Is Average Force the Same in Fast vs. Slow Weightlifting Reps?

[Dale]

the energy expenditure is identical regardless if in a minute you'll do 100 reps or 1 or even you just hold the weight.

Careful here, you are using the wrong terminology. The energy expenditure is different, you mean that the work done is identical.

The work done is the integral of the force times the differential displacement. That is always 0 over a closed loop in a conservative field like gravity.

Energy expenditure is more complicated than the work done and includes all of the "messy" biological inefficiencies. So the energy expenditure is higher for the fast reps, but not easy to calculate.

 

[waynexk8]

Sorry about my long posts, I have been told before, just can't do it any other way.

Force is not a conserved quantity. It doesn't get used up.

Yes right. However, when I lift a weight up and lower it down, I have to use a new or exert physical force once again that needs energy to do this. I mean the same weight is not going to keep being lifted up and down on its own after I just lift it once, as each and every time I need to use the same amount of force to lift and lower the weight, but every time I use this force that needs energy it’s not the first force I used.

Or maybe you call this more work ? However work will be the amount of energy transferred by a force acting through a distance. So if I lift the weight up and down a second time, I am using more energy transferred by more of the same force over a greater distance than the first lift, or you may call that more work, I call it using more of the same force, if you sort of get me.

DaleSpam, have to say to you and the rest for your time and helping me, and I know the above sounds a bit complicated to you, but you have been talking physics for many years, however I have not, so please bear with me on my layman’s way of putting thing.

No, the work done over any closed path is 0 in a conservative field like gravity.

I sort of get that; you mean it’s like dISPLACEMENT?AS IF I WALK SO AND SO MILES AND COME BACK TO THE SAME PLACE THERE IS NO dISPLACEMENT.

However if work will be the amount of energy transferred by a force acting through a distance. I have coved a distance up and down 2m in this example, thus work has been done, force has been used and energy ?

As to move a weight up and down I have to use physical force and energy in the up lift and the down lower ?

Yes, there is more muscle "activity", but not more average force nor more work done. Therefore neither average force nor work are good indicators of muscle activity.

Right we agree on more muscle activity.

But surely there must be more average force used, or should I say more of the same average force e used, as I said above, each and every time I lift the weight I have to use a new or separate force as the first effort of force will not lift the weight.

Then you say the same for work !

Ok what in physics will be a good indicators of muscle activity ? Power ? As power is the rate at which work is performed and energy is converted. But then you say no to work.

Confusing.

What we really need is with a muscle or machine, which puts the most tension on the muscles, to me and most it has to be the faster reps as of the higher high and higher peak higher done 6 times in the same time frame as the slow reps.

To me in this video, I have failed roughly 50% faster because I have used more overall or total strength up faster, but you say this is wrong ! How else would I fail if I was not using my force/strength up faster ? Or maybe that’s basically it; I am just using my force/strength up faster ?

http://www.youtube.com/user/waynerock999?feature=mhum#p/a/u/0/sbRVQ_nmhpw

Wayne

 

[JaredJames]

However if work will be the amount of energy transferred by a force acting through a distance. I have coved a distance up and down 2m in this example, thus work has been done, force has been used and energy ?

Force is a vector. It has magnitude and direction.

In the raising stage force is upwards, in the lowering stage force is downwards. The directions 'cancel out'.

Because work = force x distance, work also has this direction. So raising gives you work upwards and lowering gives you work downwards - both equal - they cancel out.

 

[waynexk8]

Careful here, you are using the wrong terminology. The energy expenditure is different, you mean that the work done is identical.

The work done is the integral of the force times the differential displacement. That is always 0 over a closed loop in a conservative field like gravity.

Energy expenditure is more complicated than the work done and includes all of the "messy" biological inefficiencies. So the energy expenditure is higher for the fast reps, but not easy to calculate.

Well said, I cannot understand why D. thinks this way, I did try to explain this to him from the start, and it’s something small I think he’s got wrong but just cannot see it, happens to us all.

DaleSpam, I have been asked to keep repeating and posting my; “Energy Expenditure and Nutrient Oxidation will and have been established in a room calorimetry.” All the time, too which you find on the page before this.

However why do you think that D. is ignoring science/physics and the way energy is and has been calculated for many years ? As every nutrition site or book states that if you do any activity twice as fast in the same time frame, as you have covered twice the distance you used twice the energy ?

