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.