Energy consumed for same work with different power?

[shivaprasadvh]

Homework Statement

When a man of 60 age and a man of 21 age cover a distance of 10 KM by walk. 21 age guy covers in 2.5 hour and 60 age guy covers in 5 hrs. How much energy is required by both to cover this journey?

Relevant Equations

E=1/2(mvsquare)

As per me the energy required by a man with 60 age and 21 age to cover same distance is same.

 

[Lnewqban]

Same amount of energy, only that delivered during different periods of time.

 

[berkeman]

Homework Statement:: When a man of 60 age and a man of 21 age cover a distance of 10 KM by walk. 21 age guy covers in 2.5 hour and 60 age guy covers in 5 hrs. How much energy is required by both to cover this journey?
Relevant Equations:: E=1/2(mvsquare)

As per me the energy required by a man with 60 age and 21 age to cover same distance is same.

The answer depends on many things. Perhaps in an idealized question the energy would be the same, but there is much more involved in a real situation.

Can you say what some of those other things might be that could make the energy different for the two different people covering that same distance?

 

[shivaprasadvh]

The answer depends on many things. Perhaps in an idealized question the energy would be the same, but there is much more involved in a real situation.

Can you say what some of those other things might be that could make the energy different for the two different people covering that same distance?

1. How flexible is the body to walk (if body is less flexible then there will be resistant force and hence more energy to over come the force).

This made be THiNK. Give me other things from you as well. I can learn...

 

Hold on, slow down, let's simplify things. A human is quite an inefficient machine, and walking is a physically complex process.

Why don't you first consider something simple, like a ball rolling along a flat frictionless surface? If the ball translates (and, rotates) more slowly, i.e. in the case that it covers a smaller distance over the same time interval, then the ball has a lower total kinetic energy. The work you would need to do on the ball to reach that smaller rolling velocity from rest would be less.

 

[shivaprasadvh]

Hold on, slow down, let's simplify things. A human is quite an inefficient machine, and walking is a physically complex process.

Why don't you first consider something simple, like a ball rolling along a flat frictionless surface? If the ball translates (and, rotates) more slowly, i.e. in the case that it covers a smaller distance over the same time interval, then the ball has a lower total kinetic energy. The work you would need to do on the ball to reach that smaller rolling velocity from rest would be less.

Okie, In the case of ball on frictionless surface. Consider that ball A takes 2 minutes to cover distance 1 m and ball B takes 5 minutes to cover distance 1 m. In this though time is different for the balls to reach the destination, but what would be energy consumed or energy required for ball A and ball B to cover 1m .It would be same.

 

[berkeman]

This made be THiNK. Give me other things from you as well. I can learn...

I can give some hints, but I think you should be able to figure most of them out.

Hint1: How much energy is consumed by a human body when it is just standing and not walking? (Ignore the age difference and any size difference for now)...

 

[shivaprasadvh]

I can give some hints, but I think you should be able to figure most of them out.

Hint1: How much energy is consumed by a human body when it is just standing and not walking? (Ignore the age difference and any size difference for now)...

When standing also there will be internal work done in the body. External work will be zero as there is not displacement but within body work happens. Okie in my query it was external work i had to mention?

 

but what would be energy consumed or energy required for ball A and ball B to cover 1m .It would be same.

No. If an idealised ball is rolling at constant velocity on a smooth horizontal surface, dissipative forces don't exist. The one that is moving faster has more kinetic energy.

When standing also there will be internal work done in the body.

By contracting muscle fibers, yes. That is required for your muscles to stay in tension, even whilst still.

 

[berkeman]

Okie in my query it was external work i had to mention?

You should definitely put some extra constraints and conditions on your question, because otherwise it becomes a very ambiguous question to answer. This was your original question:

How much energy is required by both to cover this journey?

Which could easily be interpreted to mean "How many calories of food must be consumed by these two people to carry out this journey?"

So yes, part of designing a good schoolwork-type question is to think of all of the subtle things that can enter into the answer, and be sure to explicitly state whether they apply or not. For example, that is why so many introductory projectile motion problems say explicitly to ignore the effects of air resistance.

 

[shivaprasadvh]

You should definitely put some extra constraints and conditions on your question, because otherwise it becomes a very ambiguous question to answer. This was your original question:

Which could easily be interpreted to mean "How many calories of food must be consumed by these two people to carry out this journey?"

So yes, part of designing a good schoolwork-type question is to think of all of the subtle things that can enter into the answer, and be sure to explicitly state whether they apply or not. For example, that is why so many introductory projectile motion problems say explicitly to ignore the effects of air resistance.

Thank you for this input. I will consider these while framing the questions.

 

[shivaprasadvh]

No. If an idealised ball is rolling at constant velocity on a smooth horizontal surface, dissipative forces don't exist. The one that is moving faster has more kinetic energy.



By contracting muscle fibers, yes. That is required for your muscles to stay in tension, even whilst still.

