Mixing
Kinetic energy and conservative fields
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If $W = E_1 - E_0$ for every force and a conservative field is a field such that the work on a closed path is 0, then every field should be conservative, since $W = E_0 - E_0 = 0$.
I know that this can't be true, but i can't see why
newtonian-mechanics energy work potential-energy conservative-field
edited Aug 16 at 11:17
Qmechanic♦
96.6k121631020
asked Aug 16 at 10:53
Andrea
41
add a comment |Â
up vote
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If $W = E_1 - E_0$ for every force and a conservative field is a field such that the work on a closed path is 0, then every field should be conservative, since $W = E_0 - E_0 = 0$.
I know that this can't be true, but i can't see why
newtonian-mechanics energy work potential-energy conservative-field
edited Aug 16 at 11:17
Qmechanic♦
96.6k121631020
asked Aug 16 at 10:53
Andrea
41
What do your subscripts represent here? Just positions in space?
– Aaron Stevens
Aug 16 at 11:18
It would be helpful if you clarified the meaning of $E_0$ and $E_1$. Are these potential energies in two points in space? Kinetic energies? Total energies?
– V.F.
Aug 16 at 14:33
They are kinetic energy
– Andrea
Aug 16 at 14:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $W = E_1 - E_0$ for every force and a conservative field is a field such that the work on a closed path is 0, then every field should be conservative, since $W = E_0 - E_0 = 0$.
I know that this can't be true, but i can't see why
newtonian-mechanics energy work potential-energy conservative-field
edited Aug 16 at 11:17
Qmechanic♦
96.6k121631020
asked Aug 16 at 10:53
Andrea
41
If $W = E_1 - E_0$ for every force and a conservative field is a field such that the work on a closed path is 0, then every field should be conservative, since $W = E_0 - E_0 = 0$.
I know that this can't be true, but i can't see why
newtonian-mechanics energy work potential-energy conservative-field
edited Aug 16 at 11:17
Qmechanic♦
96.6k121631020
asked Aug 16 at 10:53
Andrea
41
edited Aug 16 at 11:17
Qmechanic♦
96.6k121631020
edited Aug 16 at 11:17
Qmechanic♦
96.6k121631020
edited Aug 16 at 11:17
Qmechanic♦
96.6k121631020
96.6k121631020
asked Aug 16 at 10:53
Andrea
41
asked Aug 16 at 10:53
Andrea
41
asked Aug 16 at 10:53
Andrea
41
41
What do your subscripts represent here? Just positions in space?
– Aaron Stevens
Aug 16 at 11:18
It would be helpful if you clarified the meaning of $E_0$ and $E_1$. Are these potential energies in two points in space? Kinetic energies? Total energies?
– V.F.
Aug 16 at 14:33
They are kinetic energy
– Andrea
Aug 16 at 14:47
add a comment |Â
What do your subscripts represent here? Just positions in space?
– Aaron Stevens
Aug 16 at 11:18
It would be helpful if you clarified the meaning of $E_0$ and $E_1$. Are these potential energies in two points in space? Kinetic energies? Total energies?
– V.F.
Aug 16 at 14:33
They are kinetic energy
– Andrea
Aug 16 at 14:47
What do your subscripts represent here? Just positions in space?
– Aaron Stevens
Aug 16 at 11:18
What do your subscripts represent here? Just positions in space?
– Aaron Stevens
Aug 16 at 11:18
It would be helpful if you clarified the meaning of $E_0$ and $E_1$. Are these potential energies in two points in space? Kinetic energies? Total energies?
– V.F.
Aug 16 at 14:33
It would be helpful if you clarified the meaning of $E_0$ and $E_1$. Are these potential energies in two points in space? Kinetic energies? Total energies?
– V.F.
Aug 16 at 14:33
They are kinetic energy
– Andrea
Aug 16 at 14:47
They are kinetic energy
– Andrea
Aug 16 at 14:47
add a comment |Â
3 Answers
3
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oldest
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1
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Suppose $F$ is a resultant force acting on a system, and $A$ is a work associated with that force, then (Work-Energy Theorem)
$$
A = K_2 - K_1
$$
where K is a kinetic energy of the system. If there are conservative forces we can rewrite the equation simply as
$$
A_cons + A_other = K_2 - K_1.
$$
The potential energy $ß$ is defined, so that
$$
A_cons = ß_1 - ß_2,
$$
hence the work of non-conservative forces is equal to difference in full energy:
$$
A_other = (K_2 + ß_2) - (K_1 + ß_1)
$$
This equation applied to the closed path will give
$$
A_other = K_2 - K_1
$$
answered Aug 16 at 12:35
Ice-Nine
16114
add a comment |Â
up vote
1
down vote
Your notation presupposes that "the energy at point 0" or "the energy at point 1" is well-defined. If the work required to move an object from point 0 to point 1 is dependent on the path taken, then we can't meaningfully define a "change in energy" between the two points; we would also need to know the path taken between the two points. Only if the work required is path-independent can we even define the notion of an object's energy "at a given point".
