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Calculating Force, Work, and Power Values For A Lifting Movement

  1. Jul 5, 2014 #1
    Thank you in advance for your help!

    Consider a lifter performing a bench press. The weight of the bar is 225 lbs. Assume the duration of the entire movement from the lowest point to the highest is 1 sec., and that the bar will move a total of 2 ft. from the lifter's sternum to the height of his reach. After zero pause at the top of the lift, the bar will be lowered at a constant velocity, requiring 2 sec. to return it to the initial position.

    It would help me if we used units of foot-pounds; I will use these units and g = 32.17 ft/s^2.

    How would I calculate the force produced during BOTH the positive (upward; concentric) phase of this lift and the negative (downward; eccentric)?

    I realize that the force is essentially equal to the poundage, but it seems to me there are two components here: (1) the force necessary simply to keep the bar suspended in the gravity field and (2) the force to accelerate the bar upward. So knowing that we accelerate the bar 2 ft/1sec, it would follow that the total force would be → F = (225) + [(225) * (2/32.17)]

    So is it 225 lbs. or is it 238.9 lb ?

    Next: Work.

    I am having difficulty calculating work, especially as the combined quantity for both phases of the lift. For the upward phase, would it simply be:

    W = F x d = (either 225 or 238.9 lb)(2 ft)

    OR would it be correct to calculate work as the change in PE for the weight:

    W = ΔPE = (m)(g)(h) = (225 lb)(32.17)(2 ft)

    because then it seems to me we'd still have to add in the component of the bar being accelerated right? So we'd get:

    (m)(g)(h) + (m)(a)(ft) BUT, how would we really approximate the acceleration? The
    velocity is zero at the initial and final position of the lift. Wouldn't we hav
    to know the instantaneous acc. immediately after lift-off and just
    before peak height in order to do this?

    Lastly: Work for downward part of lift.

    I understand that the work will be negative on the downward component of the lift. The weight will be doing work on the muscle, and the muscle will be absorbing energy. But I still am not sure how to calculate the work here.

    Despite the fact the bar is descending, we are CONTROLLING the descent using muscular force, despite the fact the muscle is elongating. How would we calculate the work being done on the descending weight in order to control that descent for say perhaps 2 seconds? We are still acting with an upward force, but one which must be less than 225 lbs. I'm just not sure how to calculate that and the subsequent work done during that phase. This probably sounds stupid, because I get the work is being done BY the weight, but how is it the lifter is not doing some positive work to keep the weight from simply dropping?

  2. jcsd
  3. Jul 5, 2014 #2


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    You actually can't figure out the force, because you didn't state that the velocity was constant.

    2ft/sec is just the average velocity, it has no effect on the acceleration/force involved. To find the force you would need to know more about the acceleration throughout that 2 ft
    The simplest examples would be constant acceleration and zero acceleration (a.k.a.) constant velocity.

    This only applies for a constant force, otherwise you need to integrate the force with respect to distance. Anyway though, we can't use this method because, as said above, we don't know the force as a function of the distance (we need more info)

    But we can still use a cool little "trick" to find the work. I'll let you think about the details of why it works, but we can use conservation of energy:
    It would be correct, but the way you did it is incorrect. Remember: 225 lb is not the mass of the object, it's the weight of the object. So ΔGPE would just be 225lb*2ft
    Last edited: Jul 5, 2014
  4. Jul 5, 2014 #3


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    Since the weight comes to rest at the top and the bottom we know that the average force is equal to the weight. To determine anything more detailed than that would require detailed knowledge of the acceleration profile.

    The work done by the lifter on the weight on the way up is equal to the work done by the weight on the lifter on the way down. The net work done on the weight over a whole rep is 0.
  5. Jul 5, 2014 #4


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    I don't think this is correct. (I think that) Our muscles don't absorb energy, they exert energy. The energy exerted by the muscles would cause negative work on the weight (negative work basically meaning it would "slow it down") while gravity is doing positive work on the weight.

    The net work being done on the weight? Or the work being done on the weight by the person('s muscles)?

    But you said that the weight is moving downwards at a constant velocity. If the upward force from the muscles was less than 225 lbs there would be a net force downwards, and thus an increase in velocity.

