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Comments to OP's post (v4): OP is trying to prove via Noether's theorem that no explicit time dependence of the Lagrangian leads to energy conservation. OP's transformation seems to be a pure horizontal infinitesimal time translation $$\tag{A} t^{\prime} - t ~=:~\delta t ~=~-\epsilon, \qquad \text{(horizontal variation)}$$ $$\tag{B} q^{\prime i}(t) - ...


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The easier way of doing this is to just consider a generic transformation, G, such that the canonical co-ordinates of the Hamiltonian are shifted as below: $$ \delta p = \frac{\partial G}{\partial q} \delta \lambda$$ and $$ \delta q = - \frac{\partial G}{\partial p} \delta \lambda\,,$$ where $\lambda$ is the transformation parameter determining how much of ...


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The compression pulse that propagates through the metal spheres of Newton's cradle are not ordinary sound waves. They are approximate solitons (a nonlinear wave form that balances dispersion against nonlinearity). It is this property of soliton pulses that is responsible for the observed behavior. More Details: Newton's cradle is a physical manifestation ...


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The simple answer is that not only momentum but also energy needs to be conserved. This puts constraints on the number of balls that can be activated in the cradle. Note that this does not always give a unique solution either. But it enforces that $ n $ balls to $ n $ balls is a "stable" solution.


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Notice that in the finite approximation the vector PR is not perpendicular to the radius, so work is done on it. By the drawing, this work is of the opposite sign that that in QR, so they both compensate to zero. I'll leave to you to compute both if you are really interested. Another way to see it, the gravity force is derived from a conservative field, so ...


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Work done by a central force is Zero. At every moment the force is perpendicular to the displacement of the test particle. If you see your diagram it's very easy to see that at the final and initial position of the particle the force is not in the same direction. What he actually does is to assume that P and Q are actually arbitrarily close. So now the ...


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There are already good answers here, but I'm afraid that to the best of my knowledge, Diracology's (and indeed Halliday-Resnik-Krane's) expression of the potential energy is not correct. I would like to point to this paper by Lior M. Burko which focusses on the subtleties of the derivation of the kinetic and potential energy of the string as a whole and ...


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Take for example the double-slit experiment interpreted in the Copenhagen sense. The particle leaves as an object with mass, yet passes through the slits as a massless wave, only to collapse again as a particle. We can consider this example as a generalisation of the principle of anti-realism posited by Bohr. Where does the energy "go" when the ...


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Short answer: no. I'll give some context with the details of the simplest examples. In the context of conservation laws, "energy" refers to the Hamiltonian. In classical mechanics, a quantity without explicit time dependence is conserved iff its Poisson bracket with the Hamiltonian is 0. In quantum mechanics, quantities are promoted to operators on a ...


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The second derivation is correct, as explained by Diracology. However, the first derivation is 'sort of' correct, in the sense that the location of potential energy can be ambiguous. For example, consider the three following systems. A mass on a stretched spring. A mass sitting on a table. A charged mass next to another charge. These three systems have ...


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The energy of an element of a traveling wave is not constant. Halliday-Resnick-Krane is right. For a string of density $\mu$ and tension $T$ the kinetic energy of an element $dx$ is $$dK=\frac 12\mu dx\left(\frac{\partial \xi}{\partial t}\right)^2.$$ For the potential energy we have $$dU=Tdl,$$ where $dl$ is the stretched amount of the string. A small ...


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Energy is proportional to amplitude squared. The energy in the wave is spread out over the surface of a sphere. The area of this surface increases as the wave propagates outwards from the source and is proportional to $r^2$. So the intensity of the wave (power/area) decreases in proportion to $1/r^2$.


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I assume you are asking about a light sphere falling from infinity towards a heavy one. Well, the potential energy at $r$ is $$P = - \frac{GMm}{r}$$ And kinetic energy is $$K = \frac{mv^2}{2}$$ Total energy is zero, so $P=-K$ or $$v^2 = \frac{2GM}{r}$$


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There are different ways of stating conservation of energy and accounting for energy, which can make the issue confusing. One such statement is "the total energy of an isolated system is constant". This is true, and is the simplest way to state conservation of energy. This form of conservation of energy is the earliest taught. There's another way of stating ...