For example, imagine pushing a box across a table from point 0 to point 1. The force of friction during this process is constant and opposes the motion; so the work done is proportional to the distance travelled. But this means that the work done by friction between these two points depends on how you move the box between these two points. The quantity "$E_1 - E_0$" doesn't just depend on the positions of the points 0 and 1, and so it's not a very useful thing to keep track of.
answered Aug 16 at 13:35
Michael Seifert
13.6k12650
add a comment |Â
up vote
0
down vote
KE is not a state function, when the system returns to its initial state then KE energy is not necessarily same.Hence work can not be always zero.
answered Aug 16 at 11:04
Rajendra Pd
48728
1
If the KE is different, then the state has not returned to the initial state.
– garyp
Aug 16 at 11:33
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Suppose $F$ is a resultant force acting on a system, and $A$ is a work associated with that force, then (Work-Energy Theorem)
$$
A = K_2 - K_1
$$
where K is a kinetic energy of the system. If there are conservative forces we can rewrite the equation simply as
$$
A_cons + A_other = K_2 - K_1.
$$
The potential energy $ß$ is defined, so that
$$
A_cons = ß_1 - ß_2,
$$
hence the work of non-conservative forces is equal to difference in full energy:
$$
A_other = (K_2 + ß_2) - (K_1 + ß_1)
$$
This equation applied to the closed path will give
$$
A_other = K_2 - K_1
$$
answered Aug 16 at 12:35
Ice-Nine
16114
add a comment |Â
up vote
1
down vote
Suppose $F$ is a resultant force acting on a system, and $A$ is a work associated with that force, then (Work-Energy Theorem)
$$
A = K_2 - K_1
$$
where K is a kinetic energy of the system. If there are conservative forces we can rewrite the equation simply as
$$
A_cons + A_other = K_2 - K_1.
$$
The potential energy $ß$ is defined, so that
$$
A_cons = ß_1 - ß_2,
$$
hence the work of non-conservative forces is equal to difference in full energy:
$$
A_other = (K_2 + ß_2) - (K_1 + ß_1)
$$
This equation applied to the closed path will give
$$
A_other = K_2 - K_1
$$
answered Aug 16 at 12:35
Ice-Nine
16114
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Suppose $F$ is a resultant force acting on a system, and $A$ is a work associated with that force, then (Work-Energy Theorem)
$$
A = K_2 - K_1
$$
where K is a kinetic energy of the system. If there are conservative forces we can rewrite the equation simply as
$$
A_cons + A_other = K_2 - K_1.
$$
The potential energy $ß$ is defined, so that
$$
A_cons = ß_1 - ß_2,
$$
hence the work of non-conservative forces is equal to difference in full energy:
$$
A_other = (K_2 + ß_2) - (K_1 + ß_1)
$$
This equation applied to the closed path will give
$$
A_other = K_2 - K_1
$$
answered Aug 16 at 12:35
Ice-Nine
16114
Suppose $F$ is a resultant force acting on a system, and $A$ is a work associated with that force, then (Work-Energy Theorem)
$$
A = K_2 - K_1
$$
where K is a kinetic energy of the system. If there are conservative forces we can rewrite the equation simply as
$$
A_cons + A_other = K_2 - K_1.
$$
The potential energy $ß$ is defined, so that
$$
A_cons = ß_1 - ß_2,
$$
hence the work of non-conservative forces is equal to difference in full energy:
$$
A_other = (K_2 + ß_2) - (K_1 + ß_1)
$$
This equation applied to the closed path will give
$$
A_other = K_2 - K_1
$$
answered Aug 16 at 12:35
Ice-Nine
16114
answered Aug 16 at 12:35
Ice-Nine
16114
answered Aug 16 at 12:35
Ice-Nine
16114
answered Aug 16 at 12:35
Ice-Nine
16114
16114
add a comment |Â
add a comment |Â
up vote
1
down vote
Your notation presupposes that "the energy at point 0" or "the energy at point 1" is well-defined. If the work required to move an object from point 0 to point 1 is dependent on the path taken, then we can't meaningfully define a "change in energy" between the two points; we would also need to know the path taken between the two points. Only if the work required is path-independent can we even define the notion of an object's energy "at a given point".
