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

[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.

 

[Dale]

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.

I agree. The work done is the same (0), but more energy is expended.

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

How do you get that figure?

 

[Dale]

Getting to 1m/s in 1s uses significantly less energy than getting to 1m/s in 0.25s.

Only if the second system is less efficient. If they are equally efficient it is the same.

 

[sophiecentaur]

If no work is done then efficiency is zero. Zero for one circumstance can equal zero for another circumstance but that says nothing about the energy input in each case.
Oh God. why doesn't this all stop?
Muscles cannot be treated as simple physics systems.
Wayne, Why do you keep asking the same questions again and again? The answer just isn't there.

 

[JaredJames]

Only if the second system is less efficient. If they are equally efficient it is the same.

Just to clarify here, you are telling me that to accelerate an object at 4m/s² uses equal energy as doing so at 1m/s²? Despite the fact the force required is four times larger?

I'm not arguing the final imparted energy, only the energy required to gain said acceleration.

 

[Dale]

Just to clarify here, you are telling me that to accelerate an object at 4m/s² uses equal energy as doing so at 1m/s²? Despite the fact the force required is four times larger?

Yes, provided they are equally efficient. A good example to work out is a mass accelerated by a spring.

 

[JaredJames]

Yes, provided they are equally efficient. A good example to work out is a mass accelerated by a spring.

This is with respect to time of course?

I'm really not seeing this.

If you accelerate something at 1m/s² and something at 100m/s² they don't use the same energy, unless it is brought in respect to time to provide you with equal final velocities. Correct?

 

[JaredJames]

Another thing, take an acceleration of 4m/s².

You will travel 1m in 0.25s, which gives you a velocity of 4m/s.

But, if you stop accelerating after 0.25s your final velocity is 1m/s.

So which value is correct for energy calcs?

 

[Dale]

This is with respect to time of course?

I'm really not seeing this.

If you accelerate something at 1m/s² and something at 100m/s² they don't use the same energy, unless it is brought in respect to time to provide you with equal final velocities. Correct?

It sounds like you are confusing energy and power. Assuming 100% efficiency, if you accelerate an object to a given speed then you will use the same amount of energy regardless of how large the force is. The power for the larger force will be higher, the duration and distance will be shorter, and the energy will be the same.

 

[Dale]

Another thing, take an acceleration of 4m/s².

You will travel 1m in 0.25s, which gives you a velocity of 4m/s.

But, if you stop accelerating after 0.25s your final velocity is 1m/s.

So which value is correct for energy calcs?

Work is force times distance, not force times time.

 

[douglis]

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.

jarednjames...I understood exactly from the first time what you say.Now try to understand me.

Regardless the energy that is spent in the acceleration phase(0.5J, 2J or 8J) the work at the end of the lifting will always be mgh.So mgh is the theoretical minimum of energy that's required to lift a weight assuming 100% efficiency.
It's obvious that if you lift the weight in 1sec your muscles will work a lot more efficiently than if you lift it in 5sec so the mgh can't tell us something about the total energy expenditure.

In one of your above examples...you compared lifting 100 bags in a minute or 1 bag constantly lifted it for a whole minute.
In both cases you use exactly the same average force for 1 minute(a lot more efficiently in the 100 bags case since you produce 100 times more work).
Now to agree with you that the 100 bags require more energy all you have to do is to prove me that the force-energy relation isn't linear or else that the higher fluctuations of force when you lift the 100 bags increase the energy requirement.

 

[JaredJames]

It sounds like you are confusing energy and power. Assuming 100% efficiency, if you accelerate an object to a given speed then you will use the same amount of energy regardless of how large the force is.

I understand that, but you have to generate that force in the first place. Which takes more energy to create the larger force.

If I want my slingshot to have a greater acceleration I must input more energy to give a greater initial force. If I then only allow the projectile to the same velocity on launch, the force will be applied for a shorter time, that is all. The force must still be generated.

Work is force times distance, not force times time.

