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

In summary, when an object is raised, its gravitational potential energy increases. That just confused me even more. I have two Physics texts that give me two different formulas, but they seem to work the same way except things are arranged on different sides of the equation.
  • #1
flyingpig
2,579
1

Homework Statement




This is the conservation of energy in the absence of external forces


ΔKE + ΔPE + ΔTE + ΔU= 0

TE is friction, U is other internal energy, like the spring

So expand and disregard U and TE for a moment


½m(v² - v₀²) + mg(h - h₀) = 0

Now comes the question, why is + mg(h - h₀) and not -mg(h - h₀)? Is gravitational potential energy not the same thing as the work done by gravity? Because gravity pulls down and if you go up, that's negative work. I've used the equation (with the positive mgh) many times and have not failed me.
 
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  • #2
flyingpig said:

Homework Statement




This is the conservation of energy in the absence of external forces


ΔKE + ΔPE + ΔTE + ΔU= 0

TE is friction, U is other internal energy, like the spring

So expand and disregard U and TE for a moment


½m(v² - v₀²) + mg(h - h₀) = 0

Now comes the question, why is + mg(h - h₀) and not -mg(h - h₀)? Is gravitational potential energy not the same thing as the work done by gravity? Because gravity pulls down and if you go up, that's negative work. I've used the equation (with the positive mgh) many times and have not failed me.
Well you may have just been lucky :wink:. The work done by a conservative force, like gravity, is the negative of its potential energy change (Wg = - ΔPE). For example, if an object is dropped from rest from a building of height h, then ½m(v² - 0) + mg(0 - h) = 0, or
½m(v²) = mgh, solve for v, and note the change in potential energy is negative, and the work done by gravity is positive.

If, on the other hand, an object is thrown upward from the ground at an initial speed v, to a maximum height of h, then
½m(0 - v²)+ mg(h - 0) = 0, or
- ½m(v²) = -mgh, solve for v, and note the change in potential energy is positive, and the work done by gravity is negative. Gotta love those plus and minus signs...one of the more difficult concepts of Physics to grasp. :uhh:
 
  • #3
PhanthomJay said:
Well you may have just been lucky :wink:. The work done by a conservative force, like gravity, is the negative of its potential energy change (Wg = - ΔPE). For example, if an object is dropped from rest from a building of height h, then ½m(v² - 0) + mg(0 - h) = 0, or
½m(v²) = mgh, solve for v, and note the change in potential energy is negative, and the work done by gravity is positive.

If, on the other hand, an object is thrown upward from the ground at an initial speed v, to a maximum height of h, then
½m(0 - v²)+ mg(h - 0) = 0, or
- ½m(v²) = -mgh, solve for v, and note the change in potential energy is positive, and the work done by gravity is negative. Gotta love those plus and minus signs...one of the more difficult concepts of Physics to grasp. :uhh:

No, I am questioning about my formula, shouldn't it be -mg(h - h₀) instead of mg(h - h₀)?


½m(v² - v₀²) + mg(h - h₀) = 0 <==== this one works

or

½m(v² - v₀²) - mg(h - h₀) = 0 <==== this one does not, but it is correct
 
  • #4
Remember that g is -9.81m/s2. Thus, when you plug it into your equation, positive mgh has a negative value.
 
  • #5
p21bass said:
Remember that g is -9.81m/s2. Thus, when you plug it into your equation, positive mgh has a negative value.

My book did not use the -9.8m/s^2 and I thought we consider g only in its magnitude.
 
  • #6
flyingpig said:
No, I am questioning about my formula, shouldn't it be -mg(h - h₀) instead of mg(h - h₀)?
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.)

p21bass said:
Remember that g is -9.81m/s2. Thus, when you plug it into your equation, positive mgh has a negative value.
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.)
 
  • #7
Oh, right. Oops... Sorry about that.
 
  • #8
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.)

That just confused me even more. I have two Physics texts that gives me two different formulas, but they seem to work the same way except things are arranged on different sides of the equation.

[tex]\Delta KE = \Sigma W_{other} - \Delta E_{int}[/tex]

Ex. Problem using this problem using the equation given above

A crate of mass m = 10.7 kg is pulled up a rough incline with an initial speed of vi = 1.59 m/s. The pulling force is F = 97 N parallel to the incline, which makes an angle of θ = 19.7° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled d = 5.09 m.