    Does it take any work to hold an object stationary in the air (this isn't entirely relevant, it's just an example of how keeping an object from dropping doesn't necessarily take work)
  6. Jul 5, 2014 #5
    Thanks, Nathaniel.

    I'll try to elaborate a bit. I'm much more of a biology/biochemistry guy who is only just starting to get into biomechanics out of pure curiosity/interest. And its been a couple of years since I got through the first two semesters of physics. So I'm trying to keep my model as simple as possible, ignoring the fact we're dealing with a system of multiple muscles and rotation about joints. I'm considering the arms simply like elevators moving in a straight line from the bottom of the lift to the top.

    The acceleration is one aspect that is really confusing me, because its difficult to keep this model as simple as I'd like to. When someone performs a bench press, the acceleration isn't constant. It isn't zero either. Its an explosive movement and I'm not sure how to describe the acceleration quantitatively, but we can assume it changes very quickly in the short time span of the lift. Is there an acceleration (constant) you could suggest that we might just assume for the sake of this example and go from there? Then go again through the work calculations with this new figure? (I'm not really sure how to calculate what the actual acceleration would be for an actual bench press)

    And regarding the last part, I believe the muscle has to absorb energy in some capacity. The energy absorbed during an eccentric movement translates into sarcoplasmic/myofibrilic micro-trauma, essentially little "tears" in the tissue. The problem I'm having is really frame of reference here. In most protocols for calculating the work/power of a workout, I don't believe the calculations include work done during this "negative" part of the rep. That is what was confusing me. Because the in the cartesian/vector way of viewing it, all of the work should be happening in the opposite direction on the way down correct? But in the context of the body, there is some value in not treating these figures as negative. So I'm basically trying to get right with what the forces actually look like on the downward part so I can somehow grasp how the work being done by the muscles would be a net positive, even if I'm taking the absolute values here. I think its practically useful for assessing the work done overall in a resistance program. I'm not sure that even makes sense haha.

    But yes, I was wrong. We can't be acting - on the downward phase - with a force under 225 lbs. I'm assuming the lifter is doing what it takes to lower the weight in a controlled fashion and keeping the velocity of the descent constant.
    Last edited: Jul 5, 2014
  7. Jul 5, 2014 #6
    Overall, my point of this whole exercise is to find a way to relatively accurately (but not that accurately) assess the work being done by a lifter performing a move like this, but also in a way that incorporates the work it requires to lower a weight in a controlled way using muscular contraction. Like I said, I don't want this to get too complex because I'd like something I can just apply to virtually any lift and get a quick idea about the work/power output for that lift. But most of the methods I've found out there only calculate these figures for the upward part of the motion, which I think ignores the fact a lifter benefits from the exertion on the downward aspect.

    And to Dave's point, yes that is exactly what I am trying to get around. The net comes to zero. But what I'm trying to do is find a way to calculate the work done BY the muscle on the way up, and the work done by the muscle on the way down to just prevent the weight from experiencing free fall. I am not really interested in how the work nets for the weight itself. But from a practical standpoint, with reference to the muscle, it seems work done in both the concentric and eccentric phase of the lift, and I'm not sure how the net plays out for the muscle itself, but if it nets zero for the numbers, it isn't netting zero in terms of physical change to the tissue, which is why I'm attempting to find a way of measuring work that reconciles this, i.e. the change in direction of motion does not cancel out the fact that the muscle is - for all other purposes - doing work or having work done on it that is worth tallying.
    Last edited: Jul 5, 2014
  8. Jul 6, 2014 #7


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    It's certainly an interesting aspect of the situation, I would like to hear more of your thoughts on that side of things if any come. Even if I don't entirely understand the ideas, it's very interesting.

    It is difficult to know the acceleration in a situation like this, but the model can still be simplified, a bit. We do know one interesting piece of information; that the average acceleration is always zero. If you imagine a person lifting a weight, it would accelerate at the beginning, and decelerate towards the end. No matter how it specifically accelerates, the acceleration and deceleration must always add to zero (wrt time).