In both cases, the force is applied for a distance of 1m to a mass of 1kg.

At 1m/s², force = 1*1 = 1N.

At 4m/s², force = 4*1 = 4N.

Both forces are applied for 1m:

So the work for the first is Fd = 1*1 = 1J

And the work for the latter is Fd = 4*1 = 4J

The difference is, the former is applied for 1 second, the latter for 0.25 seconds. The resultant velocity is 1m/s in both cases with a final KE of 0.5J in both cases.

 

[douglis]

...or else prove me the following.

To lift a weight in 1 sec you need to provide initial acceleration equal with g+x and final deceleration equal with g-x.On average the acceleration you provide is g.Just like the acceleration you need to provide in order to hold the weight for 1 sec.
Prove me that the greater energy that's required when you provide acceleration g+x isn't balanced by the less energy that's required when you provide g-x.

 

[JaredJames]

...or else prove me the following.

To lift a weight in 1 sec you need to provide initial acceleration equal with g+x and final deceleration equal with g-x.On average the acceleration you provide is g.Just like the acceleration you need to provide in order to hold the weight for 1 sec.
Prove me that the greater energy that's required when you provide acceleration g+x isn't balanced by the less energy that's required when you provide g-x.

This goes back over the last few pages, I'm fed up of repeating posts now.

 

[waynexk8]

Dear members, I would like to try to clear a small issue on this debate up please.
If you move a weight up at any speed at all, and for any distance, and then move the weight down, your said no work has been done. However, I am/was not on about physics work, what I was talking about when I move a weight up and then down was physical work, physical work has been done, lifting the weight up, and lower it down, right ? And I am sure you all will say yes physical work has been done.

So next, could tell me what no work in physics has not been done ? Because as far as I know, work in physics means the amount of the energy transferred by a force acting through a distance. But what does work now mean in physics ?

Wayne

 

[JaredJames]

If you move a weight up at any speed at all, and for any distance, and then move the weight down, your said no work has been done.

Correct. Assuming the force upwards = force downwards and distance upwards = distance downwards.

However, I am/was not on about physics work

We're not interested in your own personal definitions. You have been told about this over and over. Work has a very specific meaning in physics.

EDIT: Your body does chemical work which is converted to mechanical work.

So next, could tell me what no work in physics has not been done ? Because as far as I know, work in physics means the amount of the energy transferred by a force acting through a distance. But what does work now mean in physics ?

Work is the force applied multiplied by the distance it is applied for: http://en.wikipedia.org/wiki/Work_(physics)

Posts 123, 126 and 143 from myself explain to you why work done in opposing directions cancels out.

 

[waynexk8]

Next question, and all these numbers are just for the debates sake.

We are still moving the persons 80% RM, which is at this time 80 pounds, thus the most force he can use is 100 pounds. Distance of reps up and down = 1m each way. Fast rep is done six times = 6 seconds, slow rep is done one time = 6 seconds. We have now separated the concentric/positive on the rep in five segments.

The faster rep with be the second rep, as this will be for the conceding reps with have the peak forces, these are the forces coming out of the transition from negative to positive. As the force/tension on the muscles will be far higher {someone could work it out for me if they could please ?} if the 80 pounds has been traveling down at .5 of a second per meter, as the weight will have taken on acceleration components, and be far far far harder to slow down, stop, and reverse direction in Milly seconds, than just moving it up from a still start. After the first segment will be the higher high forces.

Fast rep,
140, 100, 100, 40, 20.

Slow rep,
80, 80, 80, 80, 80.

Question one,

We all agree that the faster reps have used more energy than the slow reps, as of two factors. First the acceleration forces of the faster reps energy used is not linear, its more like the air drag equations. If this was running, {and please say if you think running and its numbers for the amount of energy used is way out} And I was running six times faster, I would use six times the energy, however as for the most deceleration phase in the faster rep, what if we took this number down to we use three times as much energy in the faster rep ? Please I am not the physicist, that was/is a guess.

