F}||d\mathbf{x}|

and if |F| is constant, you get

W = -|\mathbf{F}|\int_C |d\mathbf{x}|

where the integral equals the total distance traveled. On the other hand, an applied force, like someone pushing on an object, is non-conservative, but the calculation wouldn't generally work out the same way as it did for friction.
 
  • #34
madah12 said:
I thought that when dealing with non conservative force we take the total distance and not only the displacement or is that wrong?
You need the integral of \vec{F}\cdot d\vec{x} along the path, where d\vec{x} is the instantaneous displacement, not the net displacement. (See vela's post for details.) Sorry if that wasn't clear. My point was that whenever you calculate the work done by a force, even friction, you must consider the direction of the object's movement--the displacement not simply the distance.
i mean even if you slide a ball along a rough surface then it hit the wall and return to it's original position wouldn't the work done by friction = N times the total distance and not zero?
You are correct. In this particular example, the integral along the path will end up equaling the Force times the total distance.

But friction does not always oppose the direction of the displacement.
 
  • #35
vela said:
The work done by force F along a path C is always given by

W=\int_C \mathbf{F}\cdot d\mathbf{x}

where x is the displacement. In the case of friction, because its direction is always opposite the displacement's, you have that

\mathbf{F}\cdot d\mathbf{x} = -|\mathbf{F}||d\mathbf{x}|

and if |F| is constant, you get

W = -|\mathbf{F}|\int_C |d\mathbf{x}|

where the integral equals the total distance traveled. On the other hand, an applied force, like someone pushing on an object, is non-conservative, but the calculation wouldn't generally work out the same way as it did for friction.

Since I can't google a symbol what does the symbol that looks like a definite integral but has only one limit mean?
 
  • #36
It's notation for a path integral.
 
  • #37
and C indicates the length of the path?
 
  • #38
C indicates the path.
 
  • #39
Ok let me try this one more time, because i think I get it now and vector calculus isn't really needed here...

ΔKE = -ΔPE = Wab = Work done by other forces

What is the most confusing thing here is terminologies.

½m(v² - v₀²) = -mg(h - h₀) <=== work done by gravity

Now here comes the problem, is that the formula for the work-energy theorem or is that the conservation of energy?
 
  • #40
flyingpig said:
Ok let me try this one more time, because i think I get it now and vector calculus isn't really needed here...

ΔKE = -ΔPE = Wab = Work done by other forces
That's not correct.
What is the most confusing thing here is terminologies.

½m(v² - v₀²) = -mg(h - h₀) <=== work done by gravity

Now here comes the problem, is that the formula for the work-energy theorem or is that the conservation of energy?
The formula, by itself, isn't either one or the other. It's can be interpreted both ways.

What you wrote, particularly with your comment, is a case of the work-energy theorem. All that theorem says is the change in kinetic energy of an object is equal to the work done on it by all forces. So if you're interpreting the RHS of the equation as the work done by gravity, you're obviously using the work-energy theorem.

You could interpret that same equation as ΔKE = -ΔPE, which would be an expression of the conservation of energy.

How you read the equation is up to you.
 
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  • #41
vela said:
That's not correct

What!? Oh man and here I thought i got it...

vela said:
The formula, by itself, isn't either one or the other. It's can be interpreted both ways.

What you wrote, particularly with your comment, is a case of the work-energy theorem. All that theorem says is the change in kinetic energy of an object is equal to the work done on it by all forces. So if you're interpreting the RHS of the equation as the work done by gravity, you're obviously using the work-energy theorem.

You could interpret that same equation as ΔKE = -ΔPE, which would be an expression of the conservation of energy.

How you read the equation is up to you.

Isn't that what I had before?
 
  • #42
flyingpig said:
Ok let me try this one more time, because i think I get it now and vector calculus isn't really needed here...

ΔKE = -ΔPE = Wab = Work done by other forces
vela said:
That's not correct.
flyingpig said:
What!? Oh man and here I thought i got it...
Just to reinforce what vela said, the work done by other forces, where by 'other' forces you mean forces other than conservative forces like gravity, That is, work done by other forces is the work done by non conservative forces, then

ΔKE + ΔPE = Wnc = Work done by other forces

This is the law of the conservation of TOTAL energy. Now when Wnc =0, then

ΔKE + ΔPE = 0, which is the conservation of mechanical energy principle which applies when only conservative forces, like gravity, act.
What is the most confusing thing here is terminologies.

½m(v² - v₀²) = -mg(h - h₀) <=== work done by gravity

Now here comes the problem, is that the formula for the work-energy theorem or is that the conservation of energy?
It looks like in this equation you are assuming only gravity forces act, in which case, using the conservation of mechanical energy principle,

ΔKE + ΔPE = 0, which exactly gives your equation noted immediately above. On the other hand, echoing more or less vela's comments, if you want to use the work energy theorem, with only gravity acting, then

Total work done = ΔKE = Wc + Wnc , or,

½m(v² - v₀²) = -mg(h - h₀) + 0
which again is your same equation.
As vela noted, the formula, by itself, isn't either one or the other. It's can be interpreted both ways.

