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
 |Â
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up vote
24
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
50
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.
â Nuclear Wang
Sep 13 at 17:10
8
... in a vacuum, -> ... in a frictionless environment, ...
â CramerTV
Sep 13 at 19:29
1
In your example, the block's energy only changed when you were pushing it. After that it has a constant unchanging speed, so a constant unchanging kinetic energy. The change in energy from before the push to after is the work done on the block.
â usul
Sep 13 at 19:39
4
Work is not force times distance. That's just a simplified version used to solve textbook problems with constant force. See Dale answer for the real definition of work.
â Bakuriu
Sep 13 at 21:21
1
Maybe the counterintuitive part is why it scales with distance, not time.
â JollyJoker
Sep 14 at 6:50
 |Â
show 2 more comments
up vote
24
down vote
favorite
up vote
24
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
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 Sep 13 at 17:09
Qmechanicâ¦
97.4k121641047
97.4k121641047
asked Sep 13 at 15:44
Dominic Roy-Stang
13815
13815
50
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.
â Nuclear Wang
Sep 13 at 17:10
8
... in a vacuum, -> ... in a frictionless environment, ...
â CramerTV
Sep 13 at 19:29
1
In your example, the block's energy only changed when you were pushing it. After that it has a constant unchanging speed, so a constant unchanging kinetic energy. The change in energy from before the push to after is the work done on the block.
â usul
Sep 13 at 19:39
4
Work is not force times distance. That's just a simplified version used to solve textbook problems with constant force. See Dale answer for the real definition of work.
â Bakuriu
Sep 13 at 21:21
1
Maybe the counterintuitive part is why it scales with distance, not time.
â JollyJoker
Sep 14 at 6:50
 |Â
show 2 more comments
50
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.
â Nuclear Wang
Sep 13 at 17:10
8
... in a vacuum, -> ... in a frictionless environment, ...
â CramerTV
Sep 13 at 19:29
1
In your example, the block's energy only changed when you were pushing it. After that it has a constant unchanging speed, so a constant unchanging kinetic energy. The change in energy from before the push to after is the work done on the block.
â usul
Sep 13 at 19:39
4
Work is not force times distance. That's just a simplified version used to solve textbook problems with constant force. See Dale answer for the real definition of work.
â Bakuriu
Sep 13 at 21:21
1
Maybe the counterintuitive part is why it scales with distance, not time.
â JollyJoker
Sep 14 at 6:50
50
50
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.
â Nuclear Wang
Sep 13 at 17:10
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.
â Nuclear Wang
Sep 13 at 17:10
8
8
... in a vacuum, -> ... in a frictionless environment, ...
â CramerTV
Sep 13 at 19:29
... in a vacuum, -> ... in a frictionless environment, ...
â CramerTV
Sep 13 at 19:29
1
1
In your example, the block's energy only changed when you were pushing it. After that it has a constant unchanging speed, so a constant unchanging kinetic energy. The change in energy from before the push to after is the work done on the block.
â usul
Sep 13 at 19:39
In your example, the block's energy only changed when you were pushing it. After that it has a constant unchanging speed, so a constant unchanging kinetic energy. The change in energy from before the push to after is the work done on the block.
â usul
Sep 13 at 19:39
4
4
Work is not force times distance. That's just a simplified version used to solve textbook problems with constant force. See Dale answer for the real definition of work.
â Bakuriu
Sep 13 at 21:21
Work is not force times distance. That's just a simplified version used to solve textbook problems with constant force. See Dale answer for the real definition of work.
â Bakuriu
Sep 13 at 21:21
1
1
Maybe the counterintuitive part is why it scales with distance, not time.
â JollyJoker
Sep 14 at 6:50
Maybe the counterintuitive part is why it scales with distance, not time.
â JollyJoker
Sep 14 at 6:50
 |Â
show 2 more comments
9 Answers
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up vote
48
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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.
add a comment |Â
up vote
51
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.
5
Now what I'm curious about a followup question. If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
â orlp
Sep 15 at 10:07
1
@orlp Why not ask that as a separate question?
â Tommi Brander
Sep 15 at 13:11
I agree. I have a good answer to that, but it wonâÂÂt fit in the constraints of a comment
â Dale
Sep 15 at 13:32
1
@orlp Asked over here: physics.stackexchange.com/questions/428952/â¦
â Pilchard123
Sep 15 at 19:50
add a comment |Â
up vote
9
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.
