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

[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

 

[Dale]

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

Not in general, no.

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

So far, so good.

The difference is, the former is applied for 1 second, the latter for 0.25 seconds.

No, the former is applied for 1.41 s and the latter for .707 s.

The resultant velocity is 1m/s in both cases with a final KE of 0.5J in both cases.

No, the former is 1.41 m/s and the latter is 2.83 m/s with a final KE of 1 J and 4 J respectively.

So, quadruple the force over the same distance results in half the time, twice the final velocity, and 4 times the kinetic energy. However, note that this example is rather different from your previous example:

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

However, you can do the same kind of analysis to confirm what I was saying.

 

[Dale]

Ok, but please could you tell me what I should say

I would say "energy expended" when you are talking about the fact that you burn more calories doing fast reps. When you say "work done" it is always 0 over one rep because you start and stop at the same point.

For a 100% efficient machine "work done" = "energy expended" but humans are notoriously inefficient. All of the energy expended in lifting weights is due to inefficiencies. If our arms were perfectly elastic springs with no losses you could lift a weight up and down all day without expending energy.

 

[Dale]

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

Yes, the forces are higher for a faster rep than for a slower rep. This at least is purely mechanical and does not require complicated biological models.

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 ?

I don't know about that. This is now strongly dependent on the details of the ineffieciencies of a human muscle. I would need to see an accurate model to be convinced.

 

[JaredJames]

No, the former is applied for 1.41 s and the latter for .707 s.

It appears the SUVAT equations yield different answers.

a = (v-u)/t = 1 = (1-0)/t which leads to t = 1/1 = 1s (what I've been working with)

v = ut+0.5at² = 1 = 0t+0.5t² which leads to t = sqrt(2) = 1.41s

Where v = 1, u = 0, a = 1 or 4, t = ? and s = 1.

A bit naughty. How do you decide which is correct? They both rely on the same numbers.

EDIT: With the 1.41s answer, you input your final velocity (v) as 1m/s and your acceleration as 1m/s², yet you have been accelerating for 1.41 seconds. Now I've always worked to final speed = acceleration x time, which in this case doesn't agree with the numbers put into the second equation because the final velocity entered doesn't match the time given.

You get the value 1.41s as the time for acceleration and as such final velocity is 1.41m/s yet the value used to attain this is a final velocity of 1m/s.

So the question is, how do you know which equation to use?

 

[Dale]

It appears the SUVAT equations yield different answers.

What is a SUVAT equation?

I just used Newton's second law, f=ma. The force is constant and the initial position and velocity is 0 so:
f=m \frac{d^2 x}{dt^2}

x = \frac{f}{2m} t^2

Then just substitute in the given values for x, f, and m and solve for t. If you got something different then you must have either used the wrong equation or made a math error.

EDIT: With the 1.41s answer, you input your final velocity (v) as 1m/s and your acceleration as 1m/s², yet you have been accelerating for 1.41 seconds.

No, your final velocity is 1.41 m/s as I stated above. If you accelerate from rest at a rate of 1 m/s² for a distance of 1 m it requires 1.41 s and at the end of the 1.41 s you are traveling 1.41 m/s.

EDIT: I googled "suvat". The SUVAT equations are fine. The problem is that you incorrectly assumed that v = 1 m/s (the final velocity) was given when it is in fact an unknown. The knowns are s = 1 m (the displacement), u = 0 m/s (the initial velocity), and a = 1 m/s² (the acceleration). Then v (the final velocity) and t (the time) are unknowns. Look at the SUVAT equations and calculate s given a = 1 m/s², u = 0 m/s, and your proposed t = 1 s. You will see that it is less than 1 m.

 

[douglis]

I would say "energy expended" when you are talking about the fact that you burn more calories doing fast reps.

Hi DaleSpam,
here's the part that I don't understand.
Let's compare two cases.At first I lift a weight in 1 sec and then I hold a weight for 1 sec.

Two things we know for sure.That at both cases I used the same average force(equal with the weight) for 1 second and that at the second case I used the force with 0% efficiency(since I didn't produce any work).

How is it possible to know which way I spent more energy?

 

[Dale]

You have to have an accurate model of muscle activity and metabolism. This is not a simple physics problem, it is a biomechanics problem.

 

[douglis]

You have to have an accurate model of muscle activity and metabolism. This is not a simple physics problem, it is a biomechanics problem.

