## Impulse/force in pounds for the time frame

 Quote by DaleSpam Hi waynexk8, the first thing that you need to correct is that there is no such thing as total force. There is impulse which is defined as: $$\mathbf{I}=\int_{t_i}^{t_f} \mathbf{f}(t) \, dt$$ Impulse has units of momentum, not units of force.
Hmm, ok, seems tricky, maybe I am saying it wrong then, as I think of me using force or strength more like this. If I can only lift 100 pounds 1 time, I can only lift 80 pounds several times, let’s say 10 times up and down at 1/1, this takes me 20 seconds, here is how I think this, say like a car, I have 20 gallons of fuel {or as I think force/strength} only, then I run out of fuel/force, BUT if I lifted slower, I would not used up my fuel/force so fast, or so fast per unit of time. So I have to use more fuel/force to lift the weight more times in the same time frame.

 Quote by DaleSpam There is also average force which is defined as: $$\overline{\mathbf{f}}=\frac{\int_{t_i}^{t_f} \mathbf{f}(t) \, dt}{t_f-t_i}$$ Average force does have units of force.
Thing is, I don’t understand that, if you have time please, is there any way you could explain this or explain it in layman’s terms, otherwise it had for me, as its just numbers.

 Quote by DaleSpam Second, you are under the mistaken impression that the impulse is higher if you do a rep faster. This is incorrect. The impulse for one rep depends only on the weight and the amount of time, so one slow rep (10 s) might have the same impulse as 5 fast reps (2 s each).
Could you clear something up before I comment on that please ??? As if I lift a weight up and then down, and then just lower the weight and then lift it up, it’s going to take far more force to stop a weight that is being lowered say .5 of a second for 1m, and then immediately lift it back up, we call this the transition from negative to positive, where the peak forces and tensions are on the muscles, the MMMTs. {Momentary Maximum Muscles Tensions}

As if you lower the weight and then lift, on using 100 pounds, you might have 150 pounds of force and tension on the muscles for .1 or .2 of a second, when just lifting the weight and lowering it, you will have nowhere near that force and tension on the muscles.

So please which repetition has the same impulse as the slow one ???

 Quote by DaleSpam Similarly, you seem to be under the impression that the average force over a rep is higher if you go faster. This is also incorrect. The average force for one rep depends only on the weight. Notice that there is a close relationship between impulse and average force.
I explain before, that average means nothing in this debate. As if I do 1 repetition in 1 second, and a 100 repetitions in a 100 seconds, they say the average is the same, but we all know the force, work, energy, velocity and accelerations were far harder doing the 100, AND far harder on the muscles.

 Quote by DaleSpam Personally, I think that the question you should be asking is the following: "What different physical quantities are in fact higher in a fast rep than in a slow rep?"
Yes you could be right there, or to put it very blunt, which repetition cadence is harder and puts more tension on the muscles. As when using 80% you fail roughly 50% faster, I think we all know it’s the fast.

Is not power the mathematical product of force and velocity ??? So more power more force and velocity, more force and velocity more power. I said this before, and someone said that more power does not equal more force and velocity, but could not prove what they thought. So work is the of a force over a distance, lifting a weight up and down is an example of work. The force is equal to the weight of the object, and the distance is equal to the height lifted {W= Fxd}
So energy capacity for doing work, so mechanical work is when an object is standing still and we force it to move.

Let’s take these scenarios.

1,
I lift a 100 pounds up and down 20 times in 20 seconds, then lift 50 pounds up and down 20 times in 20 seconds.

2,
I lift 100 pounds up and down 10 times in 40 seconds.

Which is harder ??? 1, would need more work, force, energy, velocity and acceleration used.

So maximal muscle power production is the dominant factor in movements which aim to produce maximal velocity at the point of release, takeoff, or impact. Therefore the ability to perform a large amount of mechanical work in a short period of time, or the ability to produce high force output at fast movement velocities, is critical to lifting a heaver weight more times, so if the slow and the fast have to pick a weight to do any number of repetitions, the fast with be able to use the higher poundage.

Power can be defined as the force applied multiplied by the velocity of movement (Knuttgen
and Kraemer, 1987). As the work done is equal to the force times the distance moved (Garhammer,
1993) and velocity is the distance moved divided by the time taken, power can also be expressed as
work done per unit time (i.e., the rate of doing work) (Garhammer, 1993).

work = force x distance
velocity = distance/time
power = force x velocity
therefore:
power = force x distance/time = work/time.