Question two,

a,
With the above fast and slow reps segments, 140 = 60 or 75% more than 80, then a 100 = 20 or 25% more than 80, then again a 100 = 20 or 25% more than 80.

So the higher high forces and the higher high peak force are more higher in the faster reps, right ?

b,
And do we ALL agree that the forces from the last two slow rep segments, even that they are now far higher than the faster reps force, cannot, and do not balance out the energy used over whole, right ?

C,
I best not ask that yet until I have got answers and confirmation on the above.

Wayne

 

[JaredJames]

Next question, and all these numbers are just for the debates sake.

These numbers are meaningless. You've been given your answer (at least what's possible) and yet you keep posting this repetitive non-sense and being told the same thing.

What are you talking about energy "balancing out"? Energy is used in all cases. In the faster reps, more energy is used.

That's the end of it. There's no need to drag this thread out further.

 

[douglis]

In the faster reps, more energy is used.

If we assume 100% efficiency isn't the energy always equal with mgh regardless the lifting speed?

 

[waynexk8]

Correct. Assuming the force upwards = force downwards and distance upwards = distance downwards.

Well I knew basically if I lift a weight up and then lower it under control physical work has to have been done, but just wanted to confirmed, and see we are working this debate out the same.

We're not interested in your own personal definitions. You have been told about this over and over. Work has a very specific meaning in physics.

EDIT: Your body does chemical work which is converted to mechanical work.

Ok, but please could you tell me what I should say, as I thought work was the amount of energy transferred by the force acting through a distance. And as I have used force which needed energy, and moved it though a distance I thought I was right in saying work, or should I say physical work or mechanical work ?

As basically as we now all agree, that moving up and down I have done physical work, used energy for force. This in this debate if we all agree that I have done physical work moving up and down, I do not understand why we should say no work has been done, as if I said I use 10n to move the weight up, and 8n to lower the weight down, I have still used 10n and 8n and used energy, so these cannot cancel each other out, as my body has used the force and energy.

Or maybe we should add in kinology and biomechanics to the physics ?

Sorry if I am being confusing, and if so thanks for your patience and time.

Work is the force applied multiplied by the distance it is applied for: http://en.wikipedia.org/wiki/Work_(physics)

Posts 123, 126 and 143 from myself explain to you why work done in opposing directions cancels out.

Yes get what your saying with work, but it cannot cancel the force and energy out that moved the weight up and lowered it down, that's the point I am trying to make, and I think that's reverent to this debate, as we all know and agree physical work has to be done in both directions.

Wayne

 

[JaredJames]

Yes get what your saying with work, but it cannot cancel the force and energy out that moved the weight up and lowered it down, that's the point I am trying to make, and I think that's reverent to this debate, as we all know and agree physical work has to be done in both directions.

The energy use cannot cancel out, it adds up.

The force most certainly can.

Force is a vector with magnitude and direction. 10N upwards and 10N downwards result in a net force of zero.

Because of this direction, the work gives you a net of zero.

It really doesn't matter what you think, the physics say net work is zero. You need to check the definitions and make sure you understand them. At the moment you clearly don't.

 

[JaredJames]

If we assume 100% efficiency isn't the energy always equal with mgh regardless the lifting speed?

At it's most basic level the fast reps involved more repetitions than the slow ones so even if you want to look at it like this, it still shows more energy used.

 

[waynexk8]

These numbers are meaningless. You've been given your answer (at least what's possible) and yet you keep posting this repetitive non-sense and being told the same thing.

I think that is a little unfair, as lots in this forum use numbers that thay say are not the real numbers, but they put them down to make, or try and make a point, and as we have come to a little sticking point, I see nothing wrong with adding these numbers in.

As are not the higher high force and the higher peak force higher in the faster reps ? i think we all agrere they are, thus to take this further, and as all the number do average out, I see nothing wrong with adding these numbers in ?

If you still do not like these number, could you please say if we all do agree that the higher high force and the higher peak force higher in the faster reps are higher ? If not why are they not higher, but I an sure we all agree they are higher.