The phrase "the longer you apply the force" could imply a longer time, rather than a longer distance. (At least, that's my intuitive understanding.) The original question could easily arise from a confusion of time and distance, so I think it would be worth clarifying your answer.
â craq
Sep 15 at 20:19
add a comment |Â
up vote
6
down vote
Work is a definition, so the reason is "because it is defined that way".
However, we can ask why it makes sense to define it this way. Intuitively, you'd want to think of "work" as being a measure of what you do when you push, say, a box up a ramp, which causes you to get tired. In doing this, you apply a force to the box, and you also move a distance, and if the box is heavier (i.e. you need to use more force) or the distance you have to push it (the length of the ramp) is longer, then you would like to say the work is larger. If I have to push twice as hard for the same distance or I have to push twice as long, "intuitively" I should expect to do twice the work, and thus we get
$$mathrmWork = mathrmForce cdot mathrmDistance$$
And this simple, intuitive idea, it turns out makes a lot of physical sense when we actually use it, well beyond whatever limitations the original intuition might have (e.g. the biological inefficiency of our own bodies in doing "work", for example) thus we keep it. In particular it leads us to the concept of kinetic and potential energy, and their total ends up being conserved, thus showing we hit upon a core physical concept in the Universe. There isn't really any more of a "why" than this - it's science. Science is about the application of intuition or imagination, evidence, and reasoning, together, to understand how the world works. Intuition and imagination generate ideas for what is going on from which we can reason out consequences, and then we use evidence to see if those consequences are borne out and so whether or not that our ideas connect with reality.
add a comment |Â
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6
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This is a conceptual answer for learners, not a rigorous answer.
How do you know that you moved something?
You (not something else) must have pushed it. That's force.
It must have actually gone somewhere. That's distance.
So we define "work" as the product of these two things.
Physicists soon discovered that this definition is actually useful for calculating the behavior of systems. When you do work on an object, the same amount of work is taken away from you. Thus, the total amount of work that has been done (or could be done) inside a system remains a constant amount.
From this simple concept, you can elaborate it into the more rigorous definition:
- The distance is only meaningful during the interval to which your force is applied. Any continued motion by inertia doesn't count as your work.
- Positive work means "adding" to an object's movement. Negative work means "taking away" an object's movement.
- The force can be at an angle to the distance (introduce dot product).
- The force can vary (introduce integral over distance).
- We can talk about work that has been done in the past, and the ability to do work in the future (energy).
- We can calculate the work done in various scenarios, leading to some generally-useful formulas ($mgh$, $frac12mv^2$, etc.).
- We can more rigorously examine conservation of energy.
add a comment |Â
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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.
3
The OP is thinking of an object pushed in outer space. If you apply a force to the object for 1 second it would, potentially, move forever. He's wondering how you calculate the 'forever' movement. The answer is, you only pushed for 1 second - the movement due to inertia in a frictionless environment has nothing to do with the calculation of applied work.
â CramerTV
Sep 13 at 19:33
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Think of a unit of work as what you do when you lift a kilogram to a height of a metre. How do you do 2 units of work? You lift that kilogram 2 metres. 3 units? 3 metres. Etc. If you lift something the work you do is the force you exert times the distance traveled (height raised).
For your example of pushing a body in a vacuum you are only doing work while you are actually pushing, and so making it go faster. If you let it coast you are not doing work because the force during this time is zero.
The work you put in is the change of the energy of the object. When you lift the weight the work you have done becomes potential energy (stored energy). When you push the object in space the work you do becomes kinetic energy (energy of movement). You can change the potential energy to kinetic energy by letting the thing fall, and obviously it will be falling faster after it has fallen for a greater height.
Peter, can you please remove whichever parts of this post are not actually part of your answer? A lot of this seems to consist of contributions from other people, and it's hard to follow exactly which bits are yours.
â Dawood ibn Kareem
Sep 15 at 2:20
Rahul, thank you for removing the additional material
â Peter
Sep 15 at 11:49
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3
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When a weight is sitting on the floor, the floor is applying a force to the weight (and vice versa), but no distance. And it should make intuitive sense that there's no work being done.