And that’s what I have been saying all along ???

Wayne
 Mentor Wayne, for the 3rd time now, please answer this question: Do you understand that it is physically impossible to move a weight up and down using a constant force? I need to know if that makes sense to you before we can proceed.

 Quote by DaleSpam Wayne, for the 3rd time now, please answer this question: Do you understand that it is physically impossible to move a weight up and down using a constant force? I need to know if that makes sense to you before we can proceed.
That’s an odd question.

Yes, I know that. So if you are lifting a weight up 500mm in 3 seconds, and going to reverse the direction, as in weightlifting, lifting a weight up and down. First you have to accelerate, then “try” to move it with a constant force, and then decelerate the weight.

Now with the human body it’s far harder than with a machine. As take the bench press, there are several biomechanical advantages and disadvantages thought the ROM {Range Of Motion} of the exercise, there is point in the movement, where the leverage of certain muscles acting about the shoulder joint and chest, are at a mechanical disadvantage, and thus creating a disadvantage at about half way to three quarters the way up the lift, this point will decrease the force potential and thus deceleration to the bar.

Note on a 1RM, the acceleration was for roughly 65%, so on using 80% like we are; the acceleration would be at least ??? 80%

Also, if you work out in physics how far a barbell will move if you let go when pressing it, and your using 80% for 15 inch, is does NOT go what physics say, as of the biomechanical disadvantages and biomechanical advantages of the body. But it would move the 3 inch if a machine pushed with a constant force of 80% for that force and time.

Jeff Pinter wrote, and he did say I could use this.

OK, let's see if I can reproduce this. Let's assume we're doing a bench press with 200 pounds, and our 1RM is 250 pounds (80% 1RM). Furthermore, the ROM is 15 inches.

The first thing we must do is convert pounds into the English unit of mass - the "slug". That is - 200 pounds/32 ft/sec^2=6.25 slugs.

Now, from F(net)=ma, we have 250-200=6.25 x a, or a=8ft/sec^2. This is the acceleration of the bar during the concentric.

Now, let's assume that we will push with our maximal force (250 pounds) up to the 12" point. We next need to find the velocity of the bar at this point from the equation v^2=2ad. Plugging in "a" from above and "d"=1ft, v=4 ft/sec. Note that this is not the top of the lift, but 3" from the top.

Next we want to find the time it takes to get to the 12" point, from the equation d=1/2at^2. Plugging in our values of "a" and "d", t=.7 seconds (again not the top).

Now, we assume that we stop pushing the bar at the 12" point, and let gravity slow it down, so that it comes to rest at the top. From v^2=2ad, we plug in our value of "v", but here we use a=32 ft/sec^2 (acceleration of gravity). From this we find that d=3", so that the bar comes to a perfect halt right at the top of the ROM - the 15" point.

Finally, we want to find out how long it takes gravity to stop the bar, from d=1/2at^2. Plugging in d=3" and a=32 ft/sec^2, t is found to be about .1 seconds.

Therefore, the total time for the concentric in this case would be .7 seconds (for the acceleration phase, or "onloading") plus .1 seconds (for the decceleration or "offloading" phase), for a total of .8 seconds. This is a bit faster than 1/1, but the best I could do this late at night. If your ROM were a bit longer then 15", then the speed would be closer to 1/1.

So, we see that in this case the offloading relative to the ROM is 3" out of a total of 15", or 20%. The offloading relative to the time is .1 sec out of .8 sec, or about 12%. Note that this is worst case...that is the first rep. As the set progresses, the offloading will reduce with each rep, due to fatigue, and towards the end of the set will be negligible. Therefore, you could state that the "average" offloading of the entire set relative to the ROM would be about 10%.

Note that this also assumes we can push with maximal force all the way to the 12" point. If our strength curve is such that our force output diminishes towards the top, then the offloading will be less than given above.

Jeff

So as I said before, could we not keep this to a machine moving the weight please ???

Big thx for your time and help DaleSpam, and the rest, thank you.

Now its too late and I have no time to read or answer the others.

Wayne

Recognitions:
Gold Member
 Quote by waynexk8 So as I said before, could we not keep this to a machine moving the weight please ??? Wayne
Absolutely. But you still are not accepting that the energy that a machine needs to supply can be near ZERO, the average force will be ZERO and the net Impulse will also be ZERO.
You would be right in saying that the Maximum force for a fast repetition rate would be greater than the maximum force for a slow rep rate.
But how does this machine help you to analyse what you think your muscles are doing? Where's the essential connection in your mind between the two entirely separate things?
 Either it's a muscle or a machine...one thing will always be for sure. The muscle's/machine's impulse will always be equal with gravity's impulse regardless if the lifting is fast or slow for the same duration.The net impulse is always zero.

