Energy, once again weird unclear thing

[Dale]

"in the ground frame: w = f.d = 10 N 300 m = 3 kJ"
isn't the work is still 0, because the angle between the force and the distance is 90

Thanks for putting some effort in. Sorry, my description was a bit unclear. The stewardess is holding the cart stationary by pushing horizontally so that it doesn't roll backwards towards her, so the force is approximately parallel to the displacement.

 

[Mad_Eye]

Thanks for putting some effort in. Sorry, my description was a bit unclear. The stewardess is holding the cart stationary by pushing horizontally so that it doesn't roll backwards towards her, so the force is approximately parallel to the displacement.

i see.
and I've edited my post (quite a lot :D)

 

[Dale]

i've checked up your example, and in all cases, the work is the same (19.8J), if you consider both forces of action and reaction as one.
so it is just confusing you are keep telling me work is frame variant, and giving me example in which the work is the same in all frames.

i guess i miss something, and this is what I'm trying to understand.
for example, is it alright to consider the work of both forces of action and reaction, as one work?

As you did the problem I am sure that you noticed that each force individually did a different amount of work in each frame. Thus work is frame variant. (Look at the definition of work. Work is defined for an individual force, not for a 3rd law pair of forces).

You also noticed that once you included the conservation of momentum (Netwon's 3rd law) for the system then you found that energy was also conserved for the system. In other words, if you want to exert this much force on these objects using something like a spring then you need to compress it with 19.8 J. Thus since all frames see the spring lose 19.8 J of PE and all frames see the system gain 19.8 J of KE then all frames agree that energy was conserved despite the fact that the work done by each force was frame variant.

 

[DaTario]

Let me please insert a question which I repute as pertaining to this subject. When passing from one inertial frame to another is it correct to say that we are introducing that famous additive constant to work (energy)?

Trying to be clearer: Energy concept works well even if we sum a constant to all energy levels. Does the frame variance imply sum of a constant to all possible works ?


Best Regards

DaTario

 

[Mad_Eye]

You also noticed that once you included the conservation of momentum (Newton's 3rd law) for the system then you found that energy was also conserved for the system. In other words, if you want to exert this much force on these objects using something like a spring then you need to compress it with 19.8 J. Thus since all frames see the spring lose 19.8 J of PE and all frames see the system gain 19.8 J of KE then all frames agree that energy was conserved despite the fact that the work done by each force was frame variant.

What exactly do you mean by "the energy is conserved"?
I didn't found the energy conserved, I've just seen the total amount of work is the same in all frames, but I have no idea where that energy came from.

(And if already in that, where indeed this KE (19.8J) came from?)

 

[Dale]

The kinetic energy came from the potential energy of whatever force or device pushed the objects apart.

The point is that the potential energy that was lost by that force or device, regardless of whether it was chemical potential energy in some fuel or elastic potential energy in a spring or electromagnetic potential energy in a field, has decreased by some amount. Neglecting relativistic effects, that amount is the same in all reference frames (e.g. a liter of gasoline has X joules of chemical potential energy). Therefore, for energy to be conserved all reference frames must agree on the overall change in the system's kinetic energy even if they disagree on the work done by each individual force. Thus energy is frame variant, but conserved.

 

[Mad_Eye]

The kinetic energy came from the potential energy of whatever force or device pushed the objects apart.

The point is that the potential energy that was lost by that force or device, regardless of whether it was chemical potential energy in some fuel or elastic potential energy in a spring or electromagnetic potential energy in a field, has decreased by some amount. Neglecting relativistic effects, that amount is the same in all reference frames (e.g. a liter of gasoline has X joules of chemical potential energy). Therefore, for energy to be conserved all reference frames must agree on the overall change in the system's kinetic energy even if they disagree on the work done by each individual force. Thus energy is frame variant, but conserved.

I see.
I thought maybe in other frames there is different amount of KE that transform into PE. It still possible energy is conserved is such case, but I'm glad at least this is frame invariant, and correct me if I'm wrong, the amount of transferred energy, from one kind to another.

 

[Dale]

I thought maybe in other frames there is different amount of KE that transform into PE.

This does indeed happen in relativity (in a way that maintains the conservation of energy and momentum). The math is a little more complicated but, IMO, rather elegant. If you are interested we can discuss that, but I don't want to sidetrack things if you are only interested in non-relativistic effects.

 

[Mad_Eye]

This does indeed happen in relativity (in a way that maintains the conservation of energy and momentum). The math is a little more complicated but, IMO, rather elegant. If you are interested we can discuss that, but I don't want to sidetrack things if you are only interested in non-relativistic effects.

Haha, as you see I don't exactly have quick perception.
I'd be glad to learn about modern physics, but now I know nothing about it, and quite far from mastering classic physics.

I do believe things should be learned in order (unlike all this popular science!) as much as this is fascinating.

(But don't worry, I'll learned it someday, and return here to ask weird questions)


I'll just bring again two questions,
(in classic physics of course),
Is the amount of transformed energy, is frame invariant?
I mean, not only KE<->PE but even from one kind of PE to another kind pf PE?
(I almost sure yes, but just a confirmation to close the discussion)

And the other one, do you have then any recommendation for a physics book to learn things deeply?
(how did you all learn?)

 

[Dale]

I'll just bring again two questions,
(in classic physics of course),
Is the amount of transformed energy, is frame invariant?
I mean, not only KE<->PE but even from one kind of PE to another kind pf PE?
(I almost sure yes, but just a confirmation to close the discussion)

Non-relativistically, yes. Despite the fact that the work done by any given force is frame-variant.

And the other one, do you have then any recommendation for a physics book to learn things deeply?
(how did you all learn?)

I learned by working a lot of problems as part of a traditional course. I cannot overemphasize the importance of actually doing some problems, probably more important than a good textbook. My textbook was Serway, which I liked just fine.

 

[mviswanathan]

I was going through the discussion which was very interesting to me.
I was also not clear as to why the work-done as calculated is 2J and 4J in the two cases mentioned in the first post.
Now I think, I have understood .
The basic mistake I was doing was that I was considering the force which accelerates the car is acting only on the car - neglecting that any force has to act between two objects, which results in acceleration of both the objects.
In this case, the force is acting between the ground and the car which accelerates the car in one direction and also accelerates the ground in the opposite direction. Considering the change in KE of both the objects, the work done is same whether observed from the ground or from the reference of the person walking wish some velocity with respect to the ground.

In case of a rocket, the force acts between the rocket and the products exhausted.

Thanks for this thread and the discussions.

 

[Mad_Eye]

Non-relativistically, yes. Despite the fact that the work done by any given force is frame-variant.

I learned by working a lot of problems as part of a traditional course. I cannot overemphasize the importance of actually doing some problems, probably more important than a good textbook. My textbook was Serway, which I liked just fine.

Fine. Thank you very much for the patient.
Now I still have some more questions about energy, but quite different (one is https://www.physicsforums.com/showthread.php?t=382155&page=3)

Should I open new post for the rest of them, or ask them here?