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

[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.

 

[sophiecentaur]

Wayne. I think I have spotted your problem. It's an emotional one. You find it hard to accept that lifting weights is totally inefficient, albeit good fun and making muscles big. That's all there is to it.
If you really want your exercise to be useful then hook yourself up to a generator. ;-)

 

[waynexk8]

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.

Hmm, So as I am lifting the weight and using force and energy, and sort of putting energy as in movement into the weight that’s doing work ? If am right there, as I am using force to slow the weight down on the eccentric contraction, am I not also using force and thus putting energy in the weight when lowering it, as I am reducing its natural speed, its fall to the floor ?

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

Right get that, if its not moving I could be producing force but not doing work.

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.

Not sure about that and if its contradictory, let me explain, as it can be both ways.

As I am lowering the weight under control, thus using force and energy when the weight is moving thought a distance, and also the weight is doing work on me, as its pushing down on me with a force its weight due to gravity, and the weight is again moving thought a distance, so the weight is pushing down and I am pushing up whilst the weight is moving.

Wayne

 

[JaredJames]

Hmm, So as I am lifting the weight and using force and energy, and sort of putting energy as in movement into the weight that’s doing work ? If am right there, as I am using force to slow the weight down on the eccentric contraction, am I not also using force and thus putting energy in the weight when lowering it, as I am reducing its natural speed, its fall to the floor ?

Read my example.

The fact the weight is descending indicates the force downwards is greater and as such the work done is by the weight on your arm.

Not sure about that and if its contradictory, let me explain, as it can be both ways.

As I am lowering the weight under control, thus using force and energy when the weight is moving thought a distance, and also the weight is doing work on me, as its pushing down on me with a force its weight due to gravity, and the weight is again moving thought a distance, so the weight is pushing down and I am pushing up whilst the weight is moving.

Wayne

You certainly are contradicting yourself.

Once again, you are ignoring the fact force has a direction and that 10N upwards cancels out 10N downwards.

Put extremely simply:

Let's say the weight is 10N. That is 10N of force acting continuously downwards.

So, to raise the weight 1m, you apply 15N of force upwards. This gives you 10N from the weight downwards plus 15N of force upwards. 10N down + 15N up = 5N up. So, work done = 5N upwards x 1m = 5J of work done up.

Now, to lower the weight 1m, you apply 5N upwards to slow its descent. This gives you 10N from the weight downwards plus 5N of force (from you) upwards. 10N down + 5N up = 5N down. So work done = 5N downwards x 1m = 5J of work done down.

So you now have total work done = 5J up + 5J down = 0J.

Your total work done is zero. However, you have expended 10J of energy to apply the required forces.

Do you understand now?

LEARN THE MEANINGS OF THE WORDS AND USE THEM PROPERLY

 

[waynexk8]

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.

I do not think the meaning of work, that is if you move a weight up and then down = zero, is helping this debate at all. As we all know here that force and energy have been used to lift the weight and to lower it under control, thus physical work has been done. Therefore, it seems no one can work out the force used, but whatever the force the average force is the same.

However if we take some numbers, and as numbers are pretty close, as of the study and tests I remember looking at, we find that what I said before. And as we all know that the energy is far far far higher in the faster reps, I ask again, why do you think this is ?
Fast reps,
140, 100, 100, 40, 20.

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

I say its because the overall or total force thus tensions on the muscles are higher. As with the faster rep I have more power, coved more distance, used more energy, and used a higher force more times. And it’s these higher high forces, and the higher peak force, and as there are 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.




Or if anyone does not agree with that, then why do they think I use more energy in the faster reps ?

Wayne

 

[waynexk8]

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.

Ok so your saying even thou I have used force and energy to lower the weight under control, because it’s in the negative direction I can/have not done work on it, if that is right I get what you mean about me not doing work.

Wayne

 

[JaredJames]

why do they think I use more energy in the faster reps ?