Mentor
 Quote by waynexk8 That’s an odd question. Yes, I know that. So if you are lifting a weight up 500mm in 3 seconds, and going to reverse the direction, as in weightlifting, lifting a weight up and down. First you have to accelerate, then “try” to move it with a constant force, and then decelerate the weight.
OK, I am glad you understand that. I apologize that you think it was an odd question, but from several of your previous comments it was not clear that you understood that point. I am glad that you do and it was just a miscommunication.

So, let's take the scenario of a machine or a person lifting a 50 kg weight up and down 500 mm in 3 s as smoothly as possible. I generated the attached plot to represent a typical position over time graph for such a lift. Look at it and see if it seems like a reasonable approximation to you.

The plot is generated in Mathematica using the following code:
Code:
y[t_] := -.25 Cos[2 \[Pi] t/3] + .25;
Plot[y[t], {t, 0, 3}, Frame -> True,
FrameLabel -> {Style["Time (s)", Larger],
Style["Height (m)", Larger]},
PlotLabel -> Style["Position of Weight During Lift", Larger]]
Attached Thumbnails

 Quote by sophiecentaur Absolutely.
Great. Said this as the muscle will start to complicate things with their biomechanical advantages and disadvantages thought-out the ROM.

 Quote by sophiecentaur But you still are not accepting that the energy that a machine needs to supply can be near ZERO,
You lost me there ??? If the machine is powered by the energy, diesel or electricity, to move 80% of its maximum force up and down, will take a certain amount of this energy, but as we all agree, that the faster/further an object moves in the same time frame the more energy is needed, there is no debate there, or is there ??? As something, makes for the more energy used, and I say it’s the higher forces of the accelerations that cannot be made up or balanced out by the more constant forces of the slower repetitions, when the fast is on the deceleration.

Run 1 mile = 3 minutes, walk 1 mile 9 minutes, both use the same energy. Run 3 minutes, walk 3 minutes the run uses more energy, this is what I have said all along.

 Quote by sophiecentaur the average force will be ZERO and the net Impulse will also be ZERO.
Not sure I understand that, as if I lift 80% of my 1RM, {hope we all know what 1RM means by now, if anyone does not, I have explained many times, it’s the 1 repetition maximum a person can lift} I will try to use 100% force thought the ROM/lift, the only time I will use at the beginning and end, but they do not count, as it’s the lifting we are on about, not the times after the lifting.

But thought we all agreed average means nothing in this debate, as if the average force is the same for 1 repetition and a 100 repetitions for the same speed, it means nothing.

 Quote by sophiecentaur You would be right in saying that the Maximum force for a fast repetition rate would be greater than the maximum force for a slow rep rate.
Right, sure we all agree on this, however, as I lift the same weight 6 times further in the same time frame, using 100% force to the slow using 80% force, how can the total overall muscle activity or force be the same, you might say again, but in physics you can’t measure total overall muscle force, well try this.

Set our machine to lift 80% up and down for 60 seconds set at a lifting force of 80% of the machines maximum moving force, and then set the machine to move 80% up and down for 60 seconds set at a lifting force of 100%

Fast, the machine lifted with a set force of 100% efficiency for 60 seconds.

Slow, the machine lifted with a set force of 80% efficiency for 60 seconds.

The fast lift with 20% more force efficiency for 60 seconds

 Quote by sophiecentaur But how does this machine help you to analyse what you think your muscles are doing? Where's the essential connection in your mind between the two entirely separate things?
Said this as the muscle will start to complicate things with their biomechanical advantages and disadvantages thought-out the ROM.

There has to be a reason, why the faster makes for the more energy used, I said why I think, no one else does, D, mentions, biological, that’s no answer at all, it’s says nothing.

Wayne

 Quote by sophiecentaur Absolutely. But you still are not accepting that the energy that a machine needs to supply can be near ZERO, the average force will be ZERO and the net Impulse will also be ZERO. You would be right in saying that the Maximum force for a fast repetition rate would be greater than the maximum force for a slow rep rate. But how does this machine help you to analyse what you think your muscles are doing? Where's the essential connection in your mind between the two entirely separate things?
 Quote by DaleSpam OK, I am glad you understand that. I apologize that you think it was an odd question, but from several of your previous comments it was not clear that you understood that point. I am glad that you do and it was just a miscommunication.
K great, thank you.