That's exactly my point.From physics POV you can't possibly know if it takes more energy to lift 100 bags in a minute or you just hold a bag for the same time.

 

[waynexk8]

What work are you referring to here?

Work in the miles that have been ran. As in the mechanical work, the amount of the energy transferred by a force acting through the distance, in this case the miles that have been ran.

It certainly isn't mechanical work as you haven't mentioned any forces. Confusion of terms again.

The forces that are being used to run.

Wayne

 

[waynexk8]

I would say "energy expended" when you are talking about the fact that you burn more calories doing fast reps. When you say "work done" it is always 0 over one rep because you start and stop at the same point.

Still confused on this work, as I thought mechanical work was the amount of the energy transferred by a force acting through the distance, and if I use more force, the weight travels faster and goes further in the same time frame that if I used less force. Thus I am using more or a higher force, being fuelled by my energy thought-out a distance. And as I said before, even if I do lift the weight up and then down, I still have to use a force/energy to lift it up and lower it down, and it traveled a distance in both directions.

For a 100% efficient machine "work done" = "energy expended" but humans are notoriously inefficient.

Right get that.

All of the energy expended in lifting weights is due to inefficiencies. If our arms were perfectly elastic springs with no losses you could lift a weight up and down all day without expending energy.

Would you not still need energy to lift the sprig up each time.

Wayne

 

[sophiecentaur]

Still confused on this work, as I thought mechanical work was the amount of the energy transferred by a force acting through the distance, and if I use more force, the weight travels faster and goes further in the same time frame that if I used less force. Thus I am using more or a higher force, being fuelled by my energy thought-out a distance. And as I said before, even if I do lift the weight up and then down, I still have to use a force/energy to lift it up and lower it down, and it traveled a distance in both directions.
Right get that.
Would you not still need energy to lift the sprig up each time.

Wayne

You are still determined to apply your own version of Physics to this problem. What you forget in your attempt to describe the situation is that the same force is involved in lowering the mass - completely cancelling the amount of work done on it. That is how work is defined. If you have your own definitions then that is up to you but you have to part company with Science. You will never get a satisfactory answer on this forum. I suggest you stick to conversations with people who would rather be vague about the way they define things and who believe in some kind of magic rather than Science.

No one has disagreed with you about the fact that you get more knackered when you do more of your "reps" and that more energy has been expended. But NO useful work has been done. (Rather like the most part of this thread, actually) If you still think you are correct then bully for you - but you aren't - not in the context of Science.

 

[waynexk8]

Yes, the forces are higher for a faster rep than for a slower rep. This at least is purely mechanical and does not require complicated biological models.

This is what I has said to D. from the very start of the debate. That the higher high force, and the higher peak forces in the faster reps are higher, he agrees with this, but then he goes back to the average force. However I then have said that the higher forces of the faster rep, have a higher total, or overall force, meaning the higher high forces, and higher peak forces are higher, and the mediam forces of the slower reps, can not make up for the higher force of the faster reps, when the faster reps are in their phase of lower forces as of the deceleration which will be longer then the slower reps.

Now we I think all agree that the energy’s are far far far higher for the faster reps, because the persons muscles moving the weight faster has used more higher high forces, and higher peak forces, multiple times as of the 6 reps, thus more acceleration, speed, velocity, and also work because they have moved the weight 12m to the slow reps 2m.

Thus, the faster reps will have put more tension on the muscles, because they have used their available force/strength and energy up faster. As the faster and more force you use, puts more tension on the muscles.

I don't know about that. This is now strongly dependent on the details of the ineffieciencies of a human muscle. I would need to see an accurate model to be convinced.

Hmm, maybe I am not explaining that good. We all agree that if you move the weight twice as fast you use roughly twice as much energy, this to me means that the higher forces in the first 3 segments of the faster rep, use more energy for those 3 segments, and also some energy for the last 2 segments, but when the faster reps are using less energy for the last 2 segments of the fast reps, the slow reps then still not make up for the more energy used in the first 3 segments of the faster reps. Sorry that’s a bit confusing, and to write, but it does make sense.

Or an easier way,

Fast reps force,
140, 100, 100, 40, 20.

Slow reps force,
80, 80, 80, 80, 80.