For your example of a weight in a vacuum: if you push it with force 1N for distance 1m, and then stop pushing, it will move forever at constant speed. If you push another block with force 1N for distance 2m, it will move forever at a higher constant speed. You did more work to it, so it has more kinetic energy.
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Other answers have covered the misunderstandings around the W=F*d equation. I think it's also worth noting that "distance travelled" is not a form of energy. In physics, the phrase "doing work" means converting energy from one form to another, so you don't need to do any work to travel infinite distance. You only need to do work at the beginning to get some kinetic energy.
add a comment |Â
9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
48
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.
add a comment |Â
up vote
48
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.
add a comment |Â
up vote
48
down vote
accepted
up vote
48
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.
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 Sep 13 at 15:53
EuklidAlexandria
866412
866412
add a comment |Â
add a comment |Â
up vote
51
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.
5
Now what I'm curious about a followup question. If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
â orlp
Sep 15 at 10:07
1
@orlp Why not ask that as a separate question?
â Tommi Brander
Sep 15 at 13:11
I agree. I have a good answer to that, but it wonâÂÂt fit in the constraints of a comment
â Dale
Sep 15 at 13:32
1
@orlp Asked over here: physics.stackexchange.com/questions/428952/â¦
â Pilchard123
Sep 15 at 19:50
add a comment |Â
up vote
51
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.
5
Now what I'm curious about a followup question. If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
â orlp
Sep 15 at 10:07
1
@orlp Why not ask that as a separate question?
â Tommi Brander
Sep 15 at 13:11
I agree. I have a good answer to that, but it wonâÂÂt fit in the constraints of a comment
â Dale
Sep 15 at 13:32
1
@orlp Asked over here: physics.stackexchange.com/questions/428952/â¦
â Pilchard123
Sep 15 at 19:50
add a comment |Â
up vote
51
down vote
up vote
51
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.
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.
edited Sep 15 at 19:19
Glorfindel
173129
173129
answered Sep 13 at 16:26
Dale
1,722415
1,722415
5
Now what I'm curious about a followup question. If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
â orlp
Sep 15 at 10:07
1
@orlp Why not ask that as a separate question?
â Tommi Brander
Sep 15 at 13:11
I agree. I have a good answer to that, but it wonâÂÂt fit in the constraints of a comment
â Dale
Sep 15 at 13:32
1
@orlp Asked over here: physics.stackexchange.com/questions/428952/â¦
â Pilchard123
Sep 15 at 19:50
add a comment |Â
5
Now what I'm curious about a followup question. If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
â orlp
Sep 15 at 10:07
1
@orlp Why not ask that as a separate question?
â Tommi Brander
Sep 15 at 13:11
I agree. I have a good answer to that, but it wonâÂÂt fit in the constraints of a comment
â Dale
Sep 15 at 13:32
1
@orlp Asked over here: physics.stackexchange.com/questions/428952/â¦
â Pilchard123
Sep 15 at 19:50
5
5
Now what I'm curious about a followup question. If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
â orlp
Sep 15 at 10:07
Now what I'm curious about a followup question. If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
â orlp
Sep 15 at 10:07
1
1
@orlp Why not ask that as a separate question?
â Tommi Brander
Sep 15 at 13:11
@orlp Why not ask that as a separate question?
â Tommi Brander
Sep 15 at 13:11
I agree. I have a good answer to that, but it wonâÂÂt fit in the constraints of a comment
â Dale
Sep 15 at 13:32
I agree. I have a good answer to that, but it wonâÂÂt fit in the constraints of a comment
â Dale
Sep 15 at 13:32
1
1
@orlp Asked over here: physics.stackexchange.com/questions/428952/â¦
â Pilchard123
Sep 15 at 19:50
@orlp Asked over here: physics.stackexchange.com/questions/428952/â¦
â Pilchard123
Sep 15 at 19:50
add a comment |Â
up vote
9
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.
The phrase "the longer you apply the force" could imply a longer time, rather than a longer distance. (At least, that's my intuitive understanding.) The original question could easily arise from a confusion of time and distance, so I think it would be worth clarifying your answer.
â craq
Sep 15 at 20:19
add a comment |Â
up vote
9
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.
The phrase "the longer you apply the force" could imply a longer time, rather than a longer distance. (At least, that's my intuitive understanding.) The original question could easily arise from a confusion of time and distance, so I think it would be worth clarifying your answer.