 Quote by sophiecentaur So, let's take the scenario of a machine or a person lifting a 50 kg weight up and down 500 mm in 3 s as smoothly as possible. I generated the attached plot to represent a typical position over time graph for such a lift. Look at it and see if it seems like a reasonable approximation to you. The plot is generated in Mathematica using the following code: Code: y[t_] := -.25 Cos[2 \[Pi] t/3] + .25; Plot[y[t], {t, 0, 3}, Frame -> True, FrameLabel -> {Style["Time (s)", Larger], Style["Height (m)", Larger]}, PlotLabel -> Style["Position of Weight During Lift", Larger]]
Yes that look quite ok, I think. We are lifting 80% are we not, or close too ???

Wayne

 Quote by sophiecentaur This demonstrates well how you have got this wrong from beginning to end. Your muscles are not a machine. I can design a machine that uses NO energy whilst doing 'reps' at any rate and for any length of time (except to overcome some friction - and we could reduce this to an arbitrarily small amount).

How do you work that out ??? What machine can lift say 80% up, and then lower it down using very little energy, and what energy is this ???

 Quote by sophiecentaur The way the forces vary as the weights accelerate will depend on the rate

Right, the faster you try a move/accelerate the weight, the more action reaction forces go back on your muscles as tension, move/accelerate very fast very high forces, = very high tensions on the muscles, move very slow, very low force, = very low tensions on the muscles, move very slow, very low force.

 Quote by sophiecentaur but the mean velocity is zero for each cycle, and so is the energy transfer.

Not sure what this has to do with the debate, as we need to know about the velocities themselves, that are produced by the person/machine creating these forces, as I just said, the faster you try a move/accelerate the weight, the more action reaction forces go back on your muscles as tension.

 Quote by sophiecentaur btw, when you describe your machine as having a certain "lifting force" you do not specify for how long or over what distance this force is applied - nor what happens on the way down; the model is incomplete.

Ok sorry there.

The machine has a lifting force of 100 pounds, it can lift 100 pounds up 1m in .5 of a second, and any speed slower, and can lower it 1m in .5 of a second, and any speed slower, for an unlimited time, for debates sake, the weight the machine will use will be 80% that’s 80 pounds.

The machine first lift the 80% up 1m in 3 seconds and lowers it 1m in 3 seconds, 1 time, = 2m in 6 seconds.

Then the machine lifts the 80% up 1m in .5 of a second, and lowers it 1m in .5 of a second, 6 times = 12m in 6 seconds.

On the way down it lowers it in control 1m in .5 of a second, I say in control, as if let to drop, it would get to the ground far faster. Also, as of the acceleration components, when the weight is being lowered, the faster it’s lowered the more force it puts on the machine/muscles, like in the faster it hits the ground the more impact force.

 Quote by sophiecentaur All this machine needs to consists of is a wheel, with the weights hung on the periphery. You could modify it, if you like, so that the weights are on a crank and push rod so that they are constrained to go just up and down. Over one cycle of operation, ZERO work has been done on the weights and Zero energy is expended by the energy source. All that is necessary is to start the machine up and bring it to the desired speed. This (kinetic) energy could all be reclaimed at the end of the exercise. Perhaps this will help you to see that there is absolutely no parallel between what your muscles are doing and a simple mechanical model. Anything you may 'think' that your muscles are doing is entirely subjective. However many times you repeat a story about reps and rates and what you reckon you can tell us about what your muscles feel like, there is no direct correspondence between your body and simple mechanics. I believe that is what you have been trying to show all along. __________________

Wow that sounds interesting, but its far too late to comment on that now, will get back to it tomorrow.

Wayne
 To all the other that have posted here, I will get back to them tommorrow, and thx for your time and help. Wayne

Recognitions:
Gold Member
 Quote by waynexk8 How do you work that out ??? What machine can lift say 80% up, and then lower it down using very little energy, and what energy is this ??? Wayne
Plus all the rest
That comment shows that yo don't get the mosgt basic part of all this thread and others.

