Is the Olympic running race fair? [closed]What's the difference between running up a hill and running up an inclined treadmill?Effect of surface treatment on fair diceWill two trains running along the equator in opposite direction experience same wear out?Running vs. walking in slippery conditionDrifting vs. rolling when starting raceRunning or walking up stairs = same work?
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Is the Olympic running race fair? [closed]
What's the difference between running up a hill and running up an inclined treadmill?Effect of surface treatment on fair diceWill two trains running along the equator in opposite direction experience same wear out?Running vs. walking in slippery conditionDrifting vs. rolling when starting raceRunning or walking up stairs = same work?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
$begingroup$
I noticed that the 200-meter sprints are conducted on curved tracks. (See this video: world championship semifinals 2009)
Isn't that weird? I mean, just look at the curvatures of each lane!
(Source)
Since they use staggered start lines, the total track length is the same. But the person on the innermost lane would have to do more work as compared to the one on the outermost lane. He would have to put in an additional amount of work against friction to counter the extra centrifugal force.
(The extra centrifugal force is roughly 2 Newtons from my calculation).
I believe that the winners of a running race are the ones who do more work against the friction. So, don't you think the runners in the inner lanes should be given an advantage?
newtonian-mechanics friction work everyday-life centrifugal-force
edited Oct 3 at 0:20
Buzz
5,3754 gold badges20 silver badges32 bronze badges
asked Oct 2 at 14:06
Krishnanand JKrishnanand J
3,9243 gold badges10 silver badges40 bronze badges
$endgroup$
|
show 2 more comments
$begingroup$
I noticed that the 200-meter sprints are conducted on curved tracks. (See this video: world championship semifinals 2009)
Isn't that weird? I mean, just look at the curvatures of each lane!
(Source)
Since they use staggered start lines, the total track length is the same. But the person on the innermost lane would have to do more work as compared to the one on the outermost lane. He would have to put in an additional amount of work against friction to counter the extra centrifugal force.
(The extra centrifugal force is roughly 2 Newtons from my calculation).
I believe that the winners of a running race are the ones who do more work against the friction. So, don't you think the runners in the inner lanes should be given an advantage?
newtonian-mechanics friction work everyday-life centrifugal-force
edited Oct 3 at 0:20
Buzz
5,3754 gold badges20 silver badges32 bronze badges
asked Oct 2 at 14:06
Krishnanand JKrishnanand J
3,9243 gold badges10 silver badges40 bronze badges
$endgroup$
1
$begingroup$
work=force·distance. In theory he doesn't do any work against centrifugal force as long as he moves in the direction perpendicular to the force. In real life you still spend energy against static force, but you didn't estimate the work and I don't know how much that would be either.
$endgroup$
– Yi Jiang
Oct 2 at 14:28
2
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
Oct 2 at 14:41
2
$begingroup$
@JonCuster Agreed. I raised the question here so that the physical reason behind the unfairness would be rigorously confirmed. The rules will be changed when people become aware of their flaws. That's the whole point of this site, right?
$endgroup$
– Krishnanand J
Oct 2 at 15:12
2
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
Oct 2 at 17:35
3
$begingroup$
I'm voting to close this question as off-topic because it is not about physics - any physics in the issue the answer is too tied up in external sporting factors to be useful.
$endgroup$
– Emilio Pisanty
Oct 2 at 21:17
|
show 2 more comments
$begingroup$
I noticed that the 200-meter sprints are conducted on curved tracks. (See this video: world championship semifinals 2009)
Isn't that weird? I mean, just look at the curvatures of each lane!
(Source)
Since they use staggered start lines, the total track length is the same. But the person on the innermost lane would have to do more work as compared to the one on the outermost lane. He would have to put in an additional amount of work against friction to counter the extra centrifugal force.
(The extra centrifugal force is roughly 2 Newtons from my calculation).
I believe that the winners of a running race are the ones who do more work against the friction. So, don't you think the runners in the inner lanes should be given an advantage?
newtonian-mechanics friction work everyday-life centrifugal-force
edited Oct 3 at 0:20
Buzz
5,3754 gold badges20 silver badges32 bronze badges
asked Oct 2 at 14:06
Krishnanand JKrishnanand J
3,9243 gold badges10 silver badges40 bronze badges
$endgroup$
I noticed that the 200-meter sprints are conducted on curved tracks. (See this video: world championship semifinals 2009)
Isn't that weird? I mean, just look at the curvatures of each lane!
(Source)
Since they use staggered start lines, the total track length is the same. But the person on the innermost lane would have to do more work as compared to the one on the outermost lane. He would have to put in an additional amount of work against friction to counter the extra centrifugal force.