(a) What is the change in kinetic energy of the crate?

Force done by gravity = -180J

Force done by friction also the change in internal energy = 201J

Force done on the crate = 494J

[tex]\Delta KE = \Sigma W_{other} - \Delta E_{int}[/tex]

=494J + (-180J) - 201J
= 113J

Now my question is how is that formula different from the one in my first post. Also why is internal energy considered as friction? If that is the case, what is spring potential energy? I thought that is internal energy
 
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  • #9
flyingpig said:

Homework Statement




½m(v² - v₀²) + mg(h - h₀) = 0

Now comes the question, why is + mg(h - h₀) and not -mg(h - h₀)?

The total mechanical energy of a system remains constant if the internal forces are conservative and the external forces do no work. This is the principle of conservation of energy.
So K + U = constant
Ki + Ui = Kf + Uf
You can see that ' ½m(v² - v₀²) + mg(h - h₀) = 0 ' is in accordance with the above equation. If you use - mg(h - h₀), it will violate the law.
 
  • #10
flyingpig said:
Now my question is how is that formula different from the one in my first post. Also why is internal energy considered as friction? If that is the case, what is spring potential energy? I thought that is internal energy

The total mechanical energy K + U is not constant if non-conservative forces like friction etc act between the parts of the system. Hence we can't apply the principle of conservation of energy in presence of non-conservative forces.
However, the work-energy theorem is still valid. So according to it,

work done by all the forces equals the change in kinetic energy.
Wc + Wnc + Wext = Kf-Ki
where c=conservative, nc = non-conservative, ext=external forces


Also why is internal energy considered as friction?
energy = friction :confused:
May be you meant work done by friction is considered as internal energy.
Well if friction is an internal force in the system you consider, work done by it will count under internal energy of the system. If it is an external force, work done by it will come under external energy of the system.
 
  • #11
Abdul Quadeer said:
The total mechanical energy of a system remains constant if the internal forces are conservative and the external forces do no work. This is the principle of conservation of energy.
So K + U = constant
Ki + Ui = Kf + Uf
You can see that ' ½m(v² - v₀²) + mg(h - h₀) = 0 ' is in accordance with the above equation. If you use - mg(h - h₀), it will violate the law.

But - mg(h - h₀) is the work done by gravity
 
  • #12
flyingpig said:
But - mg(h - h₀) is the work done by gravity

Yes it is.
Using the equation
Ui + Ki = Uf + Kf,
Ki - Kf = Uf - Ui
½m(v₀² - v²) = Uf - Ui

Now the change in potential energy of a system corresponding to a conservative internal force is defined as Uf - Ui = -W, where W is the work done on the system by the internal force (here gravity is the internal force if you consider Earth as a part of your system) as the system passes from its initial configuration i to final configuration f.
So
½m(v₀² - v²) = -[- mg(h - h₀)]
½m(v² - v₀²) + mg(h - h₀) = 0
Again in accordance with the equation I gave.
 
  • #13
All this internal and external system really confuses me...
 
  • #14
flyingpig said:
All this internal and external system really confuses me...

To apply conservation of energy principle in questions, you have to select your system first. Here is an example-

A block of mass 4 kg starts falling with an initial velocity of 4m/s from a height of 10m. What is is velocity when it falls through a height of 2 m?

Take the block + the Earth as the system. Only the block falls, so only the work done on the block will contribute to the gravtitational potential energy. As is descends through a height 8m, the potential energy decreases by mgh. No external force is acting on the system (here gravitational force is an internal force as we already took Earth in our system).

So using conservation of energy,
increase in kinetic energy = decrease in potential energy
0.5m(vf2 - vi2) = mg(hi - hf)

Which is same as

mg10 + 0.5m(4)2 = mg2 + 0.5mv2 (taking potential energy on the surface of Earth as 0)
On solving, v=13.27m/sIf you still don't get the concept, try reading some good books like Resnick Halliday Walker Fundamentals of Physics, Sears and Zemansky University Physics etc to develop your fundamentals. Moreover, you can't understand physics unless you think about the subject.
 
  • #15
flyingpig said:
why is internal energy considered as friction? If that is the case, what is spring potential energy? I thought that is internal energy
A spring force is a conservative force. You can most of the time just use the general conservation of energy equation:

Ki +Ugi + Usi + Wnc = Kf +Ugf + Usf

Where the subscripts are defined as i = initial, f = final, g = gravitational, s =spring, and where Wnc is the all encompassing work done by non-conservative external forces, like friction forces, applied forces, tension forces, normal forces, etc.
 
