# What Work Is.

This is Liao Hui, not doing any work. He did work to get the weight over his head, but despite the tremendous force he’s applying to this 348 kilogram [Update: Thanks to commenter Ducklike for correcting this to 158 kg] weight he’s not doing any work in this picture. The weight is stationary.

Work in physics is a term of art meaning force through a distance. The reason for that definition of work is that this definition coincides perfectly with the energy transferred in the process. Work results in a change in energy. When Liao Hui did work to move the weight to its highest position, chemical energy from his body was transferred into his muscles which did work to the weight. As the work against gravity caused the weight to increase in height, it gained gravitational potential energy. And finally when Liao Hui let go of the weight, it fell back to earth: gravity did work in the opposite direction which increased the kinetic energy of the weight until it hit the ground, releasing the energy as heat and the sound of the impact.

Force through a distance. Mathematically that means work in joules is the force in newtons multiplied by the distance in meters.

I’d figure out how much work was done by the weightlifter but I don’t know how far it was lifted. Probably between 1 and 2 meters, so a few thousand joules.

As always in physics, we can take the basic concept and make it more general. What if the force isn’t pushing in the same direction as the motion? For instance, say your friends are pushing a car that has run out of gas. You decide to be a joker and start pushing the car from the side, at right angles to the direction of motion. That won’t change the car’s speed and you aren’t adding or subtracting any energy to the car, so you can’t be doing any work. It’s because you aren’t pushing in the direction of motion. Call the angle between the force and the motion by the Greek letter theta:

The work is proportional to the cosine of the angle between the force and the displacement. If the angle is zero, we just get the Fd relationship we had initially. We can save some space by writing the relationship in vector notation, which is what the last part of that equation above means.

But even this isn’t quite enough to sum up the totality of possibility. The force might not be constant, and the displacement might not be along a straight line. You could have a force with constantly changing direction and magnitude pushing an object along some weird curvy path. So the final generalization is to say the total work is the sum of each little bit of work done along the path by whatever the force and its angle happen to be at that but of path. This is written using an integral.

Here we call the path by the label C, and we call the distance by the variable s because d is already taken by the notation of the integral itself. That allows us to find the general work done over a general path.

It can be tricky to calculate that integral. But in many cases, nature works in such a way as to help us out. If the work is being done against a conservative force, the integral is just the change in potential energy. A weightlifter can use whatever weird method he wants to get the weight up, but that integral will just turn out to be the change in potential energy mg(Δh).

This fact that mechanics can be thought of in terms of force or energy equally well turns out to be one of the most important concepts in physics. And while it might not help anyone actually pick that weight up, it’s a good mental exercise to go along with the physical exercise.

1. #1 BobbyEarle
August 18, 2008

Thanks for an interesting post. I do have a question: once the weightlifter has the weight up and stationary, if it is not work at this point, what is it? I understand force through a distance, but from a physics standpoint, what is he doing now?

2. #2 Luke O'Dell
August 18, 2008

It feels difficult to hold a weight up because of the way our muscles work. Human muscles consists of tiny fibres which move over one another, and therefore when we exert our muscles they are constantly using up energy even if they are not actually moving anything (this also creates lactic acid which results in “the burn”).

Contrast this with the muscle of a clam, which has a completely different structure and once locked in place (holding the shell closed for example), does not require any further energy to maintain position.

3. #3 Eric Lund
August 18, 2008

@Bobby: I’m not completely sure since I don’t know if anyone has done the experiment, but here is an educated guess: The energy he is expending to keep the weights at that height is being dissipated in other forms, such as heat or muscular deformation. While he was lifting the weight, most of the energy expended went into overcoming gravity (that’s why it’s harder to lift something than to carry it at that same height).

The effect, at least as I notice it is that after a while your arms get tired and you have to let the weight drop. I usually see this phenomenon with groceries (I normally walk to the grocery store) rather than weights, but it’s the same thing.

4. #4 Uncle Al
August 18, 2008

Hang the weight from a cable – does the cable do work? Support the weight on a pedestal – does the pedestal do work? Don’t confuse the engine with the transmission or the clutch.

5. #5 Mark
August 18, 2008

In reply to.. #1 thru #4.. Work = F over a distance. The distance isn’t changing, so he’s not working. *But* you now have to look what Force is. F=Mass * Acceleration. Is there a mass? yes. ~130kg. Is there any Acceleration? Yes. Gravity=9.80665 m/sē. So that’s about 1274 Newtons.

So while he’s not ‘working’ he IS applying a force. (Being forceful?)
The only way to keep him from applying any force would be to take the ground from underneath him and so make his acceleration = gravity, making the net force approximate 0.

No work, just applied force.

6. #6 hibob
August 18, 2008

I think it’s probably closer to 348 lbs he’s holding above his head-
it looks like 2 x (3 x 25kg, 1 x 15kg, 1 x something small), plus the bar.
Call it 200 kilos?

7. #7 Ducklike
August 18, 2008

It’s closer to 157 Kilos.

25 for the bar (and collars), plus 4 x 25 Kilo plates, plus 2 x 15 Kilo plates, plus 2 x 1(?) Kilo plates.

I used to lift when I was younger.

8. #8 Matt Springer
August 18, 2008

#7 is right. I misread the caption on the original photo. Liao Hui lifted 348 kilograms total, with 158kg in the snatch lift (in the picture) and 190kg in the clean and jerk.

9. #9 CCPhysicist
August 18, 2008

Regarding #5: The acceleration is, according to Matt’s assumptions, zero. You are confusing F = ma, which is the expression for the NET force (zero), and Weight = mg, which is the force of gravity on the mass. The appearance of “m” in both of these is a result of the equivalence of inertial (the first) and gravitational (the second) mass.

#2 is spot on, but could have added that the “quivering” of your muscles at their limit is an indicator of the the repeated twitching that underlies a seemingly smooth effort. The force appears smooth as long as there is enough excess capacity for the next fiber to take over as another takes a rest to return to the start position.

Nit pick: The lifter’s legs are not locked out in that photo, so he might actually be doing some work at that moment. He is either in the process of pressing up out of the squat position after ‘snatching’ the weight over his head or (equally likely) trying to catch the weight at the end of the first half of the lift.

Nit pick 2: It makes little sense to use the 6 sig fig conversion factor for g in comment #5 when the acceleration of gravity in Beijing is probably around 9.801 m/s^2.

Matt, your post reminded me that we need a bit more information to look at the most important part of weightlifting, which is POWER. Since we don’t have that info, let’s do a Fermi Question estimate: You never take the weight from the ground to overhead in a single move in either the snatch or clean and jerk. Liao Hui (LIAO is his surname, which is given first in Chinese) is a little fellow in the 69 kg class (data on the official web site says he is 68.97 kg or just under 152 pounds), so he is not very tall. He lifts that 158 kg (about 1550 N) around 1.5 m in the first part of the snatch, so lets say the work done is about 2200 J (half a Calorie!). This is done in about one second, maybe less, so it requires at least 2200 W, which is 2.9 hp. So my estimate would be about 3 hp, but probably more. It would be 6 hp if he gets the weight up there in a half second.

(Quantifying the lift time and distance would be a good application for one of those video analysis programs sold by various companies.)

That power is delivered in a controlled explosion, since the weight has to be stabilized after it gets up there. Lots of snatch lifts fail when the weight ends up going behind your head.

10. #10 IBY
August 20, 2008

I imagine the term ds is displacement. The cool thing about that integral equation is that one can derive the kinetic energy equation with that.