Wayne

 

[waynexk8]

Force is a vector. It has magnitude and direction.

In the raising stage force is upwards, in the lowering stage force is downwards. The directions 'cancel out'.

Because work = force x distance, work also has this direction. So raising gives you work upwards and lowering gives you work downwards - both equal - they cancel out.

Yes I sort of get that, but all I am saying is you need physical force/strength and energy in both direction, to lift the weight up, and to lower it under control.

Wayne

 

[JaredJames]

Yes I sort of get that, but all I am saying is you need physical force/strength and energy in both direction, to lift the weight up, and to lower it under control.

You add the energy use as you would add distance travelled.

This is independent of work.

When you calculate work you use force and distance. You make the distance into displacement.

 

[douglis]

Careful here, you are using the wrong terminology. The energy expenditure is different, you mean that the work done is identical.

The work done is the integral of the force times the differential displacement. That is always 0 over a closed loop in a conservative field like gravity.

Energy expenditure is more complicated than the work done and includes all of the "messy" biological inefficiencies. So the energy expenditure is higher for the fast reps, but not easy to calculate.

Yes...I examine the case from physics POV and I ignore the biological inefficiencies.

Do you agree with this rocket example?
''Let's say a rocket starts accelerating from the Earth upwards and after a while the engine shuts down.The rocket decelerates and reaches a maximum height before it will start falling.
Let's say it reached the max H in exactly one minute.The average acceleration for that trip is zero(starting and ending velocity zero) hence the average force is exactly equal with the weight of the rocket.
Now if the rocket was just standing still in the air for that minute again its engine would have used force equal with the rocket's weight for 1min.

In both cases the rocket will use exactly identical fuels since it used the same average thrust for 1 min.''

 

[JaredJames]

Do you agree with this rocket example?

I've just responded to this.

You are ignoring basic things and confusing the difference between acceleration and constant velocity.

Energy use is not the same when you have two different accelerations.

 

[Dale]

Do you agree with this rocket example?

I think that the rocket example is irrelevant for this thread. Muscles and rockets operate on very different principles and I don't see the value of the analogy in this context.

 

[douglis]

All this average acceleration stuff is non-sense. You are ignoring the fact that to get a rocket to accelerate at 1m/s² uses significantly less energy than to get it to accelerate at 2m/s². To increase the acceleration by even such a small amount requires a drastic energy increase.

O.K...let's start from here.
We agree that acceleration requires more energy and also the greater the acceleration the greater the energy.
So in this rocket example let's say that the rocket used force that caused initial acceleration equal with 2g(if we assume that was accelerated for 30sec) and then since the engine shut down used zero force for the final deceleretion(the last 30sec).
On average the force that was used produced average acceleration equal with g or else the average force was mg for 1 minute.
Exactly like the force that was used by the rocket that was standing still in the air.

Since in both cases the average acceleretion that the engined produced was equal I can't see how the fuels can be different.The fuels that the moving rocket spent when using 2g for 30 sec are exactly equal with the fuels that the standing rocket spent by using g for 1min.

 

[JaredJames]

Since in both cases the average acceleretion that the engined produced was equal I can't see how the fuels can be different.The fuels that the moving rocket spent when using 2g for 30 sec are exactly equal with the fuels that the standing rocket spent by using g for 1min.

They most certainly are not.

Average acceleration doesn't apply to the fuel use because it isn't linear.

As previously, fuel use is exponential. If you double your speed, the fuel use more than doubles.

So to double your speed may half your travel time but it more than doubles fuel use.

1g for a minute may use 1L of fuel. 2g for 30 seconds will use 4L. 4g for 15 seconds will use 15L etc.

Obviously, it's not as straight forward as that but that's the gist of it. 1g for a minute does not equal 2g for 30 seconds.

 

[douglis]

They most certainly are not.

Average acceleration doesn't apply to the fuel use because it isn't linear.

As previously, fuel use is exponential. If you double your speed, the fuel use more than doubles.

So to double your speed may half your travel time but it more than doubles fuel use.

1g for a minute may use 1L of fuel. 2g for 30 seconds will use 4L. 4g for 15 seconds will use 15L etc.