  • #16
flyingpig said:
That just confused me even more. I have two Physics texts that gives me two different formulas, but they seem to work the same way except things are arranged on different sides of the equation.
What confused you even more? My statement that gravitational PE increases when an object is raised?

[tex]\Delta KE = \Sigma W_{other} - \Delta E_{int}[/tex]

Ex. Problem using this problem using the equation given above

A crate of mass m = 10.7 kg is pulled up a rough incline with an initial speed of vi = 1.59 m/s. The pulling force is F = 97 N parallel to the incline, which makes an angle of θ = 19.7° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled d = 5.09 m.

(a) What is the change in kinetic energy of the crate?

Force done by gravity = -180J

Force done by friction also the change in internal energy = 201J

Force done on the crate = 494J

[tex]\Delta KE = \Sigma W_{other} - \Delta E_{int}[/tex]

=494J + (-180J) - 201J
= 113J
For this problem, internal energy is irrelevant. The change in KE of the object equals the work done on the object by all forces. There are three forces acting on the object that do work: gravity, the applied force, and friction.

Now my question is how is that formula different from the one in my first post.
One key difference is that here you are treating gravity as a force, whereas before you represented gravity using gravitational PE. Either way is fine. A conservative force such as gravity can be represented as a potential energy.
Also why is internal energy considered as friction?
Not sure what you are talking about. For this problem, all you care about is the kinetic energy of the object. (It's true that the work done by friction ends up as random thermal energy, but that's irrelevant.)
If that is the case, what is spring potential energy? I thought that is internal energy
Not sure what you're getting at here. Depending upon how you define your system, you could consider spring PE as an internal energy. But in the same way that you can represent gravity either as a force or an energy, you can represent the action of the spring as either a force or an energy (since the spring force is conservative).

You have two choices in solving problems with conservative forces:

(1) Treat them as forces: ΔKE = Work done by all forces

(2) Treat them in terms of PE: ΔKE + ΔPE = Work done by non-conservative forces

The first is called the "Work-Energy" theorem; the second is called "conservation of energy". But they are equivalent (almost).

In order to use these equations intelligently, you must realize, for example, that the work done by gravity is -mgΔh and the change in gravitational PE is mgΔh. (The sign must change since the gravity term moves from one side of the equation to the other.)
 
  • #17
Abdul Quadeer said:
Using the equation
Ui + Ki = Uf + Kf,
Ki - Kf = Uf - Ui
½m(v₀² - v²) = Uf - Ui

Now the change in potential energy of a system corresponding to a conservative internal force is defined as Uf - Ui = -W, where W is the work done on the system by the internal force (here gravity is the internal force if you consider Earth as a part of your system) as the system passes from its initial configuration i to final configuration f.
So
½m(v₀² - v²) = -[- mg(h - h₀)]
½m(v² - v₀²) + mg(h - h₀) = 0
Again in accordance with the equation I gave.

Here is here I got confused, why is there another negative sign?

Shouldn't you leave it as it is and so?

-½m(v² - v₀²) = - mg(h - h₀)

So that

-ΔKE = ΔPE

But it doesn't make sense when you have

-ΔPE - ΔKE = 0

Since I said ΔPE = - mg(h - h₀)

- ( - mg(h - h₀) ) -½m(v² - v₀²) = 0

mg(h - h₀) - ½m(v² - v₀²) = 0
 
  • #18
Doc Al said:
One key difference is that here you are treating gravity as a force, whereas before you represented gravity using gravitational PE. Either way is fine. A conservative force such as gravity can be represented as a potential energy.

But what is the difference though in the formula? In my formula mgh is still work x displacement which must obey W = |F||d|cosθ

Not sure what you are talking about. For this problem, all you care about is the kinetic energy of the object. (It's true that the work done by friction ends up as random thermal energy, but that's irrelevant.)

No, this was in the solution key, it says the internal energy is friction which I don't understand what means. I thought internal energy is like the work done by the internal system such as the spring (or so I thought)

Depending upon how you define your system, you could consider spring PE as an internal energy

So the spring is never considered as an "internal energy", am I getting the wrong definition of internal energy?