Now this time is roughly what the both reps use in energy. It’s like G-force and wind resistance, if you move say 1ms you use 1 energy, but if you move at 2ms you use 4 energy, not 2 energy like some think.

Fast reps energy,
280E, 200E, 200E, 10E, 5E = 695E

Slow reps energy,
80, 80, 80, 80, 80 = 400E.

And that’s because higher tension on the muscles take/use more energy faster.

Wayne

 

[waynexk8]

You are still determined to apply your own version of Physics to this problem. What you forget in your attempt to describe the situation is that the same force is involved in lowering the mass - completely cancelling the amount of work done on it.

First, sorry, as I know I do confuse things with the wrong terms, and my way or saying and writing and the posts that are too long.

However, it will be a lower force needed to lower the weight. One because gravity is downward, and the muscles are roughly 40% stronger at lowering, meaning if you could lift 100 pounds max, you could lower 140 pounds under control.

That is how work is defined.

Ok, if I lift something up and then down no work has been done, however physical work, force and energy has been used. And the faster you lift up and down the more of these you have to use.

If you have your own definitions then that is up to you but you have to part company with Science.

Yes I agree there, we do need to have basic rules.

You will never get a satisfactory answer on this forum. I suggest you stick to conversations with people who would rather be vague about the way they define things and who believe in some kind of magic rather than Science.

No please, I want to stick to the scainces, I am not a man of mumbo jumbo, hate all that, even at times Ifind the physics definitions hard, I want to stick with them, but as I said before, its not that easy to me, we as most know physics is not easy at first for anyone.

But I just thought as power is so easy to work out, I thought working the tension put on the muscle would be as well.

But as we all agree that the muscles use twice the energy doing the same thing twice as fast, I think this basically proves that the faster reps must be putting more tension on the muscles faster in the same time as the slower reps. But D. still does not seem to see this.

Many years before this debate, say 8, I too thought the slower reps were better, but then after debating I thought otherwise, and the simple thing I thought of, was a weight on my palm, and the faster I moved/pushed that weight up, the move it pressed into my palm, thus I was using more force and in turn putting more tension on the muscles. Now I am 40 pounds heavier.

No one has disagreed with you about the fact that you get more knackered when you do more of your "reps" and that more energy has been expended. But NO useful work has been done. (Rather like the most part of this thread, actually) If you still think you are correct then bully for you - but you aren't - not in the context of Science.

Maybe the work thing is where I am going wrong as you say, as when I think of work, I think of work, I think of it as work being the amount of the energy transferred by a force acting through a distance, thus do we all not agree that I have moved a weight thought a distance using a force fuelled by the energy, thus I have done physical work, please do we all agree on that ?

Please yet again, thank you for your time and help, but I honestly do not want to go away from science for my answers, that’s not me, I am a man of science.

Wayne

 

[Dale]

Still confused on this work, as I thought mechanical work was the amount of the energy transferred by a force acting through the distance

No energy is transferred over one rep. The kinetic energy of the weight is the same and the potential energy is the same, so no energy is transfered.

And as I said before, even if I do lift the weight up and then down, I still have to use a force/energy to lift it up and lower it down, and it traveled a distance in both directions.

As you are lowering it down you are not doing work on the weight, it is doing work on you. Your force is up, the weight's displacement is down so the work done is negative. Over one rep the net work is 0. There is no point arguing further, you are simply wrong on this point.

Would you not still need energy to lift the sprig up each time.

No.

EDIT: I should be a little more clear on this last "No." No net energy is required over each rep. It does require energy to lift the weight each time, but that energy is exactly equal to the energy recovered by lowering the weight each time. The net work over one rep is 0.

 

[Dale]

Hmm, maybe I am not explaining that good.

I think you need to worry a little less about explaining and a little more about understanding. We get your argument, your argument is wrong, do you understand why it is wrong? Your position is very clear and does not need to be re-explained. Your position is also wrong and needs to be un-learned. Why did you come to this forum if not to learn?

We all agree that if you move the weight twice as fast you use roughly twice as much energy

no, we don't all agree. I agree that you use more, but I don't see any justification for the claim that it is twice as much.

 

[douglis]

Wayne...I don't understand why you continue this discussion in a physics forum.If the energy that is expended in fast reps is greater it's totally depended on biology.Physics have nothing to do with that.

