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After reading the article, I was totally perplexed . I was reading the External forces and internal energy transfers in Principles of Physics by Resnick,Halliday,Walker. It goes like that

An external force can change the kinetic energy or potential energy of an object without doing work on the object- that is, without transferring energy to the object. Instead,the force is responsible for transfers of energy from one type to another inside the object.

Then they cited an example.

An initially ice-skater pushes away from a railing and then slides over the ice. Her kinetic energy increases because of an external force $\vec{F}$ 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.

What is going on here? First they said the KE increased due to that external force. Then they said the force did no work. If force were given and displacement had not occured then I would say yes! the work done would be 0 inspite of force being applied.

But here, displacement occured and according to the book , the kinetic energy increased due to the force but due to some cause the force did no work! Is it magic??

I am really confused. Please help me giving a strong explanation to this.

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    $\begingroup$ My feeling is that this is a very unfortunate way of formulating a non-problem, which just confuses a student. The first paragraph is just total nonsense. The rail does not have any energy before or after the push. One can not "transfer energy" this way, and the book shouldn't be talking about an unphysical scenario. All the energy in this example comes from the muscles of the skater and that is what the book should have emphasized. Dump the book and get yourself a real physics textbook, if you can. $\endgroup$
    – CuriousOne
    Commented Oct 17, 2014 at 6:40
  • $\begingroup$ @CuriousOne: The book has made me really confused. Ok I can say the skater increases her kinetic energy using her energy ie. she is applying force to herself,right? Then what is the role of that force exerted by the rail??? $\endgroup$
    – user36790
    Commented Oct 17, 2014 at 6:56
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    $\begingroup$ The energy is chemical energy stored in the skater's muscle. That's why she can exert a force on the railing, which causes the railing to create an equal but opposite reaction force (actio=reactio), which is what sets the skater in motion. The work of the skater's muscle is converted into kinetic energy of the skater. If there was no railing and the skater tried to push herself, nothing would happen, so one can't say that she is applying force to herself. If the railing wasn't fixed, then part of the skater's work would become kinetic energy of the railing's mass. That's how rockets work. $\endgroup$
    – CuriousOne
    Commented Oct 17, 2014 at 7:04
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    $\begingroup$ The ice skater example is terrible, because she is actually using muscular energy to increase her kinetic energy. You should consider a racket which is moving forward that hits a tennis ball coming at it. The speed of the tennis ball increases, increasing its kinetic energy, but because the collision was perfectly elastic, the internal configuration of neither of the bodies changes. $\endgroup$
    – Peter Shor
    Commented Oct 17, 2014 at 15:02

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I agree with CuriousOne that the example is more confusing than helpful, but this is the way I would explain it.

Suppose you take a spring, place it on the ground then compress it. If you now suddenly let go of the spring it will rebound and bounce upwards off the ground:

Spring

The spring clearly has work done on it because its kinetic energy increases and that increase must have come from somewhere. However the ground can't have done any work on the spring because the ground hasn't moved. It should be obvious that the potential energy in the compressed spring has been converted into kinetic energy of the uncompressed spring - in effect the spring has done work on itself. This is what your book means by:

transfers of energy from one type to another inside the object

i.e. potential energy of the compressed spring has been converted into kinetic energy of the uncompressed spring.

In the case of the skater the skater's arms correspond to the spring and the rail corresponds to the ground. The skater's arm isn't a spring, of course, because it's chemical energy not potential energy being converted to kinetic energy by the skater's muscles. Nevertheless the same principle applies.

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  • $\begingroup$ could you clarify how something does work on itself? would we be able to actually see work done by internal forces over some displacement among component parts of the spring? such as in the spacings of the coils or in the actual molecular lattice comprising it? $\endgroup$
    – MattGeo
    Commented Jan 5, 2022 at 3:18
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I think this is a good example, and should be studied and understood carefully. But I don't know where the book goes after making it. Whether or not it is poorly stated depends on the surrounding context.

It focuses attention on two things:

1.) the skater is a deformable body. The center of mass is not fixed in the body.

2.) work is defined as a force times the displacement of the point of application.

(And, arguably a third aspect: the skater carries a store of chemical energy. But this fact is aside the point being made.)

In this case, the point of application does not move. There is no external work from that force. Both of these points are important to understand and remember. Typically, up to this point in a course, problems are implicitly formulated in terms of a point-particle model of an object. No internal structure. When deformable bodies are considered, certain things that work for point-particles no longer work. Additional analysis is needed.

Note that there is internal work being done. Chemical reactions contract muscle fibers. The fibers apply forces to tendons. Now there is a force and a displacement of the application of the force.

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  • $\begingroup$ So, with your point 2), you mean to say that when a car crashes into a tree, the tree doesn't do any work on the car because the point of application of its force is fixed? $\endgroup$
    – gatsu
    Commented Dec 17, 2015 at 6:41
  • $\begingroup$ Absolutely correct. $\endgroup$
    – garyp
    Commented Sep 25, 2016 at 2:13
  • $\begingroup$ as the car crumples while hitting the tree, the "work done" comes from the car doing work on itself? As in, the internal forces set up inside the car as it crunches are acting over the displacement brought about by compression of the car? if there is some imperceptibly small elastic deformation in the tree, will it do some amount of work on the car? $\endgroup$
    – MattGeo
    Commented Jan 5, 2022 at 3:24
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An initially ice-skater pushes away from a railing and then slides over the ice. Her kinetic energy increases because of an external force F⃗ 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.