You have two choices in solving problems with conservative forces:

(1) Treat them as forces: ΔKE = Work done by all forces

(2) Treat them in terms of PE: ΔKE + ΔPE = Work done by non-conservative forces

(1) I thought the "Work done by all forces" mean external forces like someone pulling, I've always thought gravity is work done by an internal force

(2) Wait, shouldn't it be the negative of work done by all the non-conservative forces?
 
  • #19
flyingpig said:
But what is the difference though in the formula? In my formula mgh is still work x displacement which must obey W = |F||d|cosθ
No, the work done by gravity is F.d = mg.h = mgh cos theta = +mgh if the gravity force acts in the direction of the vertical displacement, and -mgh if the gravity force acts opposite to the direction of the displacemnt.
No, this was in the solution key, it says the internal energy is friction which I don't understand what means. I thought internal energy is like the work done by the internal system such as the spring (or so I thought)
There are all sorts of definitions of intetrnal energy, depending upon what you choose as the system, and which textbook you use. Sometimes internal energy refers to energy gong on at the molecular scale (moving molecules), and sometimes energy due to friction is referred to as intenal energy, where really it is heat, sound, light, chemical energy, etc. I find it best to consider the change in potential energy equal to the negative of the work done by conservative forces, like gravity and springs.
So the spring is never considered as an "internal energy", am I getting the wrong definition of internal energy?
see above
(1) I thought the "Work done by all forces" mean external forces like someone pulling, I've always thought gravity is work done by an internal force

(2) Wait, shouldn't it be the negative of work done by all the non-conservative forces?
No, work done by all forces includes work done by both conservative forces (springs, gravity) and non conservative forces (friction, applied pulling/pushing forces, external tension forces, etc). In a nutshell:

Wc + Wnc = ΔKE, the work energy theorem, and
Wnc = ΔKE + ΔPE, the conservation of total energy theorem, from which
Wc = -ΔPE

And if your'e still confused, i don't fault you. The negative signs are killers. :devil:
 
  • #20
flyingpig said:
Here is here I got confused, why is there another negative sign?

Shouldn't you leave it as it is and so?

-½m(v² - v₀²) = - mg(h - h₀)

So that

-ΔKE = ΔPE

But it doesn't make sense when you have

-ΔPE - ΔKE = 0

Since I said ΔPE = - mg(h - h₀)

- ( - mg(h - h₀) ) -½m(v² - v₀²) = 0

mg(h - h₀) - ½m(v² - v₀²) = 0

You got confused because you don't know what is the definition of potential energy of a conservative force. Read what I posted again -


"½m(v₀² - v²) = Uf - Ui

Now the change in potential energy of a system corresponding to a conservative internal force is defined as Uf - Ui = -W, where W is the work done on the system by the internal force (here gravity is the internal force if you consider Earth as a part of your system) as the system passes from its initial configuration i to final configuration f.
So
Uf - Ui = -[- mg(h - h₀)]
 
  • #21
Conservative force = the negative of a potential function, that's all I know

Is the fundamental difference here is work vs energy?
 
  • #22
flyingpig said:
Conservative force = the negative of a potential function, that's all I know
A conservative force is the negative of the gradient of a potential function:uhh:. That's a bit advanced. By now, I thought you would know that the work done by a conservative force is the negative of its change in potential energy.
Is the fundamental difference here is work vs energy?
Yes. Work and Energy are not the same. Work is a change in the mechanical energy of a system , per equations previously tendered. If you want to leave work out of the equation, then
ΔKE + ΔPE + ΔEother = 0 (energy can neither be created or destroyed) , where
ΔEother is the change in energy transferred into or out of the system in the form of heat, sound, light, or other forms of energy. So let's say that an oject is dropped from rest from a tall building where air resistance is a factor, with gravity being the conservative force acting and air drag is the non-conservative force acting, then, using the work-energy theorem, Wc + Wnc = ΔKE, or using the conservation of energy equation, Wnc = ΔKE + ΔPE, or forgetting work and using energy terms only, then, ΔKE + ΔPE = - ΔEother . . They all lead to the same result. Note that by comparing the last 2 equations, Wnc = - ΔEother . Do you see the difference between work and energy and the darn plus and minus signs??
 
  • #23
Yes. Work and Energy are not the same. Work is a change in the mechanical energy of a system , per equations previously tendered. If you want to leave work out of the equation, then
ΔKE + ΔPE + ΔEother = 0

Wait, then why is it called the conservation of energy when it should be the conservation of work?

