Dynamics and thermodynamic problem

[Maxpin]

Homework Statement

A 80 kg man is running and each second he needs a energy of 812.77 J. Find:
a) The speed (in m/S) of the runner.
b) How many kcal will be burn in one hour with the speed calculated in a)?

Homework Equations

Kinetic energy; kcal = 4184 J

The Attempt at a Solution

a) Ek = (1/2) m v²

b) Convert to kcal with kcal = 4184 J and multiply by 3600 s.

Is it correct or is there another solution?

Thanks in advance.

 

[Andrew Mason]

Homework Statement

A 80 kg man is running and each second he needs a energy of 812.77 J. Find:
a) The speed (in m/S) of the runner.
b) How many kcal will be burn in one hour with the speed calculated in a)?

Homework Equations

Kinetic energy; kcal = 4184 J

The Attempt at a Solution

a) Ek = (1/2) m v²

b) Convert to kcal with kcal = 4184 J and multiply by 3600 s.

Is it correct or is there another solution?

We need more information. The relationship between speed and energy consumption is not something you can figure out from the information given. Unless you know the forces acting on the runner and the efficiency of the human body, you can't provide an answer.

The power output is P = F v where F = forces opposing the runner (eg. air resistance, friction). Power output = Power input x efficiency. If you are given the power consumption per unit of speed, you could work this out. But power consumption alone is not enough. Can you give us the exact wording of the question?

AM

 

[Maxpin]

We need more information. The relationship between speed and energy consumption is not something you can figure out from the information given. Unless you know the forces acting on the runner and the efficiency of the human body, you can't provide an answer.

The power output is P = F v where F = forces opposing the runner (eg. air resistance, friction). Power output = Power input x efficiency. If you are given the power consumption per unit of speed, you could work this out. But power consumption alone is not enough. Can you give us the exact wording of the question?

AM

Ok thanks. I know there is not an easy relationship between speed and consumption energy in a biomechamical system. It would be necessary to know some forces acting no the system. However considering an average calculation and if we suppose all the burned energy is only employed to develop kinetic energy, we would do this:

E = (1/2) mv² = 812.77 = (1/2) 80 v²

It gives: v = 4.50 m/s = 16.22 km/h

b) 812.77 J each second are, multiplying by 3600 s, 2926000 J in one hour and using the relation kcal = 4184 J it results: 700 kcal in one hour.

The speed and the burned calories obtained in this problems have reasonable values.

Of course, this calculation would be only an average.

I think it would be correct, wouldn't it?

Thanks.

 

[Andrew Mason]

Ok thanks. I know there is not an easy relationship between speed and consumption energy in a biomechamical system. It would be necessary to know some forces acting no the system. However considering an average calculation and if we suppose all the burned energy is only employed to develop kinetic energy, we would do this:

E = (1/2) mv² = 812.77 = (1/2) 80 v²

It gives: v = 4.50 m/s = 16.22 km/h

b) 812.77 J each second are, multiplying by 3600 s, 2926000 J in one hour and using the relation kcal = 4184 J it results: 700 kcal in one hour.

The speed and the burned calories obtained in this problems have reasonable values.

Of course, this calculation would be only an average.

I think it would be correct, wouldn't it?

No. You are confusing the kinetic energy of the moving runner with the power needed to maintain a speed. The two are not related. It may take 812.77 J for a 100% efficient runner facing no friction and air resistance to get up to a speed of 4.5 m/sec. But once he has achieved that speed, he would not need to keep outputting energy at that rate. In fact, on those facts he does not need to expend any energy to maintain that speed.

The question is unanswerable as given. You need to give us the exact wording of the question if you want some help on this.

AM

 

[Maxpin]

No. You are confusing the kinetic energy of the moving runner with the power needed to maintain a speed. The two are not related. It may take 812.77 J for a 100% efficient runner facing no friction and air resistance to get up to a speed of 4.5 m/sec. But once he has achieved that speed, he would not need to keep outputting energy at that rate. In fact, on those facts he does not need to expend any energy to maintain that speed.

The question is unanswerable as given. You need to give us the exact wording of the question if you want some help on this.

AM

I have given the exact wording of the problem. I know that once he has achieved that speed, he would not need to keep outputting energy at that rate. But if we look at the problem data we must believe that he burns 812.77 J per second as indicated for a 100% efficacy.

If these considerations are assumed we could equal burned energy to kinetic energy although we know that this is not an exact resolution.

The question is if my calculation could be considered correct as an average in view of the problem data and the feasibility of the obtained results. (700 kcal/h is a good quantity for a speed of 16 km/h).

