- #1

ando

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- #1

ando

- #2

Integral

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Your body must convert chemical potential energy to kinetic energy in order to run. Assume a flat track. Since kinetic energy is

E_{k}=.5mv^{2}

since your your mass is pretty much constant, we only need be concerned with velocity, clearly, your velocity is higher if you run, therefore it requires more energy to run a mile then to walk.

E=mc^{2} does not enter into the problem.

E

since your your mass is pretty much constant, we only need be concerned with velocity, clearly, your velocity is higher if you run, therefore it requires more energy to run a mile then to walk.

E=mc

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- #3

ando

Ok that makes sense. What about if two Cars travelled the same distance, but one went twice as fast. Would E=MC2 come into play then? If not, would you still use the Ek=.5mv2 equation to calculate the ammount of energy it takes, and if so wouldn't that have the same answer (yes the faster car would use/take more energy to get it from A to B)?

- #4

chroot

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The faster you move through air, the more drag you suffer -- and drag (wind resistance) depends on the square of the velocity. If you move twice as fast, you experience four times as much wind resistance. You have to provide the power to balance that wind resistance and continue moving at your desired pace.

I will say that muscles are not very ideal physical things. For example, if you hold a 20 lb weight over your head in mid-air, exactly in position, you are doing no work -- in the physical sense. The position of the weight is not changing, therefore no work is being done. If you put the weight down on the table, you'll see that the table doesn't seem to have to do anything to support the weight -- the table is investing no energy. However, to hold the weight over your head is a very difficult thing, and it will eventually get you to huff and puff and feel tired. Why? Because your mucles are not rigid! The individual cells constantly contract and then release and then contract again. Each contraction takes further energy.

So it's difficult to calculate how much energy you'll burn in running a mile -- very many physiological things come into play that make it a hard problem. For an ideal physical system, though, moving an object a mile down the road uses E = (force used to overcome friction) * (distance). The drag forces and so on are lumped into the friction force.

So what does E=mc

1) Anything moving at c (light, for example) appears to be moving at c for any observer who cares to look at it.

2) c appears to be the ultimate limit of velocities. No matter what you do, you'll never be able to concoct a situation in which an observer will measure something going faster than c.

E=mc

E = m.

Energy and mass then, in fact, are one and the same. The consequences of this relationship are far too numerous to list, but here are a few:

1) An atomic bomb weighs a little less after it explodes.

2) Magnets weigh a little more when they're separated than when they're together.

3) The Sun is slowly losing mass, as it converts it to light.

- Warren

- #5

ando

Ok Warren. Thanks for the post, but I'm not sure this really settles the bet. Would object X, consume more, less, or the same ammount of energy when travelling from point A (let's assume these objects are moving through a vacuum - drag was not intended to be part of the equation), to point B than object Z if object Z travelled twice as fast?

- #6

chroot

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- #7

chroot

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However, your E=mc

As a result, you win the bet, but not because you correctly proved him wrong.

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ando

- #9

chroot

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Yes, this is wrong. c is notOriginally posted by ando

If you choose to use "natural units," like the second and light-second, the conversion factor is just one, and you're left with just E = m.

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- #10

russ_watters

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- #11

chroot

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That is absolutely incorrect. It seems correct because only nuclear reactions demonstrate the effect strongly enough to be easily measurable -- but in fact, E = mcOriginally posted by russ_watters

Ie, it only applies to nuclear reactions.

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Originally posted by ando

no, in a sence you are absolutly correct. the problem is that the effect is in no way noticable to objects traveling at running or driving speeds. it is only as one increases near the speed of light that he would be burduned down by the effects of E=MC2.

- #13

chroot

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Uh... thanks Maximus. Unless you mean "the effects of relativity," you are wrong.Originally posted by maximus

it is only as one increases near the speed of light that he would be burduned down by the effects of E=MC2.

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Originally posted by chroot

Uh... thanks Maximus. Unless you mean "the effects of relativity," you are wrong.

- Warren

the effects of relativity and e=mc2 are very similar and in some cases dependant on each other. as one approaches the speed of light the energy inherent in their movement

- #15

chroot

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- #16

Alexander

You mean, result from equation (due to equation).

- #17

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Originally posted by chroot

I took offense to your wording

i am sorry. i, of course, meant no offense.

- #18

Integral

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To summerize what has been said.

1. This is not a problem in relativity, because we are assuming speeds which we encounter in our daily lives, walking, running and driving.

2. We are not converting mass to energy so E= mc

3. This IS a problem of Newtonian Forces and Energy. So the important relationships are.

Kinetic Energy

i) E

and Newtons 3rd (?)

ii) F = ma

As well as conservation of Energy (to a certian extent)

We will so need to understand

iii) v = at (assume stating from v=0

in these equations

v=velocity

a=acceleration

t=time

m=mass

F=Force

E

So, let us start with some body at rest (v=0) To increase the velocity to something greater then 0 we must apply a force. By equation ii, this force will cause an acceleration, over time by equation (iii) the acceleration will increase the velocity.

Now by (i) we can see that the Kinetic energy of a body depends on the mass and the velocity, so we can see that how much energy is required to accelerate a body depends on the final velocity. This of course is a MINIMUN number, if there is any friction in the system it represents lost energy.