Could you actually show how to derive that equation? There have been many explanations on this, but I am still not convinced.

I watched a video by Walter Lewin where he explains

Wab = ΔKE

Suppose a block was at rest and then is moved up on a ramp (frictionless)

Then

-mgh = ½mv²
 
  • #24
flyingpig said:
Wait, then why is it called the conservation of energy when it should be the conservation of work?
Why? Perhaps let's get rid of the 'delta' terms and rewrite the conservation of energy law as
(KE +PE +Eother)initial = (KE +PE +Eother)final
That is, the initial energy of a system is always equal to the final energy of the system, in different forms. Energy is conserved (never lost or gained, just transferred within or without the system to different forms).
Could you actually show how to derive that equation?
If I knew it once, i don't know it now. It is a principle that has never been to known to have been violated (at least in the classical sense).
There have been many explanations on this, but I am still not convinced.
I am sure that many have different interpretations of this law, but they all lead to Rome.
I watched a video by Walter Lewin where he explains

Wab = ΔKE
Brilliant man, Prof Lewin. This is the work energy theorem that has been alluded to more than once in this thread.
Suppose a block was at rest and then is moved up on a ramp (frictionless)

Then

-mgh = ½mv²
Well, no, how could it move up against gravity? Another force must be involved , like a pushing applied force P parallel to the ramp. In which case
(KE +PE +Eother)initial = (KE +PE +Eother)final

0 + 0 + Ph/sin theta = 1/2mv2 + mgh + 0, where Ph/sintheta is the energy
transferred into the system by the applied force P (chemical, biological, or other form of energy). Or use any variation of this equation you would like to use, like the work energy theorem, or the conservation of energy law where ΔE other can be written as equal to the negative of the work done by the force P , and entered on the left side of the equation, eliminating the Eother initial and final terms. In any case, you end up with the same equation.
 
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  • #25
Sorry, the minus sign stung me:blushing:. I edited my above post to correct this. The Eother initial term represents the chemical/biological/etc energy stored in the person applying the force P. I was trying to show that the conservation of energy law can be written in terms of all energy terms instead of work and energy terms, but it is usually almost always best to use the alternate forms
Wnc = ΔKE + ΔPE , or Wc + Wnc = ΔKE
and please watch plus and minus signs:smile:
 
  • #26
Oh my god, this is still seriously confusing me and we have been going through circles all over this thread and inspite of all the effort you guys have put in, it just isn't hitting my head at all.

Going back to the ramp problem

0 + 0 + Ph/sin theta = 1/2mv2 + mgh + 0

This is the problem, this is work x distance. The mass of the block times gravity (force) times the distance. Their dot product yields 180 and hence -1, so the god darn + sign! I got a final tomorrow and after this if I ever go on to second yr physics, my new profs are going to raise their eyebrows and look down on me for asking something so stupid.
 
  • #27
No, nothing you ask is stupid.

The only conservative forces you are going to encounter (generally) in Intro Physics -Mechanics-- are gravity forces (mg) and spring forces (kx). Every other force, especially friction and applied forces, are non-conservative.

There are 2 equations you can use when both conservative forces and nonconservative forces are acting.

1. KEinitial + PEinitial + Wnc = KEfinal + PEfinal , the conservation of total energy principle, OR
2. Wc + Wnc = KEfinal - KEinitial, the Work-Energy theorem.

So with the ramp problem inclined at an angle theta with the horizontal , if a block of mass m is pushed from rest up a ramp by an applied force P parallel to the incline , reaching a height h at the top of the ramp, then using equation 1,
0 + 0 + Ph/sintheta = 1/2mv2 + mgh, solve for v at the top of the ramp. Now using equation 2,
- mgh + Ph/sintheta = 1/2mv2 - 0, solve for v at the top of the ramp.

The 2 equations are identical
 
  • #28
PhanthomJay said:
No, nothing you ask is stupid.

The only conservative forces you are going to encounter (generally) in Intro Physics -Mechanics-- are gravity forces (mg) and spring forces (kx). Every other force, especially friction and applied forces, are non-conservative.

There are 2 equations you can use when both conservative forces and nonconservative forces are acting.

1. KEinitial + PEinitial -Wnc = KEfinal + PEfinal , the conservation of total energy principle, OR
2. Wc - Wnc = KEfinal - KEinitial, the Work-Energy theorem.