Pick one!:wink:
 
  • #43
flyingpig said:
What!? Oh man and here I thought i got it...

Isn't that what I had before?
If you're asking if writing

ΔKE = -ΔPE = Wab = Work done by other forces

and

ΔKE = -ΔPE

mean the same thing, the answer is no, they don't. (Of course, I am assuming here that when you wrote "by other forces," you meant the non-conservative forces, like friction.) PhantomJay explained how ΔKE, ΔPE, and the work done by non-conservative forces are related.
 
  • #44
Perhaps I was a bit vague, I mean in the absence of friction, my equation holds and my interpretation is correct right?
 
  • #45
Well, no. If there is any non-conservative force, like an applied force, your first equation doesn't hold. If there are no non-conservative forces, then there are no other forces, energy is conserved, and you have ΔKE = -ΔPE.

So why would you refer to "other forces" in the first equation? This isn't a matter of being vague. Clearly, you were referring to non-conservative forces. If you have non-conservative forces, you have, as PhantomJay noted, ΔKE + ΔPE = Wnc. If you move ΔPE to the other side of the equation, you get ΔKE = -ΔPE + Wnc, which isn't the same as what you wrote.

Perhaps what you meant was: when there are only conservative forces,

ΔKE = -ΔPE = Wab = work done by all (not other) forces

Then that would be correct. This is what people meant when they said the change in potential energy is the negative of the work done by the conservative force.
 
  • #46
Doc Al said:
That mysterious red stuff--the mgh term--is called potential energy. It's the negative of the work done by gravity, which is why it's on the opposite side of the equation.

-(-mgh)? Are there any other physical meaning to this?
 
  • #47
flyingpig said:
-(-mgh)? Are there any other physical meaning to this?
What did you have in mind?
 
  • #48
I am getting the feeling that mechanical energy is not the same thing as the work done by something.
 
  • #49
flyingpig said:
I am getting the feeling that mechanical energy is not the same thing as the work done by something.
Please explain what you mean as fully as you can. Give a specific example.
 
  • #50
As in like a definition. Like potential energy has nothing to do with work x displacement. It's just means how much energy something has at a certain height.
 
  • #51
flyingpig said:
I am getting the feeling that mechanical energy is not the same thing as the work done by something.
That is right. It's the change in mechanical energy (the change in kinetic plus potential energies) that is the same as the work done by non-conservative forces. It's the change in gravitational potential energy that is equal to the negative of the work done by gravity. It's the change in kinetic energy that is equal to the work done by all forces.

When a 10 kg object is 5 meters above the ground, it has a potential energy of 490 J with respect to the ground. When it falls to the ground, it has no potential energy with respect to the ground. The work done by gravity (that is,the earth) on the object is 490 J during this fall [- (0 - 490) = +490 J).

Energy is not work. Energy is the capacity to do work.
 
  • #52
flyingpig said:
As in like a definition. Like potential energy has nothing to do with work x displacement. It's just means how much energy something has at a certain height.
As PhanthomJay explained, mechanical energy is not the same as the work done by something. (Although sometimes you can use the work done by something to figure out the change in mechanical energy.)

But to say that potential energy 'has nothing to do' with force x displacement (I assume you meant force, not work) is not accurate. Gravitational potential energy does equal the work done against gravity.
 
  • #53
Doc Al said:
No. As a mass is raised, the gravitational PE increases. (Gravitational PE is minus the work done by gravity. It's the work done against gravity.)


No, g stands for the magnitude of the acceleration due to gravity; it's always positive. (Taking down as negative, the acceleration due to gravity is -g.)

this depends on your assumptions you can take g as positive but in that case many signs of other things also changed like velocity in downward direction which is usually taken as negative will positive.
Most of all taking g as +ve or -ve depends on assumptions in general we take it as -ve.
 
  • #54
Doc Al said:
...
No, g stands for the magnitude of the acceleration due to gravity; it's always positive. (Taking down as negative, the acceleration due to gravity is -g.)

I agree with Doc Al on this point, (and on all other points, as well as I can remember). In any Physics textbook with which I am familiar, the symbol 'g' stands for the magnitude of the acceleration due to gravity, as Doc Al said, so g is positive.
 
  • #55
Just use netwons third law to connect the forces. Here's how to start... The force exerted onto object one from from object two is positive 10 Newtons. The force exerted onto object two by object one is negative 10 Newtons. Afterwards, connect the definition of work = force times displacement while making sure you keep in mind the laws of thermodynamics. It takes practice, but you will succeed oh grasshopper.
 

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