jawab

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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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Oct 17 '14 at 9:29

John Rennie

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Oct 17 '14 at 14:16

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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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Oct 17 '14 at 12:43

garyp

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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? –

gatsu

Dec 17 '15 at 6:41

Absolutely correct. –

garyp

Sep 25 '16 at 2:13

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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.

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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. I