For example, imagine pushing a box across a table from point 0 to point 1. The force of friction during this process is constant and opposes the motion; so the work done is proportional to the distance travelled. But this means that the work done by friction between these two points depends on how you move the box between these two points. The quantity "$E_1 - E_0$" doesn't just depend on the positions of the points 0 and 1, and so it's not a very useful thing to keep track of.
answered Aug 16 at 13:35
Michael Seifert
13.6k12650
add a comment |Â
up vote
1
down vote
Your notation presupposes that "the energy at point 0" or "the energy at point 1" is well-defined. If the work required to move an object from point 0 to point 1 is dependent on the path taken, then we can't meaningfully define a "change in energy" between the two points; we would also need to know the path taken between the two points. Only if the work required is path-independent can we even define the notion of an object's energy "at a given point".
For example, imagine pushing a box across a table from point 0 to point 1. The force of friction during this process is constant and opposes the motion; so the work done is proportional to the distance travelled. But this means that the work done by friction between these two points depends on how you move the box between these two points. The quantity "$E_1 - E_0$" doesn't just depend on the positions of the points 0 and 1, and so it's not a very useful thing to keep track of.
answered Aug 16 at 13:35
Michael Seifert
13.6k12650
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your notation presupposes that "the energy at point 0" or "the energy at point 1" is well-defined. If the work required to move an object from point 0 to point 1 is dependent on the path taken, then we can't meaningfully define a "change in energy" between the two points; we would also need to know the path taken between the two points. Only if the work required is path-independent can we even define the notion of an object's energy "at a given point".
For example, imagine pushing a box across a table from point 0 to point 1. The force of friction during this process is constant and opposes the motion; so the work done is proportional to the distance travelled. But this means that the work done by friction between these two points depends on how you move the box between these two points. The quantity "$E_1 - E_0$" doesn't just depend on the positions of the points 0 and 1, and so it's not a very useful thing to keep track of.
answered Aug 16 at 13:35
Michael Seifert
13.6k12650
Your notation presupposes that "the energy at point 0" or "the energy at point 1" is well-defined. If the work required to move an object from point 0 to point 1 is dependent on the path taken, then we can't meaningfully define a "change in energy" between the two points; we would also need to know the path taken between the two points. Only if the work required is path-independent can we even define the notion of an object's energy "at a given point".
For example, imagine pushing a box across a table from point 0 to point 1. The force of friction during this process is constant and opposes the motion; so the work done is proportional to the distance travelled. But this means that the work done by friction between these two points depends on how you move the box between these two points. The quantity "$E_1 - E_0$" doesn't just depend on the positions of the points 0 and 1, and so it's not a very useful thing to keep track of.
answered Aug 16 at 13:35
Michael Seifert
13.6k12650
answered Aug 16 at 13:35
Michael Seifert
13.6k12650
answered Aug 16 at 13:35
Michael Seifert
13.6k12650
answered Aug 16 at 13:35
Michael Seifert
13.6k12650
13.6k12650
add a comment |Â
add a comment |Â
up vote
0
down vote
KE is not a state function, when the system returns to its initial state then KE energy is not necessarily same.Hence work can not be always zero.
answered Aug 16 at 11:04
Rajendra Pd
48728
1
If the KE is different, then the state has not returned to the initial state.
– garyp
Aug 16 at 11:33
add a comment |Â
up vote
0
down vote
KE is not a state function, when the system returns to its initial state then KE energy is not necessarily same.Hence work can not be always zero.
answered Aug 16 at 11:04
Rajendra Pd
48728
1
If the KE is different, then the state has not returned to the initial state.
– garyp
Aug 16 at 11:33
add a comment |Â
up vote
0
down vote
up vote
0
down vote
KE is not a state function, when the system returns to its initial state then KE energy is not necessarily same.Hence work can not be always zero.
answered Aug 16 at 11:04
Rajendra Pd
48728
KE is not a state function, when the system returns to its initial state then KE energy is not necessarily same.Hence work can not be always zero.
answered Aug 16 at 11:04
Rajendra Pd
48728
answered Aug 16 at 11:04
Rajendra Pd
48728
answered Aug 16 at 11:04
Rajendra Pd
48728
answered Aug 16 at 11:04
Rajendra Pd
48728
48728
1
If the KE is different, then the state has not returned to the initial state.
– garyp
Aug 16 at 11:33
add a comment |Â
1
If the KE is different, then the state has not returned to the initial state.
– garyp
Aug 16 at 11:33
1
1
If the KE is different, then the state has not returned to the initial state.
– garyp
Aug 16 at 11:33
If the KE is different, then the state has not returned to the initial state.
– garyp
Aug 16 at 11:33
add a comment |Â
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What do your subscripts represent here? Just positions in space?
– Aaron Stevens
Aug 16 at 11:18
It would be helpful if you clarified the meaning of $E_0$ and $E_1$. Are these potential energies in two points in space? Kinetic energies? Total energies?
– V.F.
Aug 16 at 14:33
They are kinetic energy
– Andrea
Aug 16 at 14:47