This is what I has said to D. from the very start of the debate. That the higher high force, and the higher peak forces in the faster reps are higher, he agrees with this, but then he goes back to the average force. However I then have said that the higher forces of the faster rep, have a higher total, or overall force, meaning the higher high forces, and higher peak forces are higher, and the mediam forces of the slower reps, can not make up for the higher force of the faster reps, when the faster reps are in their phase of lower forces as of the deceleration which will be longer then the slower reps.

Same average force means exactly that.The peak high fluctuations of force are exactly balanced by the peak lows.

 

[Dale]

Wayne, before proceeding, can you agree to the following:

1) the average force on the weight is equal and opposite to the weight
1a) the average force is the same for fast and slow reps

2) the work done on the weight over one rep is 0
2a) the work done is the same for fast and slow reps

3) the energy expended depends on the inefficiency of the musculoskeletal system
3a) the energy expended is different for fast and slow reps
3b) the energy expended has no easy relation to the average force or the work done

 

[waynexk8]

No energy is transferred over one rep. The kinetic energy of the weight is the same and the potential energy is the same, so no energy is transfered.

Maybe this is my fault again for not either using the right terms. However what I was meaning was, if you lift a weight up, you have to use force from your muscles, and in turn, they have to use energy, as in calories. Therefore, energy in the form of calories was transferred from the stores in the muscles, to and from the muscles to use their force to move the weight up. I was not referring to kinetic energy at all, as the kinetic energy has basically no time to be used, as I do not let the weight go, and I actually do the opposite, I have to slow down the weight very fast in Milly seconds for the deceleration and transition from the concentric up portion of the lift to the eccentric, the down portion of the lift. So maybe we have been on cross threads. All along this thread when I say energy I mean in energy as in calories as for the muscles to be able to perform the movements.

As you are lowering it down you are not doing work on the weight, it is doing work on you. Your force is up, the weight's displacement is down so the work done is negative. Over one rep the net work is 0. There is no point arguing further, you are simply wrong on this point.

Ok maybe I am wrong; however, I must be doing some work in lowering the weight down, as I “have” to use “a” “force” to lower the weight down, thus I “have” to have used some energy as in calories to move it a distance ?

No.

Sorry, wrote the below before IO read your edit.

I will bet you it’s a big yes, as you set your video, put a spring on the table, and show me how that spring is going to move and stretch or open up on its own ? You “must” have either a machine that uses force that in turn needs some form of energy, as in diesel, or a human muscle that uses force that in turn needs some form of energy, as in calories. How can any spring stretch and open and move without a force that has to have energy to do this ?

EDIT: I should be a little more clear on this last "No." No net energy is required over each rep. It does require energy to lift the weight each time, but that energy is exactly equal to the energy recovered by lowering the weight each time. The net work over one rep is 0.

Not sure what you mean be recovered ? s if in above I said you either need a machine or muscle to move the weight/spring up in the first place, when you let the weight/spring {not sure why we are talking of this, as I never let the weight just drop, it is a controlled lowering, that needs force and energy} drop, the machine nor the muscles recover any energy back at all, no energy goes back into the machine or muscles ?

Wayne

 

[waynexk8]

PLEASE I will get back to the other posts.

Wayne...I don't understand why you continue this discussion in a physics forum.If the energy that is expended in fast reps is greater it's totally depended on biology.Physics have nothing to do with that.

No time to answer that, but its a good point.

Same average force means exactly that.The peak high fluctuations of force are exactly balanced by the peak lows.

D. you agreed that the energy was more in the faster reps, and you said something like when the faster reps are hitting their lower force, that this does not balance out the higher energy’s they used when they were moving at their higher forces, as the higher forces were not linear. This is what happens with the forces, HOW can it be any other way ? if the energy’s are not linear, then the force cannot be, that’s when in the whole the faster reps use more energy, otherwise why do they use more energy if they are not putting an overall more tension on the muscles ?

{The faster reps have 140, 100, 100, and the average 80 of the slow reps if NOT as higher as 140, 100 or a 100 again, HOW can 80 be as high as 140 or 100, just tell me that ? The higher high force, and the higher peak forces of the faster reps, are higher than the slower reps, and when the faster reps are on their low forces, the 80 average forces of the slow reps will and cannot make up for the higher high force, and the higher peak forces of the faster reps, which was 140, 100, 100. Just tell me how do you think 80 can make up to 140, 100, 100 ? 140 is 60 or 75% higher, and 100 is 20 or 25% higher. I do not get why you say that my total or overall force is not higher when I just showed it is 140 is 60 or 75% higher, and 100 is 20 or 25% higher ?

