# Weightlifting & Physics

1. Dec 19, 2006

### methodman

I just was wondering if you could help me to count how much power these weightlifters are generating?

Clean and jerk 215kg -85 weight class
Snatch 180kg -85 weight class

2. Dec 19, 2006

### Staff: Mentor

methodman, what do you know about energy, work and power?

Last edited: Dec 19, 2006
3. Dec 19, 2006

### methodman

What should I learn? I don't know much about them. But I Learn fast.

I just want to know how to count how much power I'm producing during snatch and clean and jerk! I should know how heigh i'm lifting the weight over my head and how fast i'm lifting it also. So let's say I know these facts how can I count the power of the bar I'm using?

4. Dec 19, 2006

### methodman

Look at this video

pyrros dimas lifted that snatch in two seconds and the weight is 175kg

175kg*2m/2= 175w but that is theoretically right?

Last edited by a moderator: Sep 25, 2014
5. Dec 19, 2006

### Staff: Mentor

EDIT -- I posted a response that was full of math errors. Let me try again.

Power is the amount of energy expended per unit time. The easiest way to deal with lifting weights is in terms of the potential energy change (PE) plus any acceleration work done on the weights.

So the work done in lifting a 100kg mass 0.1 meter is the combination of its change in PE, plus any work put into accelerating the mass.

The change in PE is just the mass multiplied by the acceleration of gravity and the height change:

$$\Delta PE = m * g * h = 100kg * 9.8 m/s^2 * 0.1m = 98 Joules$$

So the power invested in the height change part is that PE divided by how long it took to move the mass. If it took 1 second in this example, that would be 98 Watts. 1W = 1J/s.

And the extra power that you put in if you are accelerating the mass while lifting it is the mass multiplied by the acceleration and the distance, all divided by the time interval.

$$P_accel = \frac{m * a * d}{t}$$

It's probably easiest for you to deal with the average power in different time intervals. To get the average acceleration for a time interval, just subtract the final velocity from the initial velocity and divide by the time interval.

I'm not sure that my rambling is going to help, but I need to get back to work. Sorry if all that I've done is confuse you.

6. Dec 19, 2006

### Staff: Mentor

I'm going to move this back to General Physics, since it doesn't appear to be a homework question after all.

7. Dec 19, 2006

### cesiumfrog

What about whether it's practically right? You should estimate the uncertainty in each of your three measurements, and deduce what the error bars are for your result (I mean.. obviously the power wasn't exactly 175.0000... watts, and there's a world of difference between 175+/-0.2w and 175+/-200w). Then google for some other measurements (of power exerted by rowing, cycling, etc) to compare whether your result seems reasonable. Are the error bars small enough to separate your power output from that of a champion weightlifter, or are both measurements consistant? Then get back to us (because once you have documented your method and produced results with known level of uncertainty, they becomes results other scientists can safely make use of in the future).

Last edited: Dec 19, 2006
8. Dec 19, 2006

### methodman

cesiumfrog I think it's not reasonable at all because that is too weak even without googling! I'm sure even without much training I can produce much more power

Last edited: Dec 19, 2006
9. Dec 21, 2006

### KingNothing

What makes you think that? As a whole, yes, your body produces as much power running up stairs. Let's do a simpler lift, like the bench press, where the speed is usually constant:

When the rate of energy transfer is constant:
Average Power = Change in Energy / Time

So if you lift a 50kg weight 0.5 meters over 2 seconds:
Energy change=50*9.8*0.5=245 Joules
Average Power = 245/2 = 122.5 Watts

10. Dec 21, 2006

### methodman

I think theorically this is right but it's really wrong. I am trying to measure my power output could you rell me how it's done? Do I need to have devices to measure my power accurately or can I do it with simple physics calculations?

11. Dec 21, 2006

### Jeff Ford

If you're a weightlifter, you don't need to worry about how much power you're producing until you're at a fairly high level. Generally, lifters who are at that level use a machine, like the Tendo, to measure bar speed and power output. They're trying to determine how efficient their lifts are, to correct technical errors.

But like I said, pretty much anyone who isn't a national level lifter doesn't need to be worrying about such things.

12. Dec 21, 2006

### methodman

I have just have this thing that I need to know:( Can you show me a picture of the machine or any machine that measures this kind of stuff??

Last edited: Dec 21, 2006
13. Dec 21, 2006

### Staff: Mentor

14. Dec 21, 2006

### cesiumfrog

Foolhardy. First result for "human power output" shows you're already talking about a quarter of a horsepower. A fit person should be able to produce a little more power for such a short period of time, but the body isn't suited to output maximum power while weightlifting (rather than cycling perhaps - weightlifting relies too much on the little arms).

Think about this: a car can be lifted with almost zero power, if it's done slowly enough. If the limitation was power (rather than "strength" somehow) you wouldn't want to jerk a weight up in just seconds. If we had a weightlifting competition where the prize went to the person with most power output, it might favour lifting incredibly light weights in an instant, and disadvantage stereotypical slow bulky weightlifters (even if they do more "work" per lift). To train you'd practice fencing, martial arts, or air hockey.

Last edited: Dec 21, 2006
15. Mar 28, 2007

### methodman

I founded out that the power that weightlifters are producing are unbelieveable. In the Clean heavy elite are producing 3,877watt and in the snatch light elite are producing 2,821watt that is amazing.

16. Apr 8, 2007

### ace21

Power

I don't find it hard to believe that they are producing that type of wattage, because they are doing so for a very short period of time. I believe you will agree after I explain.

1 Hp = 33000 (ft*lb)/min and 1 Hp = 746 watts. So at first glance their wattage output may seem impressive, but not so if you dive deeper.

1 Kwh = 2655223 (ft*lb)/hr
The average "Power" consumed in an American home is 10656 Kwh/year.
Which works out to be about 29.2 Kwh/day or 77517966 ((ft*lb)/hr)/day.

I'm begining to ramble so I will wrap it up. If 2500 Watts are produced over a 3 second time frame, then theoretically this would power a single 60 watt light bulb for a little over 2 hours.

It's been awhile since I've worked physics problems so if I've made any mistakes someone please let me know.