(The extra centrifugal force is roughly 2 Newtons from my calculation).
I believe that the winners of a running race are the ones who do more work against the friction. So, don't you think the runners in the inner lanes should be given an advantage?
Closed. This question is off-topic. It is not currently accepting answers.
Closed. This question is off-topic. It is not currently accepting answers.
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newtonian-mechanics friction work everyday-life centrifugal-force
newtonian-mechanics friction work everyday-life centrifugal-force
edited Oct 3 at 0:20
Buzz
5,3754 gold badges20 silver badges32 bronze badges
asked Oct 2 at 14:06
Krishnanand JKrishnanand J
3,9243 gold badges10 silver badges40 bronze badges
edited Oct 3 at 0:20
Buzz
5,3754 gold badges20 silver badges32 bronze badges
asked Oct 2 at 14:06
Krishnanand JKrishnanand J
3,9243 gold badges10 silver badges40 bronze badges
edited Oct 3 at 0:20
Buzz
5,3754 gold badges20 silver badges32 bronze badges
edited Oct 3 at 0:20
Buzz
5,3754 gold badges20 silver badges32 bronze badges
edited Oct 3 at 0:20
Buzz
5,3754 gold badges20 silver badges32 bronze badges
5,3754 gold badges20 silver badges32 bronze badges
asked Oct 2 at 14:06
Krishnanand JKrishnanand J
3,9243 gold badges10 silver badges40 bronze badges
asked Oct 2 at 14:06
Krishnanand JKrishnanand J
3,9243 gold badges10 silver badges40 bronze badges
asked Oct 2 at 14:06
Krishnanand JKrishnanand J
3,9243 gold badges10 silver badges40 bronze badges
3,9243 gold badges10 silver badges40 bronze badges
1
$begingroup$
work=force·distance. In theory he doesn't do any work against centrifugal force as long as he moves in the direction perpendicular to the force. In real life you still spend energy against static force, but you didn't estimate the work and I don't know how much that would be either.
$endgroup$
– Yi Jiang
Oct 2 at 14:28
2
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
Oct 2 at 14:41
2
$begingroup$
@JonCuster Agreed. I raised the question here so that the physical reason behind the unfairness would be rigorously confirmed. The rules will be changed when people become aware of their flaws. That's the whole point of this site, right?
$endgroup$
– Krishnanand J
Oct 2 at 15:12
2
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
Oct 2 at 17:35
3
$begingroup$
I'm voting to close this question as off-topic because it is not about physics - any physics in the issue the answer is too tied up in external sporting factors to be useful.
$endgroup$
– Emilio Pisanty
Oct 2 at 21:17
|
show 2 more comments
1
$begingroup$
work=force·distance. In theory he doesn't do any work against centrifugal force as long as he moves in the direction perpendicular to the force. In real life you still spend energy against static force, but you didn't estimate the work and I don't know how much that would be either.
$endgroup$
– Yi Jiang
Oct 2 at 14:28
2
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
Oct 2 at 14:41
2
$begingroup$
@JonCuster Agreed. I raised the question here so that the physical reason behind the unfairness would be rigorously confirmed. The rules will be changed when people become aware of their flaws. That's the whole point of this site, right?
$endgroup$
– Krishnanand J
Oct 2 at 15:12
2
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
Oct 2 at 17:35
3
$begingroup$
I'm voting to close this question as off-topic because it is not about physics - any physics in the issue the answer is too tied up in external sporting factors to be useful.
$endgroup$
– Emilio Pisanty
Oct 2 at 21:17
1
1
$begingroup$
work=force·distance. In theory he doesn't do any work against centrifugal force as long as he moves in the direction perpendicular to the force. In real life you still spend energy against static force, but you didn't estimate the work and I don't know how much that would be either.
$endgroup$
– Yi Jiang
Oct 2 at 14:28
$begingroup$
work=force·distance. In theory he doesn't do any work against centrifugal force as long as he moves in the direction perpendicular to the force. In real life you still spend energy against static force, but you didn't estimate the work and I don't know how much that would be either.
$endgroup$
– Yi Jiang
Oct 2 at 14:28
2
2
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
Oct 2 at 14:41
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
Oct 2 at 14:41
2
2
$begingroup$
@JonCuster Agreed. I raised the question here so that the physical reason behind the unfairness would be rigorously confirmed. The rules will be changed when people become aware of their flaws. That's the whole point of this site, right?