Now by the conservation of energy we must be converting some form of energy to kinetic energy, for the human body and the car this is chemical energy stored in our muscles or in gasoline, we must expend an amount of energy at least equivilent to the kinetic energy of the moving body.

Our body does not regain such energy when we slow down, if fact just the opposite is true, slowing down is every bit as much an acceleration as speeding up so we must spend more energy to stop.

The human body is a very complex chemical engine, it is better to learn the fundamental physics by considering simpler systems.

FURTHER POSTS SHALL NOT MENTION RELATIVITY! THERE IS NO NEED, ONLY CONFUSION THERE!

- #19

russ_watters

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Well actually, you gave 2 examples of nuclear reactions and one I wasn't familiar with. Point taken though. Nuclear reactions are by far the most commmon use for e=mc^2.Originally posted by chroot

That is absolutely incorrect. It seems correct because only nuclear reactions demonstrate the effect strongly enough to be easily measurable -- but in fact, E = mc^{2}applies to a great many systems not involving any nuclei. I gave a couple of examples.

- Warren

- #20

pmb

Originally posted by ando

I doubt that what you're friend said is true. The human body is a pretty complicated machine and as such does not work like a simple machine. When jogging the legs are doing more work since there is a lifting action which is absent in walking. Then there is other effects like cooling and the energy require to evaporate water etc. Far too complicated to derive in simple equations. I have a text on exercise phisiology - I'll look this up later tonight and get back tommorow.

E = mc^2 only means that the total amount of energy you do results in a loss of total rest mass by the amount m = E/c^2. The relationship between mass and energy is a linear one. I.e. A doubling of the mass results in a doubling of the energy.\

Pete

- #21

schwarzchildradius

The conservation equation of relativity does indeed enter. When the temperature rises in the body, the particles within the body are moving at higher velocity than at lower temperature. The higher velocity gives them a slightly larger mass, with respect to the same body at rest. It takes a little more energy to move a larger mass, but it is not significant unless the temperatures are greatly different.

- #22

HallsofIvy

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Another way to phrase this is: "A friend of mine and I are debating about "energy" and "E=MC2" and neither of us has any idea what we are talking about!"

Are you at all aware that "E= Mc

You may be confusing this with "E= (1/2)M v

I don't see any way for either of you to win this argument since neither of you knows what you are talking about!

- #23

jeff

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Most of the work done is against gravity.

Think of the simpler case of jumping in place on one leg. The knee bends moving the body's CG (centre of gravity) downwards, and hopping accelerates the CG through some vertical distance h. If the person's mass is M, then the energy expended per hop is approximately Mgh where g is the acceleration due to gravity. To approximate this vertical component of work done while running, just multiply it by the number of strides taken.

The other component involves the acceleration of your body forwards, but once you're running the work done in this direction should be much smaller than the vertical component.

Obviously a number of simplifying assumptions relating to the precise mechanics of running are implicit in my remarks, but the main point is that the largest portion of one's energy is expended working against gravity.

Think of the simpler case of jumping in place on one leg. The knee bends moving the body's CG (centre of gravity) downwards, and hopping accelerates the CG through some vertical distance h. If the person's mass is M, then the energy expended per hop is approximately Mgh where g is the acceleration due to gravity. To approximate this vertical component of work done while running, just multiply it by the number of strides taken.

The other component involves the acceleration of your body forwards, but once you're running the work done in this direction should be much smaller than the vertical component.

Obviously a number of simplifying assumptions relating to the precise mechanics of running are implicit in my remarks, but the main point is that the largest portion of one's energy is expended working against gravity.

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- #24

pmb

Originally posted by schwarzchildradius

The conservation equation of relativity does indeed enter. When the temperature rises in the body, the particles within the body are moving at higher velocity than at lower temperature. The higher velocity gives them a slightly larger mass, with respect to the same body at rest. It takes a little more energy to move a larger mass, but it is not significant unless the temperatures are greatly different.

This isn't quite right. First off the difference can b e neglected for all practical purposes regarding running etc. It's very very ver much less than the mass of a single strand of hair.

As far as the accuracy of the statement: Yes. If all other things remain the same then an increase in temperature will imply an increase in mass. However not all things remain the same. Where did that energy come from to raise the temperature? It came for Adinosine Tri-Phosphate (ATP). There is a chemical change involving ATP in which potential energy of the molecules becomes kinetic energy of other molecules. So while there is a drop in one there is an increase the other.

Also - E=mc^2 applies to *all* forms of energy - not just nuclear energy.

Pete

- #25

ando

Originally posted by HallsofIvy

Another way to phrase this is: "A friend of mine and I are debating about "energy" and "E=MC2" and neither of us has any idea what we are talking about!"

Are you at all aware that "E= Mc^{2}" has nothing at all to do with what YOUR speed is? It only says that an object of (rest) mass m contains intrinsic energy equal to Mc^{2}.

You may be confusing this with "E= (1/2)M v^{2}", the formula for kinetic energy, but that also has nothing to do with the amount of work actually done.

I don't see any way for either of you to win this argument since neither of you knows what you are talking about!

Uh thanks for the valuable input. I hope this makes you feel good about yourself.

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