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Why does work depend on distance?
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So the formula for work is$$
left[textworkright] ~=~ left[textforceright] , times , left[textdistanceright]
,.
$$
I'm trying to get an understanding of how this represents energy.
If I'm in a vacuum, and I push a block with a force of $1 , mathrmN,$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block.
I must be missing something, but I can't really pinpoint what it is.
It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed.
newtonian-mechanics forces work definition distance
edited 51 mins ago
Qmechanic♦
96.8k121631023
asked 2 hours ago
Dominic Roy-Stang
233
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up vote
4
down vote
favorite
So the formula for work is$$
left[textworkright] ~=~ left[textforceright] , times , left[textdistanceright]
,.
$$
I'm trying to get an understanding of how this represents energy.
If I'm in a vacuum, and I push a block with a force of $1 , mathrmN,$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block.
I must be missing something, but I can't really pinpoint what it is.
It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed.
newtonian-mechanics forces work definition distance
edited 51 mins ago
Qmechanic♦
96.8k121631023
asked 2 hours ago
Dominic Roy-Stang
233
New contributor
Dominic Roy-Stang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
The key is that distance in that formula refers not to the distance moved by the object, but to the distance over which the force is applied. You can push an object and have it coast due to inertia, but the formula only cares about how far you actively pushed it.
– Nuclear Wang
50 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
So the formula for work is$$
left[textworkright] ~=~ left[textforceright] , times , left[textdistanceright]
,.
$$
I'm trying to get an understanding of how this represents energy.
If I'm in a vacuum, and I push a block with a force of $1 , mathrmN,$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block.
I must be missing something, but I can't really pinpoint what it is.
It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed.
newtonian-mechanics forces work definition distance
edited 51 mins ago
Qmechanic♦
96.8k121631023
asked 2 hours ago
Dominic Roy-Stang
233
New contributor
Dominic Roy-Stang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
So the formula for work is$$
left[textworkright] ~=~ left[textforceright] , times , left[textdistanceright]
,.
$$
I'm trying to get an understanding of how this represents energy.
If I'm in a vacuum, and I push a block with a force of $1 , mathrmN,$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block.
I must be missing something, but I can't really pinpoint what it is.
It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed.
newtonian-mechanics forces work definition distance
newtonian-mechanics forces work definition distance
edited 51 mins ago
Qmechanic♦
96.8k121631023
asked 2 hours ago
Dominic Roy-Stang
233
New contributor
Dominic Roy-Stang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 51 mins ago
Qmechanic♦
96.8k121631023
asked 2 hours ago
Dominic Roy-Stang
233
New contributor
Dominic Roy-Stang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 51 mins ago
Qmechanic♦
96.8k121631023
edited 51 mins ago
Qmechanic♦
96.8k121631023
edited 51 mins ago
Qmechanic♦
96.8k121631023
96.8k121631023
asked 2 hours ago
Dominic Roy-Stang
233
New contributor
Dominic Roy-Stang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Dominic Roy-Stang
233
asked 2 hours ago
Dominic Roy-Stang
233
233
New contributor
Dominic Roy-Stang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dominic Roy-Stang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dominic Roy-Stang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
The key is that distance in that formula refers not to the distance moved by the object, but to the distance over which the force is applied. You can push an object and have it coast due to inertia, but the formula only cares about how far you actively pushed it.
– Nuclear Wang
50 mins ago
add a comment |Â
3
The key is that distance in that formula refers not to the distance moved by the object, but to the distance over which the force is applied. You can push an object and have it coast due to inertia, but the formula only cares about how far you actively pushed it.
– Nuclear Wang
50 mins ago
3
3
The key is that distance in that formula refers not to the distance moved by the object, but to the distance over which the force is applied. You can push an object and have it coast due to inertia, but the formula only cares about how far you actively pushed it.
– Nuclear Wang
50 mins ago
The key is that distance in that formula refers not to the distance moved by the object, but to the distance over which the force is applied. You can push an object and have it coast due to inertia, but the formula only cares about how far you actively pushed it.