â craq
Sep 15 at 20:19
add a comment |Â
up vote
9
down vote
up vote
9
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.
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 Sep 13 at 16:04
Farcher
45k33388
45k33388
The phrase "the longer you apply the force" could imply a longer time, rather than a longer distance. (At least, that's my intuitive understanding.) The original question could easily arise from a confusion of time and distance, so I think it would be worth clarifying your answer.
â craq
Sep 15 at 20:19
add a comment |Â
The phrase "the longer you apply the force" could imply a longer time, rather than a longer distance. (At least, that's my intuitive understanding.) The original question could easily arise from a confusion of time and distance, so I think it would be worth clarifying your answer.
â craq
Sep 15 at 20:19
The phrase "the longer you apply the force" could imply a longer time, rather than a longer distance. (At least, that's my intuitive understanding.) The original question could easily arise from a confusion of time and distance, so I think it would be worth clarifying your answer.
â craq
Sep 15 at 20:19
The phrase "the longer you apply the force" could imply a longer time, rather than a longer distance. (At least, that's my intuitive understanding.) The original question could easily arise from a confusion of time and distance, so I think it would be worth clarifying your answer.
â craq
Sep 15 at 20:19
add a comment |Â
up vote
6
down vote
Work is a definition, so the reason is "because it is defined that way".
However, we can ask why it makes sense to define it this way. Intuitively, you'd want to think of "work" as being a measure of what you do when you push, say, a box up a ramp, which causes you to get tired. In doing this, you apply a force to the box, and you also move a distance, and if the box is heavier (i.e. you need to use more force) or the distance you have to push it (the length of the ramp) is longer, then you would like to say the work is larger. If I have to push twice as hard for the same distance or I have to push twice as long, "intuitively" I should expect to do twice the work, and thus we get
$$mathrmWork = mathrmForce cdot mathrmDistance$$
And this simple, intuitive idea, it turns out makes a lot of physical sense when we actually use it, well beyond whatever limitations the original intuition might have (e.g. the biological inefficiency of our own bodies in doing "work", for example) thus we keep it. In particular it leads us to the concept of kinetic and potential energy, and their total ends up being conserved, thus showing we hit upon a core physical concept in the Universe. There isn't really any more of a "why" than this - it's science. Science is about the application of intuition or imagination, evidence, and reasoning, together, to understand how the world works. Intuition and imagination generate ideas for what is going on from which we can reason out consequences, and then we use evidence to see if those consequences are borne out and so whether or not that our ideas connect with reality.
add a comment |Â
up vote
6
down vote
Work is a definition, so the reason is "because it is defined that way".
However, we can ask why it makes sense to define it this way. Intuitively, you'd want to think of "work" as being a measure of what you do when you push, say, a box up a ramp, which causes you to get tired. In doing this, you apply a force to the box, and you also move a distance, and if the box is heavier (i.e. you need to use more force) or the distance you have to push it (the length of the ramp) is longer, then you would like to say the work is larger. If I have to push twice as hard for the same distance or I have to push twice as long, "intuitively" I should expect to do twice the work, and thus we get
$$mathrmWork = mathrmForce cdot mathrmDistance$$
And this simple, intuitive idea, it turns out makes a lot of physical sense when we actually use it, well beyond whatever limitations the original intuition might have (e.g. the biological inefficiency of our own bodies in doing "work", for example) thus we keep it. In particular it leads us to the concept of kinetic and potential energy, and their total ends up being conserved, thus showing we hit upon a core physical concept in the Universe. There isn't really any more of a "why" than this - it's science. Science is about the application of intuition or imagination, evidence, and reasoning, together, to understand how the world works. Intuition and imagination generate ideas for what is going on from which we can reason out consequences, and then we use evidence to see if those consequences are borne out and so whether or not that our ideas connect with reality.
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up vote
6
down vote
up vote
6
down vote
Work is a definition, so the reason is "because it is defined that way".