    I ignorantly said in my other post that constant and zero would be the two simplest scenarios. (I was thinking too generally, ignoring this specific problem.) Both of these cases are impossible.
    The zero is obvious why it's impossible, it would never move. You would need a slight acceleration and then you could maintain a constant velocity and then you'd need a slight deceleration (in a way which combines to zero).
    The constant acceleration is of course impossible too, as the weights will never stop until the acceleration adds to zero.
  9. Jul 6, 2014 #8


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    Do you know if absorbed energy is dependent on the rate of work? Or just the total work? Perhaps the energy is absored in a different way if the power is higher?

    All of the work happening on what? It can't be true for the entire system, (because net work is zero,) so you must specifiy what is being worked on.

    All of the work on the weights is happening in the opposite direction, on the way down.

    That's because in the context of the work being done on the muscles, the figures aren't negative.
    If you imagine, like you said, two elevators lifting the weight, you can see that on the way up, the elevators are transferring some of their energy (and momentum) to the weight (positive work on the weight, negative work on the elevators). But, on the way down, the elevator slows down the weight by "stealing" some of it's momentum (and therefore energy) (negative work on the weight, positive on the elevator).

    I don't think the absolute value of the net work on the muscles would be positive. I think the net work on them would be zero. If I had to guess; the act of lifting the weight somehow transforms one form of (potential) energy into another form of biological energy. But (from the essence of the elevator analogy) I think that the net work would be zero.
    Last edited: Jul 6, 2014
  10. Jul 6, 2014 #9


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    This would be mgh
    You can generalize it further than that. The only limitation is that the person stops the weight (in other words the final velocity is zero).
    But I don't think that the muscles do any work over this period, although, it definitely seems (intuitively) like the muscles are doing some kind of work (I'll admit I was thinking like this during my other posts).
    Indeed. The work done on the muscles (on the way down) is the same as the work done by the muscles (on the way up) which is mgh.

    I know nothing about biology, but I think the power (=work per time) might come into play for the actual physical changes. I base this on my experience as a bundle of muscles; there seems to be a potential energy that is steady and long lasting (endurance-type) and a potential energy that is quick and strong (powerful-type).

    I think that taking 60 minutes to do one repetition would result in different physical changes than taking 3 seconds to do the same rep, even though the maximum work out (or in) would be the same between both cases.

    The average power (mgh/T) would possibly be a good indicator of the details of the physical changes.
    Last edited: Jul 6, 2014
  11. Jul 6, 2014 #10
    Great thoughts, Nathaniel. I think rather than trying to sum the different forms of work being done by the muscle, and on them, it is probably more useful to just analyze this in terms of power output. If we consider the work being done by the muscle on the way up, and the work done on the muscle on the way down, both have an inherent value but its difficult to try to look at it in some additive way, given each is having a differential effect on what's occurring physically in the muscle. If I had to guess the work load performed by the muscle to accelerate the weight upward creates more of a myofibrillar trauma; we are recruiting motor units to do the work and it results in a shearing force on the heads of the filaments in the fiber. That's one form of trauma that can create an adaptive response during recovery.

    On the downward portion of the movement, the muscles are resisting the downward force of gravity acting on the weight, rather than trying to accelerate it upward, and they're doing this while elongating rather than shortening. I believe (could be wrong here) the adaptation would be more sarcoplasmic, an increase in the volume and capillarization of the tissue. This is why muscle size does not always correlate with strengh, and why bodybuilders do not have as much of an inherent focus on the speed of a lift, so much as they do time under tension, the latter which has more of an effect on sarcoplasmic volume.

    I would still like a simple method for estimating the amount of work done ON the muscle by the weight on the way down. If we assume the muscles (arms in general) produced NO resistance, and the weight simply experiences free fall, the weight will eventually just transfer all of its momentum to the lifters body when it collides. Assuming there is resistance provided to lower the weight slowly, the weight would be working on the muscle, but I don't believe we can assume it is the reciprocal of the work performed on the weight on the way up correct? This would be because we aren't accelerating the weight upward like we do in the actual "lift", but are simply slowing the acceleration that would be achieved by gravity.

    So I guess my question becomes: can we approximate how much work is performed on the muscle on the downward aspect or what the power output would have to be in that phase of lift? If the power output is BY the weight, then that energy has to be being absorbed - at least partially - by the muscle correct? I mean in a general sense.