– Nuclear Wang
50 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
10
down vote
accepted
You have to put in the distance on which the force acts. If you release the force, there will be no work done since there is no force acting on the body.
answered 2 hours ago
EuklidAlexandria
522210
add a comment |Â
up vote
4
down vote
If I'm in a vacuum, and I push a block with a force of 1N, it will move forwards infinitely
and accelerate the block ie change the block's velocity and hence change the kinetic energy of the block.
The longer you apply the force, the n more work the force does resulting in a greater change in the kinetic energy of the block.
answered 1 hour ago
Farcher
44.5k33388
add a comment |Â
up vote
2
down vote
Often it is important to know if a given formula is a simplification of a more general equation and, when you encounter a conceptual problem, check the general formula. In this case it is a simplification of this formula:
$$W=int_S Fcdot ds $$
Where S is the path over which we are interested in the work and ds is an infinitesimally small segment of S.
So back to your question, wherever F=0 the integrand is 0 regardless of how long that segment of the path is. So it is only that first segment where you are applying the 1N that work is done. Once you stop pushing the distance increases but the work does not.
answered 1 hour ago
Dale
62718
add a comment |Â
up vote
0
down vote
I see several answers that all seem to explain it, but for someone trying to understand the why, perhaps it's best answered simply.
It is going to be a lot more "work" for me to push a heavy trashcan out the door and down the driveway than the amount of "work" for me to just push the trashcan out of the house.
answered 56 mins ago
Brad
1011
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Brad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
You have to put in the distance on which the force acts. If you release the force, there will be no work done since there is no force acting on the body.
answered 2 hours ago
EuklidAlexandria
522210
add a comment |Â
up vote
10
down vote
accepted
You have to put in the distance on which the force acts. If you release the force, there will be no work done since there is no force acting on the body.
answered 2 hours ago
EuklidAlexandria
522210
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
You have to put in the distance on which the force acts. If you release the force, there will be no work done since there is no force acting on the body.
answered 2 hours ago
EuklidAlexandria
522210
You have to put in the distance on which the force acts. If you release the force, there will be no work done since there is no force acting on the body.
answered 2 hours ago
EuklidAlexandria
522210
answered 2 hours ago
EuklidAlexandria
522210
answered 2 hours ago
EuklidAlexandria
522210
answered 2 hours ago
EuklidAlexandria
522210
522210
add a comment |Â
add a comment |Â
up vote
4
down vote
If I'm in a vacuum, and I push a block with a force of 1N, it will move forwards infinitely
and accelerate the block ie change the block's velocity and hence change the kinetic energy of the block.
The longer you apply the force, the n more work the force does resulting in a greater change in the kinetic energy of the block.
answered 1 hour ago
Farcher
44.5k33388
add a comment |Â
up vote
4
down vote
If I'm in a vacuum, and I push a block with a force of 1N, it will move forwards infinitely
and accelerate the block ie change the block's velocity and hence change the kinetic energy of the block.
The longer you apply the force, the n more work the force does resulting in a greater change in the kinetic energy of the block.
answered 1 hour ago
Farcher
44.5k33388
add a comment |Â
up vote
4
down vote
up vote
4
down vote
If I'm in a vacuum, and I push a block with a force of 1N, it will move forwards infinitely
and accelerate the block ie change the block's velocity and hence change the kinetic energy of the block.
The longer you apply the force, the n more work the force does resulting in a greater change in the kinetic energy of the block.
answered 1 hour ago
Farcher
44.5k33388
If I'm in a vacuum, and I push a block with a force of 1N, it will move forwards infinitely
and accelerate the block ie change the block's velocity and hence change the kinetic energy of the block.
The longer you apply the force, the n more work the force does resulting in a greater change in the kinetic energy of the block.
answered 1 hour ago
Farcher
44.5k33388
answered 1 hour ago
Farcher
44.5k33388
answered 1 hour ago
Farcher
44.5k33388
answered 1 hour ago
Farcher
44.5k33388
44.5k33388
add a comment |Â
add a comment |Â
up vote
2
down vote
Often it is important to know if a given formula is a simplification of a more general equation and, when you encounter a conceptual problem, check the general formula. In this case it is a simplification of this formula:
$$W=int_S Fcdot ds $$
Where S is the path over which we are interested in the work and ds is an infinitesimally small segment of S.