However, we can ask why it makes sense to define it this way. Intuitively, you'd want to think of "work" as being a measure of what you do when you push, say, a box up a ramp, which causes you to get tired. In doing this, you apply a force to the box, and you also move a distance, and if the box is heavier (i.e. you need to use more force) or the distance you have to push it (the length of the ramp) is longer, then you would like to say the work is larger. If I have to push twice as hard for the same distance or I have to push twice as long, "intuitively" I should expect to do twice the work, and thus we get
$$mathrmWork = mathrmForce cdot mathrmDistance$$
And this simple, intuitive idea, it turns out makes a lot of physical sense when we actually use it, well beyond whatever limitations the original intuition might have (e.g. the biological inefficiency of our own bodies in doing "work", for example) thus we keep it. In particular it leads us to the concept of kinetic and potential energy, and their total ends up being conserved, thus showing we hit upon a core physical concept in the Universe. There isn't really any more of a "why" than this - it's science. Science is about the application of intuition or imagination, evidence, and reasoning, together, to understand how the world works. Intuition and imagination generate ideas for what is going on from which we can reason out consequences, and then we use evidence to see if those consequences are borne out and so whether or not that our ideas connect with reality.
Work is a definition, so the reason is "because it is defined that way".
However, we can ask why it makes sense to define it this way. Intuitively, you'd want to think of "work" as being a measure of what you do when you push, say, a box up a ramp, which causes you to get tired. In doing this, you apply a force to the box, and you also move a distance, and if the box is heavier (i.e. you need to use more force) or the distance you have to push it (the length of the ramp) is longer, then you would like to say the work is larger. If I have to push twice as hard for the same distance or I have to push twice as long, "intuitively" I should expect to do twice the work, and thus we get
$$mathrmWork = mathrmForce cdot mathrmDistance$$
And this simple, intuitive idea, it turns out makes a lot of physical sense when we actually use it, well beyond whatever limitations the original intuition might have (e.g. the biological inefficiency of our own bodies in doing "work", for example) thus we keep it. In particular it leads us to the concept of kinetic and potential energy, and their total ends up being conserved, thus showing we hit upon a core physical concept in the Universe. There isn't really any more of a "why" than this - it's science. Science is about the application of intuition or imagination, evidence, and reasoning, together, to understand how the world works. Intuition and imagination generate ideas for what is going on from which we can reason out consequences, and then we use evidence to see if those consequences are borne out and so whether or not that our ideas connect with reality.
answered Sep 14 at 3:38
The_Sympathizer
3,030721
3,030721
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up vote
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This is a conceptual answer for learners, not a rigorous answer.
How do you know that you moved something?
You (not something else) must have pushed it. That's force.
It must have actually gone somewhere. That's distance.
So we define "work" as the product of these two things.
Physicists soon discovered that this definition is actually useful for calculating the behavior of systems. When you do work on an object, the same amount of work is taken away from you. Thus, the total amount of work that has been done (or could be done) inside a system remains a constant amount.
From this simple concept, you can elaborate it into the more rigorous definition:
- The distance is only meaningful during the interval to which your force is applied. Any continued motion by inertia doesn't count as your work.
- Positive work means "adding" to an object's movement. Negative work means "taking away" an object's movement.
- The force can be at an angle to the distance (introduce dot product).
- The force can vary (introduce integral over distance).
- We can talk about work that has been done in the past, and the ability to do work in the future (energy).
- We can calculate the work done in various scenarios, leading to some generally-useful formulas ($mgh$, $frac12mv^2$, etc.).
- We can more rigorously examine conservation of energy.
add a comment |Â
up vote
6
down vote
This is a conceptual answer for learners, not a rigorous answer.
How do you know that you moved something?
You (not something else) must have pushed it. That's force.
It must have actually gone somewhere. That's distance.
So we define "work" as the product of these two things.
Physicists soon discovered that this definition is actually useful for calculating the behavior of systems. When you do work on an object, the same amount of work is taken away from you. Thus, the total amount of work that has been done (or could be done) inside a system remains a constant amount.
From this simple concept, you can elaborate it into the more rigorous definition:
- The distance is only meaningful during the interval to which your force is applied. Any continued motion by inertia doesn't count as your work.
- Positive work means "adding" to an object's movement. Negative work means "taking away" an object's movement.
- The force can be at an angle to the distance (introduce dot product).
- The force can vary (introduce integral over distance).
- We can talk about work that has been done in the past, and the ability to do work in the future (energy).
- We can calculate the work done in various scenarios, leading to some generally-useful formulas ($mgh$, $frac12mv^2$, etc.).
- We can more rigorously examine conservation of energy.
add a comment |Â
up vote
6
down vote
up vote
6
down vote
This is a conceptual answer for learners, not a rigorous answer.