$endgroup$
– Krishnanand J
Oct 2 at 15:12
$begingroup$
@JonCuster Agreed. I raised the question here so that the physical reason behind the unfairness would be rigorously confirmed. The rules will be changed when people become aware of their flaws. That's the whole point of this site, right?
$endgroup$
– Krishnanand J
Oct 2 at 15:12
2
2
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
Oct 2 at 17:35
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
Oct 2 at 17:35
3
3
$begingroup$
I'm voting to close this question as off-topic because it is not about physics - any physics in the issue the answer is too tied up in external sporting factors to be useful.
$endgroup$
– Emilio Pisanty
Oct 2 at 21:17
$begingroup$
I'm voting to close this question as off-topic because it is not about physics - any physics in the issue the answer is too tied up in external sporting factors to be useful.
$endgroup$
– Emilio Pisanty
Oct 2 at 21:17
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).
answered Oct 2 at 14:20
Ben51Ben51
5,58211 silver badges35 bronze badges
$endgroup$
7
$begingroup$
They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
$endgroup$
– puppetsock reinstate Monica
Oct 2 at 16:06
$begingroup$
@puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
$endgroup$
– Mark
Oct 2 at 23:16
add a comment
|
$begingroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F cdot vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.
Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.
All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.
answered Oct 2 at 22:09
BowlOfRedBowlOfRed
21.5k2 gold badges38 silver badges57 bronze badges
$endgroup$
$begingroup$
I think you meant "scalar" and not "cross" product.
$endgroup$
– hyportnex
Oct 2 at 23:11
$begingroup$
In short track speed skating, you output more useful power when turning because you can keep pushing repeatedly. But in skating even on the straights you push mostly to the side and convert that to forward momentum by angling your blades. In running you generate forward momentum by driving your legs / feet backward during ground contact so it's not clear it would be helpful.
$endgroup$
– Peter Cordes
Oct 3 at 14:09
add a comment
|
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).
answered Oct 2 at 14:20
Ben51Ben51
5,58211 silver badges35 bronze badges
$endgroup$
7
$begingroup$
They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
$endgroup$
– puppetsock reinstate Monica
Oct 2 at 16:06
$begingroup$
@puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
$endgroup$
– Mark
Oct 2 at 23:16
add a comment
|
$begingroup$
Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).
answered Oct 2 at 14:20
Ben51Ben51
5,58211 silver badges35 bronze badges
$endgroup$
7
$begingroup$
They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
$endgroup$
– puppetsock reinstate Monica
Oct 2 at 16:06
$begingroup$
@puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
$endgroup$
– Mark
Oct 2 at 23:16
add a comment
|
$begingroup$
Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).
answered Oct 2 at 14:20
Ben51Ben51
5,58211 silver badges35 bronze badges
$endgroup$
Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).
answered Oct 2 at 14:20
Ben51Ben51
5,58211 silver badges35 bronze badges
answered Oct 2 at 14:20
Ben51Ben51
5,58211 silver badges35 bronze badges
answered Oct 2 at 14:20
Ben51Ben51
5,58211 silver badges35 bronze badges
answered Oct 2 at 14:20
Ben51Ben51
5,58211 silver badges35 bronze badges
5,58211 silver badges35 bronze badges
7
$begingroup$
They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
$endgroup$
– puppetsock reinstate Monica
Oct 2 at 16:06
$begingroup$
@puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
$endgroup$
– Mark
Oct 2 at 23:16
add a comment
|
7
$begingroup$
They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
$endgroup$
– puppetsock reinstate Monica
Oct 2 at 16:06
$begingroup$
@puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
$endgroup$
– Mark
Oct 2 at 23:16
7
7
$begingroup$
They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
$endgroup$
– puppetsock reinstate Monica
Oct 2 at 16:06
$begingroup$
They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
$endgroup$
– puppetsock reinstate Monica
Oct 2 at 16:06
$begingroup$
@puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
$endgroup$
– Mark
Oct 2 at 23:16
$begingroup$
@puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
$endgroup$
– Mark
Oct 2 at 23:16
add a comment
|
$begingroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F cdot vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.
Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.
All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.
answered Oct 2 at 22:09
BowlOfRedBowlOfRed
21.5k2 gold badges38 silver badges57 bronze badges
$endgroup$
$begingroup$
I think you meant "scalar" and not "cross" product.
$endgroup$
– hyportnex
Oct 2 at 23:11
$begingroup$
In short track speed skating, you output more useful power when turning because you can keep pushing repeatedly. But in skating even on the straights you push mostly to the side and convert that to forward momentum by angling your blades. In running you generate forward momentum by driving your legs / feet backward during ground contact so it's not clear it would be helpful.