This is simply confused: This example is nothing but an elastic collision between the girl and the rail. The example given by John is appropriate but I do not agree with him that the arms behave differently from a spring. Of course the ultimate source of energy is chemical, but the girl as a system behaves exactly like a steel spring or a football hitting the rail.

enter image description here

If the skater (m = 30 Kg) had a spring and was travelling at v = 4 m/s she would have KE = 240 J and would bounce back at (roughly) the same speed. If she stretches her arms and pushes at the rail (like a spring) she can acquire the same speed of 4 m/s producing net work of 240 J. Because of human muscle inefficiency she'll have to burn 4-5 times that amount of calories $\approx 1080 J$

The body is compressed against the rail like an elastic ball, then it stretches back (in a ball the whole body, in a skater the arms, in a swimmer in a pool or in a high-jumper the legs) giving acceleration to the whole body. In the case of a xollision the compression is given by the initial v and KE, in the case of the push by the girl, by the calories burned in the muscles. The rail transfers no energy just exert a (passive) reaction to the action of the skater (it might budge a little), as per third law. The energy does not increase because of an external force acts on her.

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    $\begingroup$ Sir, if the railing is always stationary, how will the linear momentum of the girl-railing system be conserved?? $\endgroup$
    – user36790
    Commented Oct 30, 2014 at 10:57
  • $\begingroup$ What does it mean "the energy does not increase because an external force acts on her"? What energy are you talking about. As far as the work-energy theorem alone is concerned, the only energy form that exists is kinetic energy; and the kinetic energy of the free end of a spring does increase due to being accelerated by the reaction of the wall. From simple mechanics this energy seems to be provided by the wall. That does not imply, however, that the total energy of the system {spring + wall} has increased. $\endgroup$
    – gatsu
    Commented Dec 17, 2015 at 6:52
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At this point, we simply need to remember Newton's 1st law. For every force, there is an equal and opposite force. First off, make no mistake, the railing did provide a force for her. The railing pushed her at exactly the same amount that she pushed it. However, since it is cemented into the ground, the force was not enough to move it whatsoever. Work is contingent on two factors; Force, $F$, and Distance $d$: $$ W=Fd \\ W=F*0 \\ W=0 $$ The Force applied is completely irrelevant if the railing does not move. She performed no work on the railing, nor did the railing provide work on the skater. If there is no distance over which the Force is moving, no work is done.

The reason Kinetic Energy increases is because of the nature of the formula: $$ KE = \frac{mv^2}{2} $$ And: $$\Sigma F=ma$$ Kinetic Energy is not solely contingent upon Work energy. A force is still applied, just not over a distance.

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  • $\begingroup$ Is this example not like: a body is moved upward;the gravity opposes the motion by applying force & by Newton's 3rd law the body also exerts force on the earth and will do positive work... $\endgroup$
    – user36790
    Commented Oct 17, 2014 at 5:48
  • $\begingroup$ ...on the earth but instead of losing energy to the earth, that energy is converted to potential energy. Why?? How can gravity return the work to the body itself??? Coming to the skater,according to the book, the force is responsible for the energy increase,but doesn't do work ;why? Still +1 $\endgroup$
    – user36790
    Commented Oct 17, 2014 at 5:52
  • $\begingroup$ When the girl pushes herself off of the wall, she has already converted chemical energy (from eating) to me mechanical energy (her arms pushing) which is then converted into kinetic energy. If she were in a closed vacuum where there was no gravity, she would never stop. She would push off the wall and continue going at that speed. However, there is friction, even on ice, and she will be slowed by this friction. It will be converted, not to gravitational potential energy, but to heat energy due to the friction on the ice. $\endgroup$
    – Goodies
    Commented Oct 17, 2014 at 6:10
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    $\begingroup$ The railing did move. Or rather, the railing is connected to the earth, and she has very slightly changed the angular momentum of the earth (which changes the energy of the earth, but only by a minuscule amount; 99.99...9% of the energy ended up with the ice skater). If the railing didn't move, this would violate Newton's first law. $\endgroup$
    – Peter Shor
    Commented Oct 17, 2014 at 15:06
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I agree with garyp, this is an important case but I think there is an important fact which has not been noticed in the book nor in the above answers:

In this example (skater, also the example of the car which moves due to the static friction) the external force does not do work and this has an important physical meaning that is, no energy is being transferred from outside into the system (person or car).

Therefore this is a case for which the work-energy theorem ($W_{external}=\Delta K$) doesn't apply. In other words, kinetic energy has changed but external work is zero. So we should reformulate the work-energy theorem in order to include systems with internal structure. That is why in the book another term is introduced in the right-hand side of the energy equation: $W=\Delta K+\Delta U+ \Delta E_{th}+\Delta E_{int}$. The last term has not any work associated with it. Therefore this last equation is the law of conservation of energy rather than the work-energy theorem.

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Apply the work energy theorem. viz. $\Delta K +\Delta V = \Delta W$ where $\Delta K = K_f - K_i$ etc. Now we have $\Delta W = 0 ,~V_f = 0,~ V_i >0,~ K_i = 0$. Hence it follows that $K_f = V_i (>0).$

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