In the fast reps, the work done may still equal zero, but you've still expended the energy to move the weight X times more than in the slow ones (as per above you can add up the energy use to give the total expended).

It's as simple as that for why you use more energy.

 

[Dale]

Hmm, So as I am lifting the weight and using force and energy, and sort of putting energy as in movement into the weight that’s doing work ? If am right there, as I am using force to slow the weight down on the eccentric contraction, am I not also using force and thus putting energy in the weight when lowering it, as I am reducing its natural speed, its fall to the floor ?

No, reducing its "natural speed" does not imply that you are doing work. In fact, most useful means of extracting work from gravitational potential energy involve exactly that. Think of hydroelectric power, a turbine or even a watermill slows the decent of the water, and yet the water does work on the machine, not the other way around.

Going back to the definition of work, you can see this for yourself:
a) determine the direction of the force (upwards)
b) determine the direction of the displacement (downwards)
c) determine the angle between them (180º)
d) determine the cosine of that angle (-1)
e) if it is positive then it is doing work
f) if it is negative then work is being done on it

Not sure about that and if its contradictory, let me explain, as it can be both ways.

No, it cannot. Stop explaining and start learning.

Again, go back to the definition. If you are exerting a force on the weight then by Newton's 3rd law the weight is exerting an equal and opposite force on you. That means that if one force is parallel to the displacement then by Newton's 3rd law the other must be anti-parallel. Therefore, one force will do positive work and the other force will do negative work (i.e. work will be done on it). They cannot possibly both do positive work.

Please do not continue to reassert the same mistakes. If you do not understand then ask questions. Do not make erroneous assertions, particularly after having been repeatedly corrected.

 

[waynexk8]

Wayne. I think I have spotted your problem. It's an emotional one. You find it hard to accept that lifting weights is totally inefficient, albeit good fun and making muscles big. That's all there is to it.
If you really want your exercise to be useful then hook yourself up to a generator. ;-)

Yes lifting weights is totally inefficient.

Please don’t be like that, I am not the loud Bodybuilding type, I am a very quiet person and don’t watch Bodybuilding at all, I just like strength training for many reasons. And I know it’s hard for you explaining things to me, but I am learning and your helping me.

Wayne

 

[waynexk8]

In the fast reps, the work done may still equal zero, but you've still expended the energy to move the weight X times more than in the slow ones (as per above you can add up the energy use to give the total expended).

It's as simple as that for why you use more energy.

Yes that's what I sort of said to D. in the fiorst place, distance is important, he said it was not.

Yes but you still did not sort of answer the question, as it take “MORE” of not just energy to move the weight more times and more distance in the same time frame with the faster reps, as the energy/calories must power something, and that a force to move the muscles, so is it not like I just said above ? Here it is again; just not the percentages are higher with the faster reps.

However if we take some numbers, and as numbers are pretty close, as of the study and tests I remember looking at, we find that what I said before. And as we all know that the energy is far far far higher in the faster reps, I ask again, why do you think this is ?

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

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

I say its because the overall or total force thus tensions on the muscles are higher. As with the faster rep I have more power, coved more distance, used more energy, and used a higher force more times. And it’s these higher high forces, and the higher peak force, and as there are 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.


Wayne

 

[waynexk8]

No, reducing its "natural speed" does not imply that you are doing work. In fact, most useful means of extracting work from gravitational potential energy involve exactly that. Think of hydroelectric power, a turbine or even a watermill slows the decent of the water, and yet the water does work on the machine, not the other way around.

Going back to the definition of work, you can see this for yourself:
a) determine the direction of the force (upwards)
b) determine the direction of the displacement (downwards)
c) determine the angle between them (180º)
d) determine the cosine of that angle (-1)
e) if it is positive then it is doing work
f) if it is negative then work is being done on it

No, it cannot. Stop explaining and start learning.

I will have to come back to that one, as its very late here.

Bye for now all.

Wayne