# Which force is doing the work here?

My text book (Fundamentals of Physics by Halliday, Resnick, and Walker) mentions the following about the work done in internal energy transfers:

An initially stationary ice-skater pushes away from a railing and then slides over the ice. Her kinetic energy increases because of an external force F on her from the rail. However, that force does not transfer energy from the rail to her. Thus, the force does no work on her. Rather, her kinetic energy increases as a result of internal transfers from the biochemical energy in her muscles.

This is confusing me a lot. The energy transfer is clearly internal but work must be done by the force as work done is defined as the (dot) product of force and displacement and the definition makes no reference to any transfer of energy.

I thought work done by a force just means that the force is causing a transfer of energy to (or from) an object, and gives no information about whether the energy is coming from the object exerting the force.

My confusion is not over whether work is being done or not but which force is doing the work which ends up causing the change in kinetic energy.

Let's make a simple example. A block with a compressed spring attached to it is on a frictionless horizontal surface against a stationary, immovable wall. The spring is released, and the block is then pushed away from the wall, thus gaining kinetic energy.

The relevant forces here are 1) the force between the spring and the block and 2) the force between the spring and the wall. Which force does work here? Force 1 did, because it is applied over a distance. The energy is transferred from the potential energy stored in the spring to the kinetic energy of the block.

In your example, the skater is the block, and the arms/muscles are the spring.

• Ohhh. Thankyou for this example. Perfectly cleared things up. So should i change my definition to the dot profuct of force and displacement of POINT of contact, instead of, incase of non particle like bodies, the displacement of the centre of mass? Jul 6, 2020 at 4:13
• this is the most intuitive answer here, with a nice example of how to cast a complex problem involving biology into a simple physical model. +1 Jul 6, 2020 at 6:52
• @OVERWOOTCH Technically, yes. But also in the case of rigid bodies these are usually the same thing. Even here it is the same thing when you focus on force 1. Or in the case of gravity we actually do average everything out to the center of gravity and look at its displacement. But you should always start with the essential principle that the work depends on the motion of the point of application, and then make simplifications from there. Jul 6, 2020 at 12:28
• Interestingly, the situation is the time inversion of a perfectly inelastic collision (if you "play the movie backwards" the skater comes to a standstill at the railing without rebound). That the "inelastic collision" here involves an internal spring (or muscle) and no heat adds to the confusion when used as a textbook example. Jul 6, 2020 at 14:24
• @BioPhysicist I really liked your block with compressed spring analogy showing that the wall does not do work. What I would have liked to have seen is a discussion of the role of the force between the spring and the wall, given that the wall does no work. For example, suppose the wall were not there and the block spring system was kept compressed by some internal force, say a string under tension, which is suddenly cut, so that we must consider the behavior of an isolated system. Sep 30 at 15:02

Biological systems are really tricky in physics class. Our intuition of how they work rarely lines up with how they actually do.

In this case, the answer is almost correct. Cleonis is correct that there is some work due to pushing the earth, but it's negligable. We can really say that their answer is correct unless we're talking to a rules lawyer or are solving the riddle of the Sphinx.

The trick is that you have to break the skater apart into their component bits. Consider just her hand. It doesn't move, so clearly no work was done. Now move up to her forearm. There's some movement here, so we can see that probably some work was done to the forearm.

Move up futher to the upper arm, and you see much more movement. And finally, when you consider the core of the skater, she's clearly being moved a lot. Work is clearly done on the skater.

If you were to look at just her bones, muscles and tendons, what you would see is that the bones of the hand are stationary, but the muscles and tendons pull on the bones further up her arm. It is this interaction where work is being done, putting lots of force into a movement over a distance (or, alternatively, a torque into a level arm, which you will learn is an equivalent way to think about it).