What are you talking about energy "balancing out"? Energy is used in all cases. In the faster reps, more energy is used.

You need more energys for the higher peak forces and the higher high forces in the faster reps, or the acceleration phase, and when the faster reps are decelerating, the slower reps energys even that they are higher then, just do not balance out, as the faster reps in the end, after the same time frame, use more energy.

All this and the questions seem straight forward to me, look at this running example.

1,
Time ran 1 hour, Bodyweight 130 pounds Running, 10 mph (6 min mile) 944 {calories} 10 mile ran 944 {calories}
Work done 10 miles = 10mph = 10 miles in 1 hour = 944 {energy calories}


2,
Time ran 1 hour, Bodyweight 130 Running, 5 mph (12 min mile) 472 {calories}

Work done 10 miles = 5mph = 10 miles in 2 hours = 944 {energy calories}




The only time that the slower runner, or the slower repper can use the same amount of energy is when they cover the same distance as the faster runner, repper. D. this is why distance is important, see what I mean, as the only time you will use as much energy as me {and yes I know it will be a little less in the repping as of the deceleration phase for the faster reps} is when you cover the same distance

That's the end of it. There's no need to drag this thread out further.

So my final question, which I asked before is, why does the faster reps in the same time frame use more energy, is it because the faster rep used more force thus put more tension on the muscles ?




Here is my answer, and I think there can only be one answer.

Why would the muscle moving the weight faster up and down, be using more energy in the same time frame as the muscle moving the up and down slower ? Because the higher high forces, and the higher peak forces, or the accelerations, are greater in the faster reps, and the slow reps forces when the faster reps are deceleration, does and cannot make up for this.

So the forces are higher in the faster reps, thus they put more total or overall tension on the muscles in the same time frame as the slower reps.

If anyone disagrees, please state why you think the muscles use more energy in the faster reps, as this is a physics site, there must be an answer.

Wayne

 

[waynexk8]

At it's most basic level the fast reps involved more repetitions than the slow ones so even if you want to look at it like this, it still shows more energy used.

Thats what I always said.

Wayne

 

[JaredJames]

Time ran 1 hour, Bodyweight 130 pounds Running, 10 mph (6 min mile) 944 {calories} 10 mile ran 944 {calories}
Work done 10 miles = 10mph = 10 miles in 1 hour = 944 {energy calories}

Time ran 1 hour, Bodyweight 130 Running, 5 mph (12 min mile) 472 {calories}

Work done 10 miles = 5mph = 10 miles in 2 hours = 944 {energy calories}

What work are you referring to here? It certainly isn't mechanical work as you haven't mentioned any forces. Confusion of terms again.

 

[waynexk8]

The energy use cannot cancel out, it adds up.

Right get that.

The force most certainly can.

Force is a vector with magnitude and direction. 10N upwards and 10N downwards result in a net force of zero.

Because of this direction, the work gives you a net of zero.

It really doesn't matter what you think, the physics say net work is zero. You need to check the definitions and make sure you understand them. At the moment you clearly don't.

Hmm, maybe your right, but all I am saying is that I need a force to lift the weight, and a force to lower the weight, and we all agree with that.

So when you say the up force cancels the down force out, it does not register with me, as I “have” used a force to lift the weight, and I “have” use a force to lower the weight. The down force cannot physical cancel the up force out, as I “have” to use a force to lift the weight up, and a force to lower the weight, as its imposable otherwise.

As I have so much force, I use some to lift the weight, and some to lower the weight, let's say I had 100 force, I used 50 to lift the weight and 40 to lower the weight, when I use the 50 force to lift the weight, that force is and has been used and gone, as a certain amount of energy has been used, and I only have so much of it. I when use some more force to lower the weight, but than that's all used and gone because I have used more energy. Then I cannot lift the weight again, as I have no energy to fuel my force. I do not see where any force or energy are canceled out.

Bed time here, thanks your patience and time.

Wayne