How do you know that you moved something?
You (not something else) must have pushed it. That's force.
It must have actually gone somewhere. That's distance.
So we define "work" as the product of these two things.
Physicists soon discovered that this definition is actually useful for calculating the behavior of systems. When you do work on an object, the same amount of work is taken away from you. Thus, the total amount of work that has been done (or could be done) inside a system remains a constant amount.
From this simple concept, you can elaborate it into the more rigorous definition:
- The distance is only meaningful during the interval to which your force is applied. Any continued motion by inertia doesn't count as your work.
- Positive work means "adding" to an object's movement. Negative work means "taking away" an object's movement.
- The force can be at an angle to the distance (introduce dot product).
- The force can vary (introduce integral over distance).
- We can talk about work that has been done in the past, and the ability to do work in the future (energy).
- We can calculate the work done in various scenarios, leading to some generally-useful formulas ($mgh$, $frac12mv^2$, etc.).
- We can more rigorously examine conservation of energy.
This is a conceptual answer for learners, not a rigorous answer.
How do you know that you moved something?
You (not something else) must have pushed it. That's force.
It must have actually gone somewhere. That's distance.
So we define "work" as the product of these two things.
Physicists soon discovered that this definition is actually useful for calculating the behavior of systems. When you do work on an object, the same amount of work is taken away from you. Thus, the total amount of work that has been done (or could be done) inside a system remains a constant amount.
From this simple concept, you can elaborate it into the more rigorous definition:
- The distance is only meaningful during the interval to which your force is applied. Any continued motion by inertia doesn't count as your work.
- Positive work means "adding" to an object's movement. Negative work means "taking away" an object's movement.
- The force can be at an angle to the distance (introduce dot product).
- The force can vary (introduce integral over distance).
- We can talk about work that has been done in the past, and the ability to do work in the future (energy).
- We can calculate the work done in various scenarios, leading to some generally-useful formulas ($mgh$, $frac12mv^2$, etc.).
- We can more rigorously examine conservation of energy.
edited Sep 14 at 19:51
answered Sep 13 at 21:09
Dr Sheldon
26010
26010
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up vote
4
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.
3
The OP is thinking of an object pushed in outer space. If you apply a force to the object for 1 second it would, potentially, move forever. He's wondering how you calculate the 'forever' movement. The answer is, you only pushed for 1 second - the movement due to inertia in a frictionless environment has nothing to do with the calculation of applied work.
â CramerTV
Sep 13 at 19:33
add a comment |Â
up vote
4
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.
3
The OP is thinking of an object pushed in outer space. If you apply a force to the object for 1 second it would, potentially, move forever. He's wondering how you calculate the 'forever' movement. The answer is, you only pushed for 1 second - the movement due to inertia in a frictionless environment has nothing to do with the calculation of applied work.
â CramerTV
Sep 13 at 19:33
add a comment |Â
up vote
4
down vote
up vote
4
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.
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 Sep 13 at 17:04
Brad
1412
1412
3
The OP is thinking of an object pushed in outer space. If you apply a force to the object for 1 second it would, potentially, move forever. He's wondering how you calculate the 'forever' movement. The answer is, you only pushed for 1 second - the movement due to inertia in a frictionless environment has nothing to do with the calculation of applied work.
â CramerTV
Sep 13 at 19:33
add a comment |Â
3
The OP is thinking of an object pushed in outer space. If you apply a force to the object for 1 second it would, potentially, move forever. He's wondering how you calculate the 'forever' movement. The answer is, you only pushed for 1 second - the movement due to inertia in a frictionless environment has nothing to do with the calculation of applied work.
â CramerTV
Sep 13 at 19:33
3
3
The OP is thinking of an object pushed in outer space. If you apply a force to the object for 1 second it would, potentially, move forever. He's wondering how you calculate the 'forever' movement. The answer is, you only pushed for 1 second - the movement due to inertia in a frictionless environment has nothing to do with the calculation of applied work.
â CramerTV
Sep 13 at 19:33
The OP is thinking of an object pushed in outer space. If you apply a force to the object for 1 second it would, potentially, move forever. He's wondering how you calculate the 'forever' movement. The answer is, you only pushed for 1 second - the movement due to inertia in a frictionless environment has nothing to do with the calculation of applied work.