Don't you mean

So with the ramp problem inclined at an angle theta with the horizontal , if a block of mass m is pushed from rest up a ramp by an applied force P parallel to the incline , reaching a height h at the top of the ramp, then using equation 1,
0 + 0 + Ph/sintheta = 1/2mv2 + mgh[color], solve for v at the top of the ramp. Now using equation 2,
- mgh + Ph/sintheta = 1/2mv2 - 0, solve for v at the top of the ramp.

The 2 equations are identical


What is the meaning of the red stuff? It isn't work, so what is it? What is this mysterious quantity that also takes force x distance where the force happens to be equal to the magnitude of mg, but having opposite direction?
 
  • #29
flyingpig said:
What is the meaning of the red stuff? It isn't work, so what is it? What is this mysterious quantity that also takes force x distance where the force happens to be equal to the magnitude of mg, but having opposite direction?
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.

For conservative forces such as gravity you have a choice of how to treat them. You can treat them like any other force, including the work that they do. Or you can represent them by a potential energy term. The choices are equivalent and lead to the same equation. Take your pick!
 
  • #30
*sigh*, I just had my final and there were two annoying energy questions, thankfully none of them involved this concept.

Anyways, this also brings me the attention to friction doing work, is the "d" actually displacement or distance?
 
  • #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.
 
  • #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

[tex]W=\int_C \mathbf{F}\cdot d\mathbf{x}[/tex]

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

[tex]\mathbf{F}\cdot d\mathbf{x} = -|\mathbf{F}||d\mathbf{x}|[/tex]

and if |F| is constant, you get

[tex]W = -|\mathbf{F}|\int_C |d\mathbf{x}|[/tex]

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 [tex]\vec{F}\cdot d\vec{x}[/tex] along the path, where [itex]d\vec{x}[/itex] 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

[tex]W=\int_C \mathbf{F}\cdot d\mathbf{x}[/tex]

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

[tex]\mathbf{F}\cdot d\mathbf{x} = -|\mathbf{F}||d\mathbf{x}|[/tex]

and if |F| is constant, you get

[tex]W = -|\mathbf{F}|\int_C |d\mathbf{x}|[/tex]

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?
 
<h2>1. What is the conservation of energy?</h2><p>The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant.</p><h2>2. How does the conservation of energy work?</h2><p>The conservation of energy works by following the law of conservation of energy, which states that energy can neither be created nor destroyed. This means that energy can only be transferred from one form to another, such as from potential energy to kinetic energy.</p><h2>3. What is negative work and positive work?</h2><p>Negative work refers to the situation where the force and displacement act in opposite directions, resulting in a decrease in the energy of a system. Positive work, on the other hand, occurs when the force and displacement act in the same direction, resulting in an increase in the energy of a system.</p><h2>4. How is negative work different from positive work?</h2><p>Negative work and positive work differ in terms of the direction of the force and displacement. In negative work, the force and displacement act in opposite directions, while in positive work, they act in the same direction. This results in a decrease or increase in the energy of a system, respectively.</p><h2>5. Why is the concept of negative work important in the conservation of energy?</h2><p>The concept of negative work is important in the conservation of energy because it helps us understand how energy is transferred and transformed in a system. It also allows us to account for any changes in the energy of a system, whether it is an increase or decrease, and how this affects the overall conservation of energy in the system.</p>

1. What is the conservation of energy?

The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant.

2. How does the conservation of energy work?

The conservation of energy works by following the law of conservation of energy, which states that energy can neither be created nor destroyed. This means that energy can only be transferred from one form to another, such as from potential energy to kinetic energy.

3. What is negative work and positive work?

Negative work refers to the situation where the force and displacement act in opposite directions, resulting in a decrease in the energy of a system. Positive work, on the other hand, occurs when the force and displacement act in the same direction, resulting in an increase in the energy of a system.

4. How is negative work different from positive work?

Negative work and positive work differ in terms of the direction of the force and displacement. In negative work, the force and displacement act in opposite directions, while in positive work, they act in the same direction. This results in a decrease or increase in the energy of a system, respectively.

5. Why is the concept of negative work important in the conservation of energy?

The concept of negative work is important in the conservation of energy because it helps us understand how energy is transferred and transformed in a system. It also allows us to account for any changes in the energy of a system, whether it is an increase or decrease, and how this affects the overall conservation of energy in the system.

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