    And one other question: for the upward part of the lift, can we ignore the fact that we are accelerating the weight? I understand the work is represented by the change in PE for the weight, but isn't the fact we are accelerating the weight during at least half of the upward phase a component that we'd add to the change in PE?
    Last edited: Jul 6, 2014
  12. Jul 6, 2014 #11
    Also, it may be useful to consider another lift in this scenario: the squat. Its a bit different than the bench press dynamically but still has a concentric and eccentric aspect. Instead of accelerating the weight upward initially, we are resisting gravity as we lower the weight. But the key thing I think is to consider that at the top of the lift, our muscles are in the strongest position. The muscles elongate during this eccentric (lowering) part of the lift and we grow weaker as we descend into the squat position, but in general, we are keeping the weight moving at a relatively steady rate all the way down. This suggests a greater level of force must be being generated by the muscle to maintain that velocity while going down, else the weight would accelerate as we got lower and lower. In reality, it probably does a little bit. Usually by the time we are close to parallel-squat, we sort of "drop" into the very bottom of the movement. But we can assume that for the majority of the lowering phase, we are keeping the decline at a constant speed. If the muscles are in fact elongating and becoming "weaker" going down, what is doing the work to resist the force of gravity acting on the weight. At the bottom position we then accelerate the weight upward. Its really just a change in reference for where the "lift begins".

    Maybe its that conceptually I'm off here. Can those muscles - which are lowering the weight in a controlled way against gravity - still be generating force while having work done upon them? And if so, what would be a way to calculate the work done on the muscle? Again, I can't imagine its the reciprocal of the work done ON the weight by lifting it, because on the way up we are actually accelerating the weight upward whereas we are only resisting free fall on the way down.
  13. Jul 6, 2014 #12
    I'm not sure of any way to measure the absorption of energy by the body other than in terms of metabolism of carbs/fats or maybe the SAR for radio wave absorption. But, I may have an insight as to why the work performed is different on the way down than on the way up, with respect to the muscle.

    During the eccentric (lowering) phase of the motion, the muscle is ABSORBING energy. The elongation of the muscle, under resistance from the weight it is supporting against gravity, causes the muscle to absorb energy as mechanical energy - literally as the stretching of adjacent muscle fibers.

    At that point, there is a short time window during which that energy is quickly converted, either dissipated as heat or as elastic recoil. If the muscle immediately shortens again (returns to the concentric phase), the mechanical energy is converted into energy of elastic recoil, allowing the muscle to generate more force. If this does not occur almost immediately, that energy is lost as heat.

    So it seems that the work being done on the muscle by the weight is being absorbed and converted due to the actual elongation of fibers. So the muscle is absorbing energy and "reusing" that energy to do work in the next rep of the lift. In that sense, the muscle is actually acting to conserve energy like a spring.

    But I wonder if we can ignore this work then? Its confusing because I'm getting the ideas of physical work and "effort" confused in my head. Yes, the muscles are absorbing energy which produces the recoil, but the effort is still taxing. Sure that mechanical energy is being reapplied to the production of force for the next lift, but does this indicate the muscle isn't expending energy during the downward part of the lift?

    How I'm breaking it down is this:
    - The muscle, in the downward phase, is elongating and absorbing mechanical energy which will produce elastic recoil at the bottom of the movement.
    - But the muscle is still generating force in order to resist the downward acceleration due to gravity.
    - Can a net work experienced by the muscle be calculated here?

    If I had to guess I'd say that any loss of energy from the muscle during the lowering of the weight is being overcome by the addition of mechanical energy to the stretching muscle. On the second rep of any exercise we perceive we are exerting "less effort". What I"m picturing is that energy is lost by the muscle in pushing the weight up, but that any energy that would be lost in the lowering of the weight is offset by the gain of mechanical energy, creating the recoil we experience in the second rep. Thus, over the course of the whole set (say, 10 reps) the only real relevant output of energy occurs during the concentric phase of the lift, even if there are still relevant physical changes in the muscle during the eccentric phase. Sound okay? Haha.
    Last edited: Jul 6, 2014
  14. Jul 6, 2014 #13


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    Yes, but the total work should also play a role. The reason I say this is because, doing 1 rep will have a different physical effect than doing 100 reps, (even if they're done with the same power). So both power and total work seem to be important factors. Of course, to get a meaningful value, we have to ignore the fact that it technically sums to zero by only summing the work one way (either up or down).