Obviously, it's not as straight forward as that but that's the gist of it. 1g for a minute does not equal 2g for 30 seconds.

Look...I tried to make an equivalent with weight lifting and I assumed that fuel expenditure is linear.If we don't make that assumption the discussion goes too far.

 

[JaredJames]

Look...I tried to make an equivalent with weight lifting and I assumed that fuel expenditure is linear.If we don't make that assumption the discussion goes too far.

In no case, ever, is fuel consumption linear. Weight lifting and rockets alike.

To make that assumption would completely change the physics and make it so that twice the speed with double the distance would give you the same energy use. Complete non-sense.

 

[douglis]

In no case, ever, is fuel consumption linear. Weight lifting and rockets alike.

To make that assumption would completely change the physics and make it so that twice the speed with double the distance would give you the same energy use. Complete non-sense.

I insist that speed and distance are irrelevant and keep the nonsense comments for youself.
Forget for a moment the acceleration phases.Two rockets with different constant speeds will spend identical fuels for the same duration if you exclude the air resistance.

Again at the acceleration part now...do you believe that double force will result in more than double fuel expenditure?

 

[JaredJames]

I insist that speed and distance are irrelevant and keep the nonsense comments for youself.
Forget for a moment the acceleration phases.Two rockets with different constant speeds will spend identical fuels for the same duration if you exclude the air resistance.

Constant speed yes, but it has to get to that speed which involves acceleration which, depending on the required velocity is different (as you are aware). To accelerate to 50mph does not take half the fuel it takes to accelerate to 100mph.

For simplicity you can ignore constant speed and simply have acceleration and deceleration.

If deceleration is due to gravity the rocket expends no fuel.

Leaving you only acceleration to consider for fuel use and the values for different accelerations (1m/s², 2m/s², 4m/s² etc) do not have the same fuel use and it is not simply double fuel as you double acceleration.

Again at the acceleration part now...do you believe that double force will result in more than double fuel expenditure?

There's nothing to believe. It is fact.

 

[douglis]

Constant speed yes, but it has to get to that speed which involves acceleration which, depending on the required velocity is different (as you are aware). To accelerate to 50mph does not take half the fuel it takes to accelerate to 100mph.

Yes...but the constant speed case proves that it's not about speed and distance but only about acceleration.
Anyway...I find more interesting the acceleration/force part right now.

There's nothing to believe. It is fact.

I have no reason to doubt you.
So...let's say I hold a 50 pounds weight for a minute and then a 100 pounds weight for another minute.Since the force/energy relation is not linear...in the second case I'll spend more than double energy.I guess the magnitude of the exponent is different in any case of engine/muscle.
That's really interesting and changes everything.Not that I don't believe you but can you give me a link or something I can search about it?

 

[waynexk8]

D. I did tell you distance was important, but you did not want to listen to me for some reason. Time and distance.

What D. is saying is that if you climb 10m up a robe in 100 seconds, and then climb up a rope 5m in 100 seconds that you will use the same energy.

As we know, and as science has told us for many years, when you do something slower, the only time you will use the same energy is when you travel the same distance as in the person who has done the activity faster.

Wayne

 

[waynexk8]

Also, this debate dose “NOT” only depend on ACCELERATION, as if I ran uphill at a speed of 1ms for 5 hours and also ran uphill at a speed of 5ms for 5 hours, the ONLY time that the speed of 1ms would use the same energy is when it covered the same distance.

As you can see by the example, acceleration energy would be negotiable. This is because we HAVE to take into consideration, Heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, will ALL be more, higher when moving faster for the same time frame, however the faster moving will cover more distance.

Wayne

 

[Dale]

I mean the same weight is not going to keep being lifted up and down on its own after I just lift it once

It would if our arms were elastic, like springs. The weight certainly could continue to be lifted up and down on its own, precisely because force is not conserved and work is not done. You are confusing biology with physics.

you may call that more work, I call it using more of the same force, if you sort of get me.

When you are discussing physics it is very important to use the correct terminology. These are not just words with ambiguous meanings that depend on context to "sort of get". These are precisely defined technical terms with specific and exact meanings. I think that more than half of the problem in this thread and the other is that you persistently continue to use the same incorrect terminology even after you have been corrected. The frustration that you occasionally hear directed at you stems from that consistent behavior. It is OK to not know the right word in the beginning, that is the purpose of an educational site like this, but once it has been explained to you it is not OK to continue misusing precisely defined physics terminology.