 

[Andrew Mason]

I have given the exact wording of the problem.

Then the answer to the problem is that it cannot be answered. Is the question perhaps related to the answer from a previous question?

I know that once he has achieved that speed, he would not need to keep outputting energy at that rate. But if we look at the problem data we must believe that he burns 812.77 J per second as indicated for a 100% efficacy.

If these considerations are assumed we could equal burned energy to kinetic energy although we know that this is not an exact resolution.

That is not a reasonable assumption. The human body is a thermodynamic system. It cannot be anything close to 100% efficient.

The question is if my calculation could be considered correct as an average in view of the problem data and the feasibility of the obtained results. (700 kcal/h is a good quantity for a speed of 16 km/h).

No. Why would the runner not keep increasing his kinetic energy? Where does the energy that is being consumed go?

If the question said: it takes 350 kcal/h to maintain a runner running at 8 km/h, how fast would he be have to run if he was consuming 700 kcal/h assuming the forces opposing motion are constant, you would be right. P = Fv. F = P/v. P' = Fv' = Pv'/v; v' = P'v/P = 700*8/350 = 16.

AM

 

[Maxpin]

Then the answer to the problem is that it cannot be answered. Is the question perhaps related to the answer from a previous question?

That is not a reasonable assumption. The human body is a thermodynamic system. It cannot be anything close to 100% efficient.

No. Why would the runner not keep increasing his kinetic energy? Where does the energy that is being consumed go?

If the question said: it takes 350 kcal/h to maintain a runner running at 8 km/h, how fast would he be have to run if he was consuming 700 kcal/h assuming the forces opposing motion are constant, you would be right. P = Fv. F = P/v. P' = Fv' = Pv'/v; v' = P'v/P = 700*8/350 = 16.

AM

You are right and I know this problem cannot be answered in a simple form. The runner doesn't keep increasing his kinetic energy because we assume the given data are correct in the text question. That means he needs 812.77 J each second to keep a constant speed. Of course, we know there could be energy leaks in something (friction, air resistance, etc).

The difficult task would be to know what part of this energy is spent in kinetic energy and what part is spent in other things. In view of my calculations and as kinetic energy depends on v², if kinetic energy is reduced to its half then the speed would be divided by 2^0.5. That means that if the burn energy would be 350 kcal/h then the speed would be reduced to 11.31 km/h.

This would be a big reduction in energy but not too much in speed. I mean, it is impossible to burn 700 kcal/h at 16 km/h and only to burn 350 kcal/h at 11.3 km/h.

I would be interesting to know what is the relation in a real human body between speed and burn energy at constant time (one hour). With this data we could have an idea of the energy fraction used as kinetic energy and energy leaks (friction, sweat, etc...)

Thanks again.

 

[Andrew Mason]

You are right and I know this problem cannot be answered in a simple form. The runner doesn't keep increasing his kinetic energy because we assume the given data are correct in the text question. That means he needs 812.77 J each second to keep a constant speed. Of course, we know there could be energy leaks in something (friction, air resistance, etc).

We know that since his kinetic energy is not increasing ALL that energy is being lost.

The difficult task would be to know what part of this energy is spent in kinetic energy and what part is spent in other things.

Well, none of it is being used to provide kinetic energy. All of the energy is being used just to maintain his speed. None of it is adding any new kinetic energy.

In view of my calculations and as kinetic energy depends on v², if kinetic energy is reduced to its half then the speed would be divided by 2^0.5. That means that if the burn energy would be 350 kcal/h then the speed would be reduced to 11.31 km/h.

This would be a big reduction in energy but not too much in speed. I mean, it is impossible to burn 700 kcal/h at 16 km/h and only to burn 350 kcal/h at 11.3 km/h.

Why impossible? All it would mean is that the power input per unit of speed (which is just the opposing force: P/v = F) would be less at the lower speed. That actually is not unreasonable - it certainly is the case with a car, for example.

AM

 

[Maxpin]

We know that since his kinetic energy is not increasing ALL that energy is being lost.

Well, none of it is being used to provide kinetic energy. All of the energy is being used just to maintain his speed. None of it is adding any new kinetic energy.

Why impossible? All it would mean is that the power input per unit of speed (which is just the opposing force: P/v = F) would be less at the lower speed. That actually is not unreasonable - it certainly is the case with a car, for example.

AM

For a running man it would be interesting to know what's the energy lost (friction, hot, etc). Then you say that this lost energy MUST be all the burnt energy independently of the speed. The question is why in the real life at a bigger burnt energy there is a bigger speed?