Can We Do Better Than Mechanical Energy Conservation?

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Note: It is assumed that the reader has read part I of the series.
Introduction
The ambiguity and flaws discussed in part I can be resolved using the law of conservation of energy.  In the words of Richard Feynman,
There is a fact, or if you wish, a law, governing all natural phenomena that are known to date. There is no known exception to this law—it is exact so far as we know. The law is called the conservation of energy. It states that there is a certain quantity, which we call energy, that does not change in the manifold changes that nature undergoes. That is a most abstract idea because it is a mathematical principle; it says that there is a numerical quantity that does not change when something happens. It is not a description of a mechanism or anything concrete; it is just a strange fact that we can calculate some number and when we finish watching nature go...

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So can we say that conservation of energy is one size fits all law (or the perfect law)?
 
Let's just say that one can do more with it than one might have originally thought. Near the end of section 4.2 of the reference cited in the article, Feynman shows how one can use energy conservation to solve statics problems. It fits there, but it doesn't fit when one needs to predict the force between two charged particles. Coulomb's law is an experimental result that can be used to write an expression for the change in electric potential energy when the two charged particles move relative to each other but cannot be derived from energy conservation.
 
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I have also read the first part but do not understand what the point is.
Let a point mass m be undergone with a force ##\boldsymbol F##. Assume that it is reasonable to split this vector in two parts
$$\boldsymbol F=-\nabla V+\boldsymbol F^{nonconservative}.$$
Then one has $$\frac{d}{dt}(T+V)=(\boldsymbol F^{nonconservative},\boldsymbol v).$$ If the right side is equal to zero then the energy is conserved: ##T+V=const##.
That is all
Sorry if my remark is inappropriate.
 
Hi,

a doubt about the section Work done against friction (part II). As highlighted in point 1. the system examined is '(sliding) block + surface'.

Now the (kinetic) friction force ##\vec f_k## involved in 'block+surface' system analysis is 'internal' and, as far as I can see there, is just one: namely the friction force ##\vec f_k## that the 'surface' component of the system applies on the block.

What about the other of the couple: namely the friction force ##\vec f_k## that the 'block' component applies on the 'surface' component ? It just exists in accordance with Newton III principle and I suppose it has to be taken in account in the analysis as the former
 
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All that is correct but I do not understand your "doubt". When the system is the block + surface according to Newton's 3rd law, the block exerts a force ##\vec f_k## on the surface and the force exerts a force ##-\vec f_k## on the block. The heat generated at the interface is ##Q=|f_k \cdot \Delta \vec{x}_{rel}|## where ##\Delta \vec{x}_{rel}## is the relative displacement between the surfaces. Kinetic energy is converted into heat, one system one conversion into heat.

To draw an analogy with other familiar cases, in the case of the falling flower pot, the Earth-flower pot relative displacement results in a change of potential energy of the two-component system. Kinetic energy is converted into potential energy, one system one conversion into potential energy.
 
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kuruman said:
When the system is the block + surface according to Newton's 3rd law, the block exerts a force ##\vec f_k## on the surface and the force exerts a force ##-\vec f_k## on the block.

Maybe there is a typo, it should read "...and the surface force exerts a force ##-\vec f_k## on the block"

kuruman said:
The heat generated at the interface is ##Q=|f_k \cdot \Delta \vec{x}_{rel}|## where ##\Delta \vec{x}_{rel}## is the relative displacement between the surfaces. Kinetic energy is converted into heat, one system one conversion into heat.
You're definitely right: we have to consider in the calculation the relative displacement