Work when lifting weights- fast vs slow

  • #1
I am confused about something and can't seem to put things straight in my head. Working with my students we were trying to calculate the work done in one repetition of an exercise like a bench press or a push up. Some students measured the distance the weight moved and the time it took and then used d = ViT + 0.5at2 to calculate the acceleration (by assuming the initial velocity is 0) which allowed them to calculate the force [ma], and use that force to calculate work [Fd]. For this method, how quickly you move the weight impacts the work done (shorter time means more acceleration which means more force and more work).
Others looked at the change in potential energy [mgh] and in this method how quickly you move the weight does not matter [for work, though obviously the power is different]. The second method makes sense to me, but I can't get my head around how the difference in acceleration in the first method doesn't change the work. I feel like I am missing something basic. Any help is greatly appreciated.
 

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  • #2
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The problem is that in the first method, some of the work is adding kinetic energy to the weight. The second(potential energy) method assumes that the work is done slowly enough to neglect kinetic energy.
 
  • #3
PhanthomJay
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You start from rest and end at rest . No change in kinetic energy during the rep. So the work done in the upstroke is the potential energy change. Regardless of the speed during the motion.
 
  • #4
Dale
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Some students measured the distance the weight moved and the time it took and then used d = ViT + 0.5at2 to calculate the acceleration (by assuming the initial velocity is 0)
Isn't the final velocity also zero? It should be unless you are throwing the weights.

If the final velocity is also 0, then the force cannot be constant so that formula won't apply. The students using this method made a mistake. This type of mistake comes from simply plugging numbers into formulas without checking that the assumptions were met.
 
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  • #5
First, thanks for the answers.

Isn't the final velocity also zero? It should be unless you are throwing the weights.

If the final velocity is also 0, then the force cannot be constant so that formula won't apply. The students using this method made a mistake. This type of mistake comes from simply plugging numbers into formulas without checking that the assumptions were met.

And d = Vit + 0.5at2 assumes a constant acceleration with a differing Vi and Vf, right? I think I understand that with the formula, but I am still confused. During the rep, the weight is moving and so the faster moving weight would have more kinetic energy during the rep, so does that not mean that the person doing the pressing had to use more energy? What am I missing?

Is this just as simple as the difference between 'work' in physics and energy in biology. For example, if I was using a medicine ball which I could throw when I pressed it upwards- If I press up with the enough force to throw the ball I do work equal to the mass of the ball times gravity times the height the ball climbed to, but if I do another rep where I generate the same force to press the ball, but do not let it go I have done less work [because the ball did not rise as high] but from the standpoint of the energy use in my body they would be the same because I generated the same force over the same distance?
 
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  • #6
A.T.
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Is this just as simple as the difference between 'work' in physics and energy in biology.
It's all physics. Work done externally is often different from energy transformed internally, due to inefficiency.
but from the standpoint of the energy use in my body they would be the same
You might even end up using more energy, while doing less work. Muscles need energy to do negative work.
 
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  • #7
jbriggs444
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And d = Vit + 0.5at2 For example, if I was using a medicine ball which I could throw when I pressed it upwards- If I press up with the enough force to throw the ball I do work equal to the mass of the ball times gravity times the height the ball climbed to, but if I do another rep where I generate the same force to press the ball, but do not let it go I have done less work [because the ball did not rise as high] but from the standpoint of the energy use in my body they would be the same because I generated the same force over the same distance?
With the medicine ball, if you exert the same force over the same distance while throwing but do not let go then you must exert considerable force to hold it back, slowing it down so that it does not fly out of your hands. From a physics point of view, the effort you use to slow the ball down is negative work. It counts against the total work done on the ball.

From a biological perspective, of course, you have to exert yourself to slow something down. Muscles do not regain energy when they are moved in the opposite direction that they are pulling. Physicicts would point out that you could, in principle, regain energy in such a situation, like regenerative braking systems in cars do.
 
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  • #8
A.T.
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With the medicine ball, if you exert the same force over the same distance while throwing but do not let go then you must exert considerable force to hold it back, slowing it down so that it does not fly out of your hands. From a physics point of view, the effort you use to slow the ball down is negative work. It counts against the total work done on the ball.

From a biological perspective, of course, you have to exert yourself to slow something down. Muscles do not regain energy when they are moved in the opposite direction that they are pulling. Physicicts would point out that you could, in principle, regain energy in such a situation, like regenerative braking systems in cars do.
All true, but I wouldn't frame it as "physics vs. biology". There are enough non-biological examples where the same happens: A plane during landing can use it's engines to brake, and thus do negative work while consuming energy (just like contracted muscles being stretched). On the other hand, there are elastic parts of the muscle-tendon which do regain and store some of the energy from doing negative work (just like regenerative braking systems).
 