The machine doesn't need to use use "very little energy" on the way down. IT GETS ALL THE ENERGY BACK! Unlike your muscles, which don't have Energy Recovery. So the two cannot be compared.
You insist that this problem can be solved your way and you have the nerve to hang onto the idea in the face of people who know much more basic Physics than you. The only hope you have is to do a Physics course at some level which may help you understand what you need to know in order to grasp how crazy your idea is.
If someone told you that swimming the Pacific is a no no, how long would you not believe them?
 sophiecentaur....I think that he first needs to grasp more simple things.For example...he doesn't seem to understand the meaning of average force. Otherwise he wouldn't say nonsense like "...the forces don't make up..." or "...average means nothing in this debate...". More important he would understand that same average force equates same impulse(what he calls "total/overall force") when is applied for the same duration.

Mentor
 Quote by waynexk8 Yes that look quite ok, I think. We are lifting 80% are we not, or close too ???
I don't know about percentages, but from this motion diagram we can easily use Newton's laws to calculate the force exerted on the 50 kg mass by the lifter. I have plotted the force over time in the attached image.

At each point in time, the height of this plot represents the force applied by the lifter at that time. That has units of N. On this plot, we can also see a shaded area. This shaded area represents the impulse. The shaded area has units of Ns = kg m/s which is the same units as momentum. The average force is the impulse divided by 3 s. It has units of N also.

Note, as we had discussed before, the force must change over the course of the lift, it is higher when the weight is accelerating upwards, and lower when accelerating downwards. Also, although the force gets lower when accelerating downwards, it is never 0 in this case.

Does all of this make sense, in particular the impulse and average force?

Code:
Plot[Evaluate[50 D[y[t], {t, 2}] + 50 9.8], {t, 0, 3},
Frame -> True,
FrameLabel -> {Style["Time (s)", Larger], Style["Force (N)", Larger]},
PlotLabel -> Style["Force from Lifter", Larger],
PlotRange -> {0, 560},
Filling -> 0]
Attached Thumbnails

 Recognitions: Gold Member Science Advisor I don't think that Wayne actually knows what question he is actually asking any more. The only way he can ask about anything is in the form of a two page story instead of a concise question. That shows he has not actually formulated a question but just wants to chat about Reps and all that stuff. He just feels that there must be 'some simple Physics' involved. (I'm right, aren't I, Wayne?)

 Quote by douglis DaleSpam.....it doesn't have to be for over the whole rep. Even if you examine separately the lifting and the lowering phase the average force is always the weight.In both phases the weight starts and ends at rest so the average acceleretion is always zero.
Look D. You say you will not let me apologise for repeating, but I have too, as I said over two years ago on a thread I actually called; average force means nothing here. So I say it again, if the average force is the same for 1 rep at 1/1 and 100 reps at 1/1 and 1 rep at 5/5 and 100 reps at 5/5, it means nothing in this debate.

As we are debating which rep with the same weight, in the same time frame puts the most tension on the muscles, as you fail at roughly 50% faster with this % and rep speeds on the fast, that’s MORE then obvious it’s the fast rep, or do you think you fail faster on the fast because it puts less tension on the muscle ??? If so please say why.

Also, you said there was only one way to sort this out, with EMG, I bought one, and it showed after three experts on EMG said RMS was about the best way to find this out, that there is more overall/total activity on the fast reps. But you still insist when a real World practical experiment has proved you wrong, but you can’t state why.

You keep going on about average, when as I said it means nothings, and you also cant not tell me how you or why you work out this average, what’s average got to do with this.

Wayne

 Quote by douglis No Wayne....we will not forgive you for repeating yourself.You ask again the same nonsense ignoring all the answers. For some strange reason you're unable to understand that it's impossible to use the 100% of your force for the whole set.Regardless
No D. I TOTALLY understand this, I know the force/velocity curve, and I know I can’t use 100% force on 80% that’s why I always say, I try to use 100% of my force. But the point is, for the set amount of time, let’s call it 20 seconds, I am try to use as much force as I can, that’s quite close to 100%, you on the other had for the set time of 20 seconds, are not trying to use 100% force, you “are” only using 80% of your force, 20% less than me for the set time of 20 seconds.

Fast = as much force as he can exert for 20 seconds, let’s just call that 100.

Slow, = 80% of his maximum force for 20 seconds.

How can 80% for the same time frame be as high as 100% ???

Same in a car going uphill and down, I hit the gas 100% say this speed = 100mph, in one hour I have travelled 100 miles, you hit the gas at 80%, = 80mph, in one hour you have travelled 80 miles. [b/]You “only” travelled 80 miles as you DID NOT hit the gas with ENOUGH force, as I hit the gas with MORE force for the same time frame,[/b]

 Quote by douglis if you lift fast or slow you use the same average force for the same duration.
Explained why/why this average means, and explain why/how you have worked it out ??? As I don’t know why you bring it up, or what it means. Sorry there, but I just don’t understand.