I give an initial velocity U to a ball on smooth surface and it covers say 1 m in t secs. Now for the same ball i give initial velocity of 2U. Under the same conditions as above thought it covers 1 m in half the time, to cover 1m the energy spent in both cases is same. Though ball in second case will travel more distance further to 1 m. Final KE in case 1 is lower that final KE in case 2 at 1 m. But change in KE i.e. energy consumed remains same.

 

You must first devise a simplified model for "walking". One example, that might seem ridiculous at first, could be a rolling polygon, like this:

Take an $n$ sided polygon. How much energy is required to roll it around to the highest point (standing vertically on its edge), assuming it will have some more GPE (how much more?) and some amount of kinetic energy at that top position?

Do we assume that all of that energy is dissipated away as heat when it lands on the next flat surface, and this cycle continues as it rolls? How might you adapt the model for different types of people?

I give an initial velocity U to a ball on smooth surface and it covers say 1 m in t secs. Now for the same ball i give initial velocity of 2U. Under the same conditions as above thought it covers 1 m in half the time, to cover 1m the energy spent in both cases is same. Though ball in second case will travel more distance further to 1 m. Final KE in case 1 is lower that final KE in case 2 at 1 m. But change in KE i.e. energy consumed remains same.

This is just wrong. The first ball has kinetic energy $\frac{1}{2}mU^2$, the second ball has kinetic energy $\frac{1}{2}m(2U)^2$. I have no idea what you mean by energy "spent". If the surface is smooth then neither undergoes a change in kinetic energy.

 

[berkeman]

I give an initial velocity U to a ball on smooth surface and it covers say 1 m in t secs. Now for the same ball i give initial velocity of 2U. Under the same conditions as above thought it covers 1 m in half the time, to cover 1m the energy spent in both cases is same. Though ball in second case will travel more distance further to 1 m. Final KE in case 1 is lower that final KE in case 2 at 1 m. But change in KE i.e. energy consumed remains same.

Again, I think a better problem statement could help this. If these are metal balls rolling on a hard metal surface, there is very little kinetic energy lost while rolling, so there is no energy input needed to maintain the rolling velocities once established.

There is a different energy input needed to get the two balls rolling at the different speeds, and that comes from the forces (linear and/or torques) needed to get the balls rolling at their final velocities. Those forces act over some finite time, which determines the powers needed to spin the balls up.

 

[shivaprasadvh]

You must first devise a simplified model for "walking". One example, that might seem ridiculous at first, could be a rolling polygon, like this:

View attachment 266494

Take an $n$ sided polygon. How much energy is required to roll it around to the highest point (standing vertically on its edge), assuming it will have some more GPE (how much more?) and some amount of kinetic energy at that top position?

Do we assume that all of that energy is dissipated away as heat when it lands on the next flat surface, and this cycle continues as it rolls? How might you adapt the model for different types of people?



This is just wrong. The first ball has kinetic energy $\frac{1}{2}mU^2$, the second ball has kinetic energy $\frac{1}{2}m(2U)^2$. I have no idea what you mean by energy "spent". If the surface is smooth then neither undergoes a change in kinetic energy.

Okie. Energy Spent = Final Energy - Initial Energy. So in this ideal case the energy spent will be zero as there is no change in mass and velocity over period of time. So energy spent in case 1 and energy spent in case 2 = 0 and equal.

 

Okie. Energy Spent = Final Energy - Initial Energy. So in this ideal case the energy spent will be zero as there is no change in mass and velocity over period of time. So energy spent in case 1 and energy spent in case 2 = 0 and equal.

Right. If you have something like a block that you release with some initial velocity on a surface with friction, then friction will do negative work on that block and reduce its energy. Things get a little more complicated.

 

[haruspex]

A human is quite an inefficient machine, and walking is a physically complex process.

One result of aging is loss of elasticity in the tendons. So more work is done for the same result.
But 10km taking 5 hours at 60 may leave you with an unnecessarily bleak view of your old age. I'm 70+, and still run 10km in not too much over the hour.

 

One result of aging is loss of elasticity in the tendons. So more work is done for the same result.

I suck at Biology so I'll take your word for it, and I trust you are correct, but I wondered if you could explain how a loss of elasticity results in more work being done for the same result? If we model a tendon as a spring, then if $k$ has reduced, then to stretch the spring to the same extension wouldn't we need to do less work? Maybe I misinterpret what a tendon does.

 

[berkeman]

I wondered if you could explain how a loss of elasticity results in more work being done for the same result? If we model a tendon as a spring, then if has reduced, then to stretch the spring to the same extension wouldn't we need to do less work?

I think it's better modeled as increasing damping, not loss of the spring constant...

 

[berkeman]

I think it's better modeled as increasing damping, not loss of the spring constant...

However, there are new boots available for us older folks to put the spring back in our step!

https://li0.rightinthebox.com/images/640x640/202003/zlry1585311844038.jpg?fmt=webp&v=1

 

I think it's better modeled as increasing damping, not loss of the spring constant...

Gracias, yes that would work!