And in theory, a teeny-weenie bit of work is being done by her hand on the railing to move the Earth further away from her. (or more practically, she will bend the railing ever so imperceptibly)

• I am aware that this not being a rigid body, we cant just multiply “force” and “displacement”. But which force is actually doing the work? The reaction force or the applied force itself? This is where im confused Jul 5, 2020 at 19:32
• @OVERWOOTCH For every force, its work done is force times displacement. Applied force, reaction, always the same. However, when we start talking about energy, we typically care about which body expended the energy to do the work. In this case, one body (the skater) is expending energy and one body (the rail) is not. Jul 5, 2020 at 20:05
• Oh ok. Is this just a convention or does this have anything to do with the definition of work ? Jul 5, 2020 at 20:44
• I would consider it more of a convention. Its a way of thinking about how things move. However it does have a tie in to work. Depending on our frame of reference, we may find that all of the work done is done by the "action," or all of the work is done by the "reaction" or both. Consider a case where you have two steel balls and a force drives them apart. From the perspective of an outside observer, it looks like some work was done to each ball. However, if you view it from the perspective of one ball, the work was done to the other ball, because... Jul 5, 2020 at 22:10
• ... its the other ball that moves Jul 5, 2020 at 22:10

I think I agree there is something off in that quoted section.

The purpose of that section is not practical application, the purpose is to bring an abstract concept into focus. And that means the usual simplifications for practical purposes should not be used.

There is a physics joke that goes as follows: When Arnold Schwarzenegger does his push-ups, he is actually pushing the Earth away from himself.

In the case of the skater the standard for-practical-purposes simplification is: she is in effect pushing off against the entire Earth, so we treat the railing she is pushing off against as immovable.

But: that simplification violates the third law. It is more instructive to think of a setup where the third law is visibly involved. You can have the skater push off against a sled that is, say, twice her mass. And when you have a clear concept of the third-law-visibly-involved case you can take that to the limit of pushing off against something that is vastly more massive than the skater.

• Surely you meant Chuck Norris rather than Arnold Schwarzenegger. Jul 7, 2020 at 15:31
• @DavidHammen :-) I'll give that Chuck Norris can move faster than his own shadow, but Schwarzenegger is more massive. So I vote to give this one to Arnie. Jul 7, 2020 at 21:39

The energy transfer is clearly internal but work must be done by the force as work done is defined as the (dot) product of force and displacement and the definition makes no reference to any transfer of energy.

Yes, but you must be careful. What force and what displacement?

In this case the force is the force from the wall on the skater’s hand.

The displacement is the displacement of the hand at that point of contact with the wall. That displacement is zero (neglecting the practically negligible effects like bending the wall or the utterly negligible effects like accelerating the earth). Since the displacement is zero the work is zero.

So the thermodynamic work done is zero. Thus, the external force provides momentum, but not energy. The energy comes entirely from internal sources.

Now, there is one other displacement that you might consider, and that is the displacement of the center of mass. By the work energy theorem the net force and the displacement of the center of mass give the change in KE. However, it is important to understand that the dot product of the net force and the displacement of the center of mass, sometimes confusingly called the (net work), is not generally equal to the sum of the thermodynamic work of all forces.

Non-rigid bodies like the skater are prime examples. In this case the sum of the thermodynamic work is zero even though the “net work” is non-zero b

• Oh ok. So the work done, in a rigorous sense, is 0 because there is no displacement of the point of contact, andI didn’t need to take into account where the energy came from. Also, can you pls elaborate a little on what thermodynamic work is? Im trying to do physics properly from the start and have not reached thermodynamics yet, although I am aware of what it is about, Jul 5, 2020 at 19:49
• One more thing. Why cant I just consider the skater a point particle and the displacement to be that of the centre of mass? Everything makes perfect sense then. Jul 5, 2020 at 20:46
• @OVERWOOTCH The thermodynamic work is the actual work done. I.e. a transfer of energy by any mechanism other than heat. The skater isn’t a point particle. If you just use the “net work” then you have no explanation about where the energy came from. A point particle has no internal energy and the wall provided no energy from the environment. So KE was created without any decreased energy elsewhere.
– Dale
Jul 5, 2020 at 21:03
• Oh. One other thing that comes to mind is another book mentioning that for a system of particles, (which i assume refers to anything not rigid, pls correct me if im wrong), the external forces as well as internal forces have to be taken into account. If im getting thus correctly, I am ignoring all of the spring like forces of the muscles, as Biophysicist puts it, which are actually doing the work? Jul 6, 2020 at 4:17
• Yes, that is right
– Dale
Jul 6, 2020 at 4:45