I have coved a distance up and down 2m in this example, thus work has been done

No work has been done on the weight if you lift it up 1 m and then down 1 m. As you lift it up, the force is directed up and the displacement is also up, therefore you are doing work on the weight. As you let it down, the force is still directed up but the displacement is down, therefore you are doing negative work on the weight (the weight is doing work on you). A concentric contraction does positive work, an eccentric contraction does negative work, and over one full rep the work is 0.

But surely there must be more average force used

There is not. I have proved that in our previous conversation.

Ok what in physics will be a good indicators of muscle activity ?

The energy consumed would be a good indicator of muscle activity. This is different from the work done, but it is also not a simple quantity to calculate and would require some complicated modeling similar to what Hill did, but including metabolism. It could be measured with the room calorimetry as you mention, but not easily calculated in advance.

 

[waynexk8]

You add the energy use as you would add distance travelled.

Yes agree.

This is independent of work.

So you’re saying physical force/strength in both directions, to lift the weight up, and to lower it under control, is independent of work ? If that’s true I am glad we got that out of the way.

When you calculate work you use force and distance. You make the distance into displacement.

I find this a little hard to understand. So is physics saying if I move up and down that as no displacement has been done no work has been done, but what if I move up and then down, but down on a slight angle ? Meaning I am not back in the same place ? Or have I got the wrong end of the stick.

In my way of thinking, like in the weighting lifting repping example, in .5 of a second, I have moved the weight 1m, however the slower moving rep has only moved the weight .16 of a meter, or one sixth of what I have moved in the same time frame. To me that means I have used far far far more force/strength.

Wayne

 

[waynexk8]

It would if our arms were elastic, like springs. The weight certainly could continue to be lifted up and down on its own, precisely because force is not conserved and work is not done. You are confusing biology with physics.

When you are discussing physics it is very important to use the correct terminology. These are not just words with ambiguous meanings that depend on context to "sort of get". These are precisely defined technical terms with specific and exact meanings. I think that more than half of the problem in this thread and the other is that you persistently continue to use the same incorrect terminology even after you have been corrected. The frustration that you occasionally hear directed at you stems from that consistent behavior. It is OK to not know the right word in the beginning, that is the purpose of an educational site like this, but once it has been explained to you it is not OK to continue misusing precisely defined physics terminology.

No work has been done on the weight if you lift it up 1 m and then down 1 m. As you lift it up, the force is directed up and the displacement is also up, therefore you are doing work on the weight. As you let it down, the force is still directed up but the displacement is down, therefore you are doing negative work on the weight (the weight is doing work on you). A concentric contraction does positive work, an eccentric contraction does negative work, and over one full rep the work is 0.

There is not. I have proved that in our previous conversation.

The energy consumed would be a good indicator of muscle activity. This is different from the work done, but it is also not a simple quantity to calculate and would require some complicated modeling similar to what Hill did, but including metabolism. It could be measured with the room calorimetry as you mention, but not easily calculated in advance.

Hi DaleSpam,

Just going out for the night, so will have to get back to this one, however as I have said before, a big thanks to all for your help and time. Hey, with a few more weeks of this, I will have Stephen Hawking begging me to help him with some of his harder equations ROL.

Wayne

 

[douglis]

Wayne...check BB.com.

 

[JaredJames]

So you’re saying physical force/strength in both directions, to lift the weight up, and to lower it under control, is independent of work ? If that’s true I am glad we got that out of the way.

Force applied to an object, moving it a distance gives you work. I have no idea what you're talking about now.

I find this a little hard to understand. So is physics saying if I move up and down that as no displacement has been done no work has been done, but what if I move up and then down, but down on a slight angle ? Meaning I am not back in the same place ? Or have I got the wrong end of the stick.

If you end up in the same place, the weight has been displaced but the final displacement is zero.

Distance is a scalar. This means it simply has a magnitude.
Displacement is a vector. This means it has a magnitude and a direction.

They are not the same thing.

If I move 3m North and 4m East, I have traveled a distance of 7m but my displacement is 5m North East.

Bolded: Ok, now you're just being ridiculous and adding insignificant factors trying to complicate things. For the purpose of simplicity this is irrelevant.