â CramerTV
Sep 13 at 19:33
add a comment |Â
up vote
4
down vote
Think of a unit of work as what you do when you lift a kilogram to a height of a metre. How do you do 2 units of work? You lift that kilogram 2 metres. 3 units? 3 metres. Etc. If you lift something the work you do is the force you exert times the distance traveled (height raised).
For your example of pushing a body in a vacuum you are only doing work while you are actually pushing, and so making it go faster. If you let it coast you are not doing work because the force during this time is zero.
The work you put in is the change of the energy of the object. When you lift the weight the work you have done becomes potential energy (stored energy). When you push the object in space the work you do becomes kinetic energy (energy of movement). You can change the potential energy to kinetic energy by letting the thing fall, and obviously it will be falling faster after it has fallen for a greater height.
Peter, can you please remove whichever parts of this post are not actually part of your answer? A lot of this seems to consist of contributions from other people, and it's hard to follow exactly which bits are yours.
â Dawood ibn Kareem
Sep 15 at 2:20
Rahul, thank you for removing the additional material
â Peter
Sep 15 at 11:49
add a comment |Â
up vote
4
down vote
Think of a unit of work as what you do when you lift a kilogram to a height of a metre. How do you do 2 units of work? You lift that kilogram 2 metres. 3 units? 3 metres. Etc. If you lift something the work you do is the force you exert times the distance traveled (height raised).
For your example of pushing a body in a vacuum you are only doing work while you are actually pushing, and so making it go faster. If you let it coast you are not doing work because the force during this time is zero.
The work you put in is the change of the energy of the object. When you lift the weight the work you have done becomes potential energy (stored energy). When you push the object in space the work you do becomes kinetic energy (energy of movement). You can change the potential energy to kinetic energy by letting the thing fall, and obviously it will be falling faster after it has fallen for a greater height.
Peter, can you please remove whichever parts of this post are not actually part of your answer? A lot of this seems to consist of contributions from other people, and it's hard to follow exactly which bits are yours.
â Dawood ibn Kareem
Sep 15 at 2:20
Rahul, thank you for removing the additional material
â Peter
Sep 15 at 11:49
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Think of a unit of work as what you do when you lift a kilogram to a height of a metre. How do you do 2 units of work? You lift that kilogram 2 metres. 3 units? 3 metres. Etc. If you lift something the work you do is the force you exert times the distance traveled (height raised).
For your example of pushing a body in a vacuum you are only doing work while you are actually pushing, and so making it go faster. If you let it coast you are not doing work because the force during this time is zero.
The work you put in is the change of the energy of the object. When you lift the weight the work you have done becomes potential energy (stored energy). When you push the object in space the work you do becomes kinetic energy (energy of movement). You can change the potential energy to kinetic energy by letting the thing fall, and obviously it will be falling faster after it has fallen for a greater height.
Think of a unit of work as what you do when you lift a kilogram to a height of a metre. How do you do 2 units of work? You lift that kilogram 2 metres. 3 units? 3 metres. Etc. If you lift something the work you do is the force you exert times the distance traveled (height raised).
For your example of pushing a body in a vacuum you are only doing work while you are actually pushing, and so making it go faster. If you let it coast you are not doing work because the force during this time is zero.
The work you put in is the change of the energy of the object. When you lift the weight the work you have done becomes potential energy (stored energy). When you push the object in space the work you do becomes kinetic energy (energy of movement). You can change the potential energy to kinetic energy by letting the thing fall, and obviously it will be falling faster after it has fallen for a greater height.
edited Sep 15 at 11:47
answered Sep 14 at 3:45
Peter
813
813
Peter, can you please remove whichever parts of this post are not actually part of your answer? A lot of this seems to consist of contributions from other people, and it's hard to follow exactly which bits are yours.
â Dawood ibn Kareem
Sep 15 at 2:20
Rahul, thank you for removing the additional material
â Peter
Sep 15 at 11:49
add a comment |Â
Peter, can you please remove whichever parts of this post are not actually part of your answer? A lot of this seems to consist of contributions from other people, and it's hard to follow exactly which bits are yours.
â Dawood ibn Kareem
Sep 15 at 2:20
Rahul, thank you for removing the additional material
â Peter
Sep 15 at 11:49
Peter, can you please remove whichever parts of this post are not actually part of your answer? A lot of this seems to consist of contributions from other people, and it's hard to follow exactly which bits are yours.