Impulse is when a force is applied to a rigid body it changes the momentum of that body. A small force applied for a long time can produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important.

BUT these reps are done some the same time frame, thus the only way your low force could make up, is that if your reps were or went on for a longer time frame, but they do not.

Wayne

 

[Dale]

Maybe this is my fault again for not either using the right terms. However what I was meaning was, if you lift a weight up, you have to use force from your muscles, and in turn, they have to use energy, as in calories.

The term is "energy expended" as I have pointed out a half-dozen times already. It is not the same as "energy transfered" or "work done" as has also been explained several times already. Please go back and carefully read what has already been written.

I must be doing some work in lowering the weight down, as I “have” to use “a” “force” to lower the weight down

No you don't. The weight will go down all on its own without you using a force at all. You are simply and clearly wrong on this point, stop repeating it. You do not do work lowering the weight, in fact the opposite occurs, the weight does work on you.

, thus I “have” to have used some energy as in calories to move it a distance ?

Yes, you have expended energy, you have not done work. See points 2 and 3 above.

Not sure what you mean be recovered ? s if in above I said you either need a machine or muscle to move the weight/spring up in the first place, when you let the weight/spring {not sure why we are talking of this, as I never let the weight just drop, it is a controlled lowering, that needs force and energy} drop, the machine nor the muscles recover any energy back at all, no energy goes back into the machine or muscles ?

I mean that the spring does work on the weight on the way up and the weight does work on the spring on the way down. It is, as you say, a controlled lowering that needs force, but it does not need energy. In fact, energy is obtained from it and does go back into the machine. Your statements to the contrary are wrong.

Again Wayne, please look at these three statements and make sure that you understand and agree with them:
https://www.physicsforums.com/showpost.php?p=3197286&postcount=192

 

[waynexk8]

The term is "energy expended" as I have pointed out a half-dozen times already. It is not the same as "energy transfered" or "work done" as has also been explained several times already. Please go back and carefully read what has already been written.

Ok thx, yes get you now energy expended is different to energy transpired. Too which I was on about energy expended.

No you don't. The weight will go down all on its own without you using a force at all. You are simply and clearly wrong on this point, stop repeating it.

No, I am not wrong here. As I am lowering the weight under control, meaning I am using the force/strength of my muscles to lower the weight. If I was not using a force, the weight would fall far far far faster, like if I dropped the weight from 1m it would fall or drop faster that 1m every .5 of a second to which I am lowering it. If I am not using a force controlling the downward speed, what is controlling the speed ?

You do not do work lowering the weight, in fact the opposite occurs, the weight does work on you.

Right I can see that the weight does work on me, however surely I am also doing work on the weight, as I am controlling its downward movement, by the force/strength of my muscles to lower the weight/

Yes, you have expended energy, you have not done work. See points 2 and 3 above.

Ok I have used energy. However could we clear the meaning of work up please, as I thought work was the amount of energy transferred by the force acting through a distance. So I say again, if I move the weight up I use force and energy to do this and its over a distance, so is not that work, or physical work ??/ And the same for lowering the weight, as I have used more force and energy to do this surely ?

I mean that the spring does work on the weight on the way up and the weight does work on the spring on the way down.

Ah, get you now, I did not realize that you were saying the spring had a weight on it.

It is, as you say, a controlled lowering that needs force, but it does not need energy.

But the muscle do need to expend or use energy to apply this force ? yes ?

In fact, energy is obtained from it and does go back into the machine. Your statements to the contrary are wrong.

Yes kinetic energy goes back into the machine, but not the energy I meant, which was the fuel spent by the machine, as that’s imposable.

Again Wayne, please look at these three statements and make sure that you understand and agree with them:
https://www.physicsforums.com/showpost.php?p=3197286&postcount=192

Right have a go now.

Wayne

 

[JaredJames]

No, I am not wrong here.

Ok I have used energy. However could we clear the meaning of work up please, as I thought work was the amount of energy transferred by the force acting through a distance. So I say again, if I move the weight up I use force and energy to do this and its over a distance, so is not that work, or physical work ??/ And the same for lowering the weight, as I have used more force and energy to do this surely ?