 Quote by douglis The impulse(what you stupidly call "total/overall force") is always the same.
I/we have been told now that physics can’t seem to measure overall/total force, or there can’t be one ??? But the EMG can measure this, and it states categorically you are wrong, as it take into account the higher high force on the fast as the higher peak forces of the accelerations, and what I have said all along, is that your medium force cannot make up balance this out, if so, why would the EMG state higher for my fast ???

 Quote by douglis I don't care if you don't want to accept the truth or you just don't have the intelligence to understand it. It's a fact...accept it.
That’s quite odd ??? What truth have you ???

1,
EMG states fast,

2,
You use more energy in the fast,

3,
You do more work in the fast.

4,
So that’s more power in the fast,

5,
You move the weight 6 times further in the fast,

6,
You fail with these variables, 50% faster in the fast = there MUST be more tension on the muscles per unit of time to make them fail faster, = more tension = there must be more total/overall force if there is more tension as on failing faster.

7,
More speed, velocity and acceleration on the fast.

Is that enough truth, not sure what you have ???

Wayne

 Quote by douglis The same as always!You ignore everything that has been answered.
No I do not; you are the one doing that.

[QUOTE=douglis;3808610]Let's redefine the question.You asked which lifting speed has greater effect of force over time(impulse) which in Wayne's world is defined as "total/overall force".

Everyone explained to you that the impulse is identical regardless the lifting speed but you deny that fact based on some "practical proofs".Let's see them once again.[/quyote]

Before I answer this, you NEED to state why lift you are referring to, as the two different lifts, MUST have a different impulse.

Lift 1,
You lift 80% of the ground, up 1m and then down 1m all in 1 second, .5/.5

Lift 2,
You start at the top, lower the weight down 1m, and then lift it back up 1m all in 1 second, .5/.5

On lift 2, on the transition from negative to positive, there will be huge force on the muscles and coming from the muscles.

Also, “IS” the impulse the same on 1 rep and a 100 ??? As I don’t think you are measuring the impulse with respect to time, as if you think you are, you are then saying f=ma is wrong, as your saying you don’t need a higher force to move something further in the same time frame.

Slow,
Car has a weight of 400kg and an Acceleration of 30m/s force pushing the car 400 x 30 = 12000N of force for 1 second.

Fast,
Car has a weight of 400kg and an Acceleration of 100m/s force pushing the car 400 x 100 = 40000N of force for 1 second.

Fast = 2800N more for 1 second.

“NOT” sure why I am wrong there, is it the deceleration ???

 Quote by douglis The EMG reads the Root Mean Square of the values. Does the higher RMS somehow change the fact that the impulse is the same?NO!
The EMG show the total/overall muscle activity = force = tension on the muscles, its show what we are debating about, which has the most total/overall muscle activity, = the fast says the EMG.

 Quote by douglis Does the higher rate of energy expenditure somehow change the fact that the impulse is the same?NO!
That is NOT an answer, you need to say and explain why there is more energy used in the fast, I have told you, and you need to counter.

 Quote by douglis Does the greater distance somehow change the fact that the impulse is the same?NO!
Again, that’s no answer. Yes it does, I move the weight further in the same time frame, of course that needs more force, more acceleration needs more force, I mean its simple physics, its common sense. Newton's Law that force is equal to mass times acceleration, but we know in non-relativistic limit mass is invariant so if we apply more force it causes greater acceleration. The net force acting upon the object will be equal to the rate at which its momentum/movement change. When the object's velocity increases, so does its energy and hence it’s mass equivalent. It thus requires more force to accelerate it the same amount than it did at a lower velocity. Newton's Second Law.

 Quote by douglis Does the higher rate to fatigue somehow change the fact that the impulse is the same?NO!
All these prove what you say wrong; actually it would be nice if you gave me readable equations that state the impulse is the same per time, using the same weight over different distances and with diffract velocities and accelerations ???

 Quote by douglis So Wayne....quit these nonsense and see what everybody's telling you.Your above pseudoarguments are just a sad attempt to ignore the facts and keep leaving in your own world.
Odd think to say, why 1 to 7 prove you wrong. D. for once shows me proof, I showed you on 1 to 7 of the last post.

Do you admit you only use 80% force for the set time ??? I try to use 100% force, 80 – 100 = 20 more force used, quite simple, I use more force, thus more tension on the muscle = you fail faster.

Wayne