â Dawood ibn Kareem
Sep 15 at 2:20
Peter, can you please remove whichever parts of this post are not actually part of your answer? A lot of this seems to consist of contributions from other people, and it's hard to follow exactly which bits are yours.
â Dawood ibn Kareem
Sep 15 at 2:20
Rahul, thank you for removing the additional material
â Peter
Sep 15 at 11:49
Rahul, thank you for removing the additional material
â Peter
Sep 15 at 11:49
add a comment |Â
up vote
3
down vote
When a weight is sitting on the floor, the floor is applying a force to the weight (and vice versa), but no distance. And it should make intuitive sense that there's no work being done.
For your example of a weight in a vacuum: if you push it with force 1N for distance 1m, and then stop pushing, it will move forever at constant speed. If you push another block with force 1N for distance 2m, it will move forever at a higher constant speed. You did more work to it, so it has more kinetic energy.
add a comment |Â
up vote
3
down vote
When a weight is sitting on the floor, the floor is applying a force to the weight (and vice versa), but no distance. And it should make intuitive sense that there's no work being done.
For your example of a weight in a vacuum: if you push it with force 1N for distance 1m, and then stop pushing, it will move forever at constant speed. If you push another block with force 1N for distance 2m, it will move forever at a higher constant speed. You did more work to it, so it has more kinetic energy.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
When a weight is sitting on the floor, the floor is applying a force to the weight (and vice versa), but no distance. And it should make intuitive sense that there's no work being done.
For your example of a weight in a vacuum: if you push it with force 1N for distance 1m, and then stop pushing, it will move forever at constant speed. If you push another block with force 1N for distance 2m, it will move forever at a higher constant speed. You did more work to it, so it has more kinetic energy.
When a weight is sitting on the floor, the floor is applying a force to the weight (and vice versa), but no distance. And it should make intuitive sense that there's no work being done.
For your example of a weight in a vacuum: if you push it with force 1N for distance 1m, and then stop pushing, it will move forever at constant speed. If you push another block with force 1N for distance 2m, it will move forever at a higher constant speed. You did more work to it, so it has more kinetic energy.
answered Sep 14 at 0:21
upper
311
311
add a comment |Â
add a comment |Â
up vote
1
down vote
Other answers have covered the misunderstandings around the W=F*d equation. I think it's also worth noting that "distance travelled" is not a form of energy. In physics, the phrase "doing work" means converting energy from one form to another, so you don't need to do any work to travel infinite distance. You only need to do work at the beginning to get some kinetic energy.
add a comment |Â
up vote
1
down vote
Other answers have covered the misunderstandings around the W=F*d equation. I think it's also worth noting that "distance travelled" is not a form of energy. In physics, the phrase "doing work" means converting energy from one form to another, so you don't need to do any work to travel infinite distance. You only need to do work at the beginning to get some kinetic energy.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Other answers have covered the misunderstandings around the W=F*d equation. I think it's also worth noting that "distance travelled" is not a form of energy. In physics, the phrase "doing work" means converting energy from one form to another, so you don't need to do any work to travel infinite distance. You only need to do work at the beginning to get some kinetic energy.
Other answers have covered the misunderstandings around the W=F*d equation. I think it's also worth noting that "distance travelled" is not a form of energy. In physics, the phrase "doing work" means converting energy from one form to another, so you don't need to do any work to travel infinite distance. You only need to do work at the beginning to get some kinetic energy.
answered Sep 15 at 20:26
craq
22417
22417
add a comment |Â
add a comment |Â
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50
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.
â Nuclear Wang
Sep 13 at 17:10
8
... in a vacuum, -> ... in a frictionless environment, ...
â CramerTV
Sep 13 at 19:29
1
In your example, the block's energy only changed when you were pushing it. After that it has a constant unchanging speed, so a constant unchanging kinetic energy. The change in energy from before the push to after is the work done on the block.
â usul
Sep 13 at 19:39
4
Work is not force times distance. That's just a simplified version used to solve textbook problems with constant force. See Dale answer for the real definition of work.
â Bakuriu
Sep 13 at 21:21
1
Maybe the counterintuitive part is why it scales with distance, not time.
â JollyJoker
Sep 14 at 6:50