In my way of thinking, like in the weighting lifting repping example, in .5 of a second, I have moved the weight 1m, however the slower moving rep has only moved the weight .16 of a meter, or one sixth of what I have moved in the same time frame. To me that means I have used far far far more force/strength.

I've never argued that. I don't know what you're debating now.

 

[sophiecentaur]

@Wayne
You started this thread 'outside' Physics and your last post is also way outside. You seem to have made no progress with this but see determined, somehow, to prove that you are still 'right'. Why are you bothering still.

 

[Dale]

So is physics saying if I move up and down that as no displacement has been done no work has been done, but what if I move up and then down, but down on a slight angle ? Meaning I am not back in the same place ?

If you do not return to the same place then the path is not a closed loop and the work done may be non-zero. In a gravitational field the work done on the weight depends only on the difference in height between the starting point and the stopping point.

 

[douglis]

I have no reason to doubt you.
So...let's say I hold a 50 pounds weight for a minute and then a 100 pounds weight for another minute.Since the force/energy relation is not linear...in the second case I'll spend more than double energy.I guess the magnitude of the exponent is different in any case of engine/muscle.
That's really interesting and changes everything.Not that I don't believe you but can you give me a link or something I can search about it?

Hi jarednjames,
you said that double force results in more than energy expenditure(expontential relationship) and sounded logical to me.

But I still would like to see some refferences.In the above example,after a little research I did,the relationship seems to be linear.

 

[waynexk8]

Have to say a big thank you to all the people here that have helped.

As looking at the two other forums we are debating on, seems that my friend D. is going more to the side that the faster reps must use more energy. This is what I said.




Therefore, if more energy is used on the faster reps, and in this example I will just say the reps are twice as fast. So we have 100 pounds moved up 1m and down 1m, twice in 2 seconds = 4m. We then have 100 pounds moved up 1m and down 1m, one time in 2 seconds = 2m.

I would say the faster rep uses twice the amount of energy.

I would also say, that because twice the amount of energy was used, was because there was more muscle activity, meaning the higher high forces, and the higher peak forces uses far far far more energy in the same time frame as the slower rep. As it needs to use more energy and higher high forces and higher peak forces to cover twice the distance in the same time frame.

Wayne

 

[sophiecentaur]

Is this still going on?
What is there possibly left to say except to give more and more instances of numbers of "reps" and the fact that the system is not possible to characterise as a piece of mechanics?

 

[douglis]

Therefore, if more energy is used on the faster reps, and in this example I will just say the reps are twice as fast. So we have 100 pounds moved up 1m and down 1m, twice in 2 seconds = 4m. We then have 100 pounds moved up 1m and down 1m, one time in 2 seconds = 2m.

I would say the faster rep uses twice the amount of energy.

Wayne

Wayne...the total displacement is zero so the work done is zero too.End of story.

As for the energy expenditure what only matters is if the higher fluctuations of force in fast reps expend energy in a exponential or linear mode.
That's something that can't be proved or disproved with physics.In the biology forum maybe you can find a better answer.

 

[JaredJames]

Hi jarednjames,
you said that double force results in more than energy expenditure(expontential relationship) and sounded logical to me.

But I still would like to see some refferences.In the above example,after a little research I did,the relationship seems to be linear.

OK douglis, I'm disappointed you haven't done the calcs yourself (seeing as I did them a few pages back).

Let's keep it simple. You have a 1kg object that you accelerate for 1 second.

So we accelerate it at 1m/s², 2m/s² and 4m/s² so the resulting speed after 1s is 1m/s, 2m/s and 4m/s.

This gives each one a final KE of 0.5mv². Which is 0.5J, 2J and 8J respectively.

As you can see it's an exponential increase in energy required per second of acceleration to achieve the required final velocity (and KE).

Now I can see your next question already (and I can see what you did to get a linear relationship).

You're going to point out that in the rep example you accelerate for 1s in the slow reps and only 0.25s in the fast ones (as per my previous example). Which is all well and good, at the end of those times you will have imparted equal KE on the weight (it would be moving at 1m/s in both cases and so would have the same KE). However, the energy required for it to accelerate at each rate is different. Getting to 1m/s in 1s uses significantly less energy than getting to 1m/s in 0.25s.