Dammit wayne stop this non-sense. Either learn what these words mean or just drop it.

What part of force has a direction are you not understanding?

As you clearly aren't prepared to learn these things, you're just going to have to accept the following:

If the weight gives a downwards force of 10N and you provide an upwards force of 5N (to stop it freefalling), the net force is 5N downwards and that is the force you use to calculate the work done. The work done is in the downward direction - or more correctly, it is done by the weight on you. You have expended energy to generate the counteracting force, but you haven't done any work. Period. End of story. Finito. Drop it.

 

[waynexk8]

Wayne, before proceeding, can you agree to the following:

1)the average force on the weight is equal and opposite to the weight

Yesish.

1a) the average force is the same for fast and slow reps

Yes, agree force is the same.

However do not think this means much in this debate, as I could do 6 reps to the slow reps 1 rep, or I could do 10 reps to the slow reps 1 rep, and if I did the 6 reps it would have 6 more highs than the slow rep, and if I did 10 reps it would have 10 more highs than the slow rep. Thus with the faster rep I have more power, coved more distance, used more energy, and used a higher force more times. And it’s this higher high force and the high peak force that D. does not seem to understand, as the more of these in the same time frame “WILL” put more tension on the muscles, that’s why the muscles use more energy, because they are putting out more higher high force and the high peak force, and the slow reps medium forces cannot make up for this when the fast reps forces are using low forces on the decelerating for the transition.

It would be the same if there fast reps were 3 reps done at 3/3 = 18 seconds, and the slow reps were 1 rep done at 9/9 = 18 seconds. The higher high forces and the higher peck forces would have to be higher in the faster 3/3 reps, but just not as much as in a .5/.5 set of reps.

2) the work done on the weight over one rep is 0

Yes.

However, more the physical work that uses force and energy is more my concern.

2a) the work done is the same for fast and slow reps

Not if I do more reps in the same time frame as the slower rep.

3) the energy expended depends on the inefficiency of the musculoskeletal system

Well yes. But all animals can lift things using energy and force, I would call that efficiency.

3a) the energy expended is different for fast and slow reps

Got to say a big yes for that.

efficiency3b) the energy expended has no easy relation to the average force or the work done

No not to the average force. We need to look and work out the %s of higher high forces and high peak forces in the faster rep, in their relation to the lower force in the faster rep and those lower forces in the faster rep not letting the slower reps balance out the energy usage.

Again thanks for your help and time, this is very interesting.

Wayne

 

[Dale]

No, I am not wrong here. As I am lowering the weight under control, meaning I am using the force/strength of my muscles to lower the weight.

Sure, but work is different from force. You are exerting force, that does not imply that you are doing work.

Right I can see that the weight does work on me, however surely I am also doing work on the weight

You can't have it both ways. If the work does work on you then you do not do work on the weight. Your statement here is a self-contradiction.

Ok I have used energy. However could we clear the meaning of work up please, as I thought work was the amount of energy transferred by the force acting through a distance.

Technically the meaning of work is:
W=\int \mathbf{F} \cdot \mathbf{dx}
if the force is constant then that simplifies to
W=\mathbf{F} \cdot \mathbf{d}= F d \, cos(\theta)

The cosine term is very important. When the angle between the force and the displacement is 0º (as in when you are lifting the weight) then the cosine evaluates to 1 and the work is positive, but when the angle between the force and the displacement is 180º (as in when you are lowering the weight) then the cosine evaluates to -1 and the work is negative. Thus during concentric contractions the muscles do work on the weight and during eccentric contractions the weight does work on the muscles.

 

[Dale]

2) the work done on the weight over one rep is 0

Yes.

However, more the physical work that uses force and energy is more my concern.

2a) the work done is the same for fast and slow reps

Not if I do more reps in the same time frame as the slower rep.

OK, let's look at this. You agree that the work done on the weight over one rep is 0. So if I do one rep in 10 s then that is 0 J work done. On the other hand if I do 3 reps in 10 s then that is 0 J + 0 J + 0 J = 0 J work done. So the work done is 0 J in both cases and therefore since 0 J = 0 J the work done is the same in both cases.

Again, your statements are self-contradictory. If the work done over one rep is 0 then logically the work done is the same for fast and slow reps. You cannot say "yes" to 2) and "no" to 2a) without contradicting yourself.