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Feynman knows better than that, actually, and this is one of these cases were undergrad textbooks like Feynman's come dangerously close to being not only incomplete but deceptive. Conservation of energy is a consequence of time translation symmetry (see Noether's theorem) and he mentions it in chapter 52 without even an attribution (bah!). If that holds, ...


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Energy is conserved in an isolated system. The energy of the system is constant. But it can flow from one part of the system to another. Energy is not conserved in the part. The energy in the part can increase or decrease. No energry is created or destroyed. It just moves to a place where you stop counting it.


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To begin with, $v dv= a ds$ is the differential equation that relates velocity with variable acceleration. $a=(G*\mathrm{massOfEarth})/s^2$ Substitute it in the first equation and integrate for $v$, 0 to $v$ and $s$, distance between the moon and earth to 0 Hope this is right


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Instead of forces, you work with conservation of potential and kinetic energy. In this case, the kinetic energy that gives this force is $-\frac{Gm_1 m_2}{r}$. The difference in potential energy is transformed into kinetic energy. With forces, you'd have to go the long way around (integrate the acceleration over time to get the velocity and figure out the ...


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This question is easilly answered by considering the gravitational potential of earth, and invoking conservation of energy. The potential is $V(r) = -G\frac{m_1 m_2}{r}$ and the kinetic energy of the moon will be given by $$ K = V(r_{moon}) - V(r_{impact}) $$ where $r_{moon}$ is the current distance from the center of the earth to the center of the moon, ...


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Einstein's equation does not indicate how mass can be converted into energy. It only says that mass and energy are equivalent. Nobody has as yet found a way to convert mass directly and completely into energy. The best that can be done at present is nuclear fission or fusion reactions. Fission is already well established in the nuclear energy industry. ...


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I've noticed that you have referred to energy conservation a couple of times in your question and comments. So there is something that we should make very clear. Conservation of energy in general does not mean the conservation of any given kind of energy. Potential energy can turn into kinetic energy can turn into nuclear, chemical, thermal or acoustic ...


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Without further information it is not possible to predict if the two bodies will stick together or coalesce. This depends on the nature of the materials which come into contact, and the presence of any interlocking mechanism (such as used to couple carriages of a train). Momentum is always conserved in collisions; kinetic energy is not always conserved. ...


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Energy is not conserved during an inelastic collision so you cannot find $v$ using conservation of mechanical energy concept.


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I think your problem is that you didn't change the units in the constant g. It has a value of approximately $9.8ms^{-2}$. Notice that it depends on meters. To obtain the correct result, you should use $980cms^{-2}$. Notice that this constant is off by a factor of 100, so that the result (after the square root) is off by a factor of $\sqrt{100}=10$.


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Yes you are. If a force is conservative, its work does not depend of any path between any points $A$ and $B$. Since the work integral can depend only on the initial and final points themselves, we define $$W_{A\rightarrow B}=\int_A^B\vec F\cdot d\vec r\equiv U(A)-U(B).$$ Now define the mechanical energy as $E=K+U$ so that $$dE=dK+dU.$$ Suppose there are two ...


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Yes, at the fundamental level all energy terms are normally either kinetic or potential energy. The only demonstration of this that I know of requires a tool called the Lagrangian, which you might not be familiar with. But maybe you can at least get a flavor of how it goes. The Lagrangian, very briefly, is a particularly useful way to represent all the ...


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You're not the first, nor the last, to find the phrase "power flow" somehow wrong. For example, from W J Beaty's article on electrical misconceptions: ELECTRIC POWER FLOWS FROM GENERATOR TO CONSUMER? Wrong. Electric power cannot be made to flow. Power is defined as "flow of energy." Saying that power "flows" is silly. It's as silly as saying that ...



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