$endgroup$
– Peter Cordes
Oct 3 at 14:09
add a comment
|
$begingroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F cdot vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.
Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.
All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.
answered Oct 2 at 22:09
BowlOfRedBowlOfRed
21.5k2 gold badges38 silver badges57 bronze badges
$endgroup$
$begingroup$
I think you meant "scalar" and not "cross" product.
$endgroup$
– hyportnex
Oct 2 at 23:11
$begingroup$
In short track speed skating, you output more useful power when turning because you can keep pushing repeatedly. But in skating even on the straights you push mostly to the side and convert that to forward momentum by angling your blades. In running you generate forward momentum by driving your legs / feet backward during ground contact so it's not clear it would be helpful.
$endgroup$
– Peter Cordes
Oct 3 at 14:09
add a comment
|
$begingroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F cdot vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.
Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.
All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.
answered Oct 2 at 22:09
BowlOfRedBowlOfRed
21.5k2 gold badges38 silver badges57 bronze badges
$endgroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F cdot vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.
Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.
All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.
answered Oct 2 at 22:09
BowlOfRedBowlOfRed
21.5k2 gold badges38 silver badges57 bronze badges
edited Oct 2 at 23:19
answered Oct 2 at 22:09
BowlOfRedBowlOfRed
21.5k2 gold badges38 silver badges57 bronze badges
answered Oct 2 at 22:09
BowlOfRedBowlOfRed
21.5k2 gold badges38 silver badges57 bronze badges
answered Oct 2 at 22:09
BowlOfRedBowlOfRed
21.5k2 gold badges38 silver badges57 bronze badges
21.5k2 gold badges38 silver badges57 bronze badges
$begingroup$
I think you meant "scalar" and not "cross" product.
$endgroup$
– hyportnex
Oct 2 at 23:11
$begingroup$
In short track speed skating, you output more useful power when turning because you can keep pushing repeatedly. But in skating even on the straights you push mostly to the side and convert that to forward momentum by angling your blades. In running you generate forward momentum by driving your legs / feet backward during ground contact so it's not clear it would be helpful.
$endgroup$
– Peter Cordes
Oct 3 at 14:09
add a comment
|
$begingroup$
I think you meant "scalar" and not "cross" product.
$endgroup$
– hyportnex
Oct 2 at 23:11
$begingroup$
In short track speed skating, you output more useful power when turning because you can keep pushing repeatedly. But in skating even on the straights you push mostly to the side and convert that to forward momentum by angling your blades. In running you generate forward momentum by driving your legs / feet backward during ground contact so it's not clear it would be helpful.
$endgroup$
– Peter Cordes
Oct 3 at 14:09
$begingroup$
I think you meant "scalar" and not "cross" product.
$endgroup$
– hyportnex
Oct 2 at 23:11
$begingroup$
I think you meant "scalar" and not "cross" product.
$endgroup$
– hyportnex
Oct 2 at 23:11
$begingroup$
In short track speed skating, you output more useful power when turning because you can keep pushing repeatedly. But in skating even on the straights you push mostly to the side and convert that to forward momentum by angling your blades. In running you generate forward momentum by driving your legs / feet backward during ground contact so it's not clear it would be helpful.
$endgroup$
– Peter Cordes
Oct 3 at 14:09
$begingroup$
In short track speed skating, you output more useful power when turning because you can keep pushing repeatedly. But in skating even on the straights you push mostly to the side and convert that to forward momentum by angling your blades. In running you generate forward momentum by driving your legs / feet backward during ground contact so it's not clear it would be helpful.
$endgroup$
– Peter Cordes
Oct 3 at 14:09
add a comment
|
1
$begingroup$
work=force·distance. In theory he doesn't do any work against centrifugal force as long as he moves in the direction perpendicular to the force. In real life you still spend energy against static force, but you didn't estimate the work and I don't know how much that would be either.
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– Yi Jiang
Oct 2 at 14:28
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Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
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– Jon Custer
Oct 2 at 14:41
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@JonCuster Agreed. I raised the question here so that the physical reason behind the unfairness would be rigorously confirmed. The rules will be changed when people become aware of their flaws. That's the whole point of this site, right?
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– Krishnanand J
Oct 2 at 15:12
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Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
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– BowlOfRed
Oct 2 at 17:35
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I'm voting to close this question as off-topic because it is not about physics - any physics in the issue the answer is too tied up in external sporting factors to be useful.
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– Emilio Pisanty
Oct 2 at 21:17