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  • #9
Dale
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During the rep, the weight is moving and so the faster moving weight would have more kinetic energy during the rep, so does that not mean that the person doing the pressing had to use more energy? What am I missing?
There are two things that you may be interested in. The first is the work done on the weight. The second is the energy used by the weightlifter. If the weightlifter were perfectly efficient then those two numbers would be equal, but humans are not particularly efficient so they are not equal.

The work done on the weight can be calculated fairly easily by equations like mgh and 0.5mv^2.

The energy used by the weightlifter cannot easily be calculated and would generally be measured instead by calorimetry or something similar.
 
  • #10
I think this problem would be solved by breaking the movement down into three phases.

For the case of the slow movement, for theoretical purposes, say the weightlifter lifts a weight vertically upward for a given distance (say 2 feet this for example). In order to get the weight moving, the force exerted by the lifter must be more than the gravitational force (From F = ma. Since the net acceleration is upward, the net force must also be upward.) Here, the weightlifter exerts the greatest force of the lift. Once the weight is moving upward, it reaches some intermediate velocity (the max speed at which the weight moves upward. If the lifter takes 2 seconds to lift the weight from initial to final position, then this velocity is 1 foot per second.) During this stage of movement, the net acceleration of the weight is zero, since it is moving at constant velocity, until the lifter nears the top of his movement. Now the weight will be slowed by the downward acceleration due to gravity, with the lifter no longer exerting any counteracting upward force. In this third phase, the weight would have acquired a kinetic energy of 0.5 m v2, and gravity must convert this kinetic energy to potential energy until the kinetic energy equals zero, where the weight will be at a stop (at rest). Therefore the work done by the lifter would have only been done in the first and second phases of the lift, whereas in the third phase, gravity does the work of slowing the weight to a stop.

In the case of the fast moving weight, all the things are the same, except that the initial force exerted by the weightlifter is greater (since it reaches a greater acceleration), and the time of the second phase is shorter (since it has a higher velocity but the same total distance as in the slow case), so the weightlifter exerts the force for less time. Not to mention, the weight would take longer to stop since it has a higher kinetic energy than in the slow case, so the lifter must stop his rep sooner to allow more time for the weight to come to a full stop. This means that the distance traveled by the weight in this phase would be shorter than in the slow case. So in this example, a greater force is exerted for less time, and so I suspect that the equations would work out to be the same no matter which formula was used.

To show it another way:

(top = finish)[COLOR=#black].....[/COLOR]SLOW[COLOR=#black].....[/COLOR]FAST
|[COLOR=#black]........................[/COLOR][3][COLOR=#black]...........[/COLOR][3]
|[COLOR=#black]........................[/COLOR][2][COLOR=#black]...........[/COLOR][3]
|[COLOR=#black]........................[/COLOR][2][COLOR=#black]...........[/COLOR][3]
|[COLOR=#black]........................[/COLOR][2][COLOR=#black]...........[/COLOR][2]
|[COLOR=#black]........................[/COLOR][2][COLOR=#black]...........[/COLOR][1]
|[COLOR=#black]........................[/COLOR][1][COLOR=#black]...........[/COLOR][1]
|[COLOR=#black]........................[/COLOR][1][COLOR=#black]...........[/COLOR][1]
(bottom = start)

Where phase [1] is the weightlifter pushing the weight to its max velocity (in phase [2]), so the weight is accelerating upward

In phase [2] the weight has reached a constant velocity (the speed of the weight during the lift), and so it has no net acceleration, and the force exerted by the lifter is equal to the force of gravity on the weight (which is the weight of the weight)

And in phase [3], the lifter is doing NO work and could essentially let go of the weight as the weight slows to its final velocity (zero) and its kinetic energy gets converted to potential energy by the work done by earth's gravity

In the real world, of course the force exerted by the lifter will not be constant, nor will the acceleration nor the upward velocity of the weight. But for an idealized model I believe this is correct.

The profiles of the two cases are different and so I suspect that the equations will work out to show equal work done in both cases, but I have already spent like 20 minutes typing this so I'm done and I'll leave the math to someone else :woot:
 
  • #11
rbelli1
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The profiles of the two cases are different and so I suspect that the equations will work out to show equal work done in both cases,
The work done lifting a mass in opposition to gravity in a vacuum will be the same no matter how fast you take. The only difference at normal room conditions will be air resistance. The faster you lift the more work is required. However you will find that the difference at normal human weight lifting speeds at standard room conditions will be negligible.

BoB
 
  • #12
The work done lifting a mass in opposition to gravity in a vacuum will be the same no matter how fast you take. The only difference at normal room conditions will be air resistance.
Yes except the lifter would suffocate in the vacuum because there would be no oxygen :biggrin: and I'm pretty sure we are safe neglecting air resistance in this case
 

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