So back to your question, wherever F=0 the integrand is 0 regardless of how long that segment of the path is. So it is only that first segment where you are applying the 1N that work is done. Once you stop pushing the distance increases but the work does not.
answered 1 hour ago
Dale
62718
add a comment |Â
up vote
2
down vote
Often it is important to know if a given formula is a simplification of a more general equation and, when you encounter a conceptual problem, check the general formula. In this case it is a simplification of this formula:
$$W=int_S Fcdot ds $$
Where S is the path over which we are interested in the work and ds is an infinitesimally small segment of S.
So back to your question, wherever F=0 the integrand is 0 regardless of how long that segment of the path is. So it is only that first segment where you are applying the 1N that work is done. Once you stop pushing the distance increases but the work does not.
answered 1 hour ago
Dale
62718
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Often it is important to know if a given formula is a simplification of a more general equation and, when you encounter a conceptual problem, check the general formula. In this case it is a simplification of this formula:
$$W=int_S Fcdot ds $$
Where S is the path over which we are interested in the work and ds is an infinitesimally small segment of S.
So back to your question, wherever F=0 the integrand is 0 regardless of how long that segment of the path is. So it is only that first segment where you are applying the 1N that work is done. Once you stop pushing the distance increases but the work does not.
answered 1 hour ago
Dale
62718
Often it is important to know if a given formula is a simplification of a more general equation and, when you encounter a conceptual problem, check the general formula. In this case it is a simplification of this formula:
$$W=int_S Fcdot ds $$
Where S is the path over which we are interested in the work and ds is an infinitesimally small segment of S.
So back to your question, wherever F=0 the integrand is 0 regardless of how long that segment of the path is. So it is only that first segment where you are applying the 1N that work is done. Once you stop pushing the distance increases but the work does not.
answered 1 hour ago
Dale
62718
answered 1 hour ago
Dale
62718
answered 1 hour ago
Dale
62718
answered 1 hour ago
Dale
62718
62718
add a comment |Â
add a comment |Â
up vote
0
down vote
I see several answers that all seem to explain it, but for someone trying to understand the why, perhaps it's best answered simply.
It is going to be a lot more "work" for me to push a heavy trashcan out the door and down the driveway than the amount of "work" for me to just push the trashcan out of the house.
answered 56 mins ago
Brad
1011
New contributor
Brad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
I see several answers that all seem to explain it, but for someone trying to understand the why, perhaps it's best answered simply.
It is going to be a lot more "work" for me to push a heavy trashcan out the door and down the driveway than the amount of "work" for me to just push the trashcan out of the house.
answered 56 mins ago
Brad
1011
New contributor
Brad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I see several answers that all seem to explain it, but for someone trying to understand the why, perhaps it's best answered simply.
It is going to be a lot more "work" for me to push a heavy trashcan out the door and down the driveway than the amount of "work" for me to just push the trashcan out of the house.
answered 56 mins ago
Brad
1011
New contributor
Brad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I see several answers that all seem to explain it, but for someone trying to understand the why, perhaps it's best answered simply.
It is going to be a lot more "work" for me to push a heavy trashcan out the door and down the driveway than the amount of "work" for me to just push the trashcan out of the house.
answered 56 mins ago
Brad
1011
New contributor
Brad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 56 mins ago
Brad
1011
New contributor
Brad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 56 mins ago
Brad
1011
answered 56 mins ago
Brad
1011
1011
New contributor
Brad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Brad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Brad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
Dominic Roy-Stang is a new contributor. Be nice, and check out our Code of Conduct.
Dominic Roy-Stang is a new contributor. Be nice, and check out our Code of Conduct.
Dominic Roy-Stang is a new contributor. Be nice, and check out our Code of Conduct.
Dominic Roy-Stang is a new contributor. Be nice, and check out our Code of Conduct.
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3
The key is that distance in that formula refers not to the distance moved by the object, but to the distance over which the force is applied. You can push an object and have it coast due to inertia, but the formula only cares about how far you actively pushed it.
– Nuclear Wang
50 mins ago