    (Perhaps that's the reason that "most protocols don't calculate the negative part of the rep"?)

    You can do the downward part of the rep with power. You would just let the weight fall freely and then quickly stop it towards the end. I think that, as you said, the time under tension is the important quality (whereas the downward part of the rep doesn't necessarily have to be "time-under-tension"-type, it can also be done in a powerful way)

    It wouldn't be the (mathematical) reciprocal. The work done on the muscles on the way down will always be the same as the work done on weight on the way up (assuming you successfully stop the weight and don't drop it or something).
    I think this is why it can be valid to ignore either the work on the way up or on the way down in order to find the total work done (which is important because otherwise the number of reps would be irrelevant, which is intuitively not true).

    It is still technically accelerating the weight upwards. The weight falls freely for a brief time, gaining a downwards velocity. Then we must slow down that velocity (to zero). Since the velocity is downward and we are slowing down the velocity, the change in velocity (a.k.a. acceleration) is still upwards.

    Both parts of the lift (up and down) have a part where the weight is accelerated up, and a part where the weight it accelerated down. This must be true because the net acceleration for each part (individually) is zero.

    The work, yes. We can do better than approximate it. We can say that the work done on the muscles (on the way down) is equivalent to the work done by the muscles (on the way up).
    Which happens to be pretty simple to calculate, it's just the weight multiplied by the height it's lifted to! :)
    To figure out about the power, though, we would need to know about the time.

    We can "ignore" it.
    Actually, though, we are not ignoring it, because calculating the work done via ΔPE already takes into account the acceleration. (No work could be done on the weight unless it was accelerated.)

    We wouldn't add it to the ΔPE because it's already accounted for in the ΔPE
  15. Jul 6, 2014 #14


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    Indeed, this is what I was thinking. Perhaps there are biological differences between the two, but to my level of understanding, the only difference is the "first part" of the squat is the "second part" of the bench press (and vice versa).

    I said this in my last post but I'll reiterate it. The work done ON the weight (on the way up) is exactly equal to the work done ON the muscles (on the way down).

    In both cases (up and down) there is some upwards acceleration, and some downwards acceleration.
  16. Jul 6, 2014 #15

    You are the man. Thanks for all the back and forth sir.

    My confusion was certainly because I didn't have the right idea about the acceleration aspect of the lift. But its definitely apparent that for the sake of calculation, we can just sum the work in one direction. Of course there are qualitative aspects of the downward aspect we should take into consideration when trying to control the time of the negative. We'd be fatiguing the muscle and transferring energy at a different rate which would alter the stimulus.

    And on a related note, I'd never even considered the idea of letting the weight fall a little more rapidly to snatch it at the bottom. Granted, that isn't the kind of training you'd want to subject your joints to regularly but for power output that could be useful.

    But I've definitely got a better idea of what's occurring here and what I actually need to measure. I appreciate it.

    One thing I'm left wondering, however, is how efficiently the muscles are able to act as springs. The work done by the muscles is equal to the work done on the muscles. But that work could not be perfectly transferred into mechanical energy of the muscles' elongation (for example, into the mechanical energy of the pectorals stretching out in the bottom of the lift). If it were, we'd expect metabolic fuel to be the only limit to muscular endurance, which can't be the case I'd imagine. But its possible (I don't really know how much microtrauma actually limits endurance; it seems lactic acid accumulation limits us before tissue damage does). It would be interesting to know at what percentage the energy is converted into either heat or elastic potential.
    Last edited: Jul 6, 2014
  17. Jul 6, 2014 #16


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    No problem at all! I love discussion :)
    Awesome, I'm happy to have actually helped.

    Good luck with biomechanics! It is an interesting subject that it is wayyy beyond me hahah
  18. Jul 6, 2014 #17


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    Hopefully you never run out of stuff to wonder about!

    I can't help you here, of course. But hopefully you'll find a more biomechanically-experienced person to have discussions with.

    In the mean time you'll just have to have fun imagining the possibilities :)
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