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## New answers tagged energy-conservation

2

SECTION A : Non-relativistic conservation of energy The work done by the non-relativistic force $\:\mathbf{f}\:$ per time unit, that is the power produced or consumed, on a particle moving with velocity $\:\mathbf{v}=d\mathbf{r}/dt\:$ is \begin{align} \dfrac{dW}{dt}=\mathbf{f}\circ \mathbf{v}=&\dfrac{d\mathbf{p}}{dt}\circ \mathbf{v}=\\ ...

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The general statement of the conservation of the energy is $$W_{\textrm{non-cons}}= (T_{\textrm{init}} - T_{\textrm{fin}})_{\gamma}$$ hence the work done by any non conservative force (in this case the friction) is equal to the difference in kinetic energies along the path $\gamma$. Friction is a non-conservative force, that is, by definition, there is no ...

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I think one has to be very careful when talking about "particles popping in and out of existence". This interpretation is only sort of fine in flat-spacetime QFT, where the Minkowski metric is time-invariant, so has a global timeline Killing vector. The definition of a particle depends on the notion of there existing time invariance! Since black hole ...

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Energy and momentum are conserved at every vertex of a Feynman diagram in quantum field theory. No internal lines in a Feynman diagram associated with a virtual particles violate energy-momentum conservation. It is true, however, that virtual particles are off-shell, that is, they do not satisfy the ordinary equations of motion, such as $$E^2=p^2 + m^2.$$ ...

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I suspect this is an example of the spinning-egg problem, in which a prolate spheroid (such as an egg) spun on a table about one of its "short" axes will tend to "stand up" so that it's spinning about its long axis. A few explanations have been proposed for this phenomenon, most notably: H. K. Moffatt & Y. Shimomura, "Spinning eggs — a paradox ...

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This chapter is on the conservation of energy. Ideally the machine he shows will work. If we jump ahead in the book to where we know about conservation of energy, we see that energy gained by one side is lost by the other. But in practice, some energy is lost to friction. The little extra weight is needed to overcome friction. You could also overcome ...

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In mechanics, a mass $m$ experiences a force $\textbf{F}$ along some path $C$. The work done on the mass is given by $$W = \int_C \textbf{F} \cdot d\textbf{r},$$ such that the energy of the mass increases by $W$. Positive work corresponds to energy being added to the system in question (which is inevitably taken from the surroundings). Edit: To answer ...

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I will try to give you here the proof of work-energy theorem. We have that $\bar p = {m \bar u \over \sqrt{1- {u^2 \over c^2} } }$ the relativistic momentum. We define the force as $\bar F ={dp \over dt}$. So the work is: $W= \int \bar F \cdot d \bar l =\int {d \bar p \over dt} \cdot d \bar l = \int {d \bar p \over dt} {\cdot d \bar l \over dt } dt = ... 1 My question is, are we really must say that the energy of the tide and the loss of the kinetic energy of the moon are equal? The answer is obviously "YES." It must be so. I would refer at this point these words; Nature is relentless and unchangeable, and it is indifferent as to whether its hidden reasons and actions are understandable to man or not.- ... 3 Here is the elementary particle table from which all others are built up , the standard model of particle physics. Which shows the conserved quantum numbers that characterize the particles (columns and rows have quantum numbers assigned too) plus the measured masses. The quantum numbers have to "annihilate" to have an annihilation event, i.e. they should ... -4 I wrote a paper in which I show that the electric field of a charged particle can be described as composed of two quanta. Two types of quanta are enough to describe the electric field, the magnetic field and photons. It was shown, why electron does not fall into nucleus and why annihilation happens between particle and anti-particle. A more deep explanation ... 0 Another way to look at this is to realize that whenever you move charge, you produce a magnetic field. Even a straight wire will have a certain amount of self-inductance. Initially, this will provide a limit as to how fast the current can increase. As the voltages equalize, the interaction with the magnetic field will actually cause the current to continue ... 0 Your presentation is a little confused. I would read your book again very carefully, taking care to understand the meaning of each variable. The integral that you write for potential energy is the definition of potential energy, if$F$is taken to be an internal force, that is, a force between two objects within the system. For example for free fall at ... 1 The whole energy-concept is a reformulation of Newtons laws. Starting from$\vec{F}=m.\vec{a}$, you could wonder about the effect of a force during a displacement. You call the concept 'work' and do the math $$W=\int_{\vec{x}_1}^{\vec{x}_2} \vec{F} d\vec{x} = \ldots = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$$ To save yourself some work you define $$T = ... 2 Even an ideal capacitor cannot be losslessly charged to a potential E from a potential E without using a voltage "converter" which accepts energy at Vin and delivers it to the capacitor at Vcap_current. If you connect an ideal voltage source via a lossless switch to an ideal capacitor which is charged to a lower voltage, infinite current will flow when the ... 1 For the sake of simplicity assume the binaries' motion to be circular. Then you can use the formalism of the circular restricted 3-body problem (CR3BP) to model the motion of the test particle m_3. Your Lagrangian will be time-independent and the conserved quantity (Jacobi constant) can be evaluated at infinity to give an equation for conditions of escape, ... 0 Energy wise everything works out fine if you choose to identify energy with mass. What is confusing is the identity of the particles. Mass is an identity, and if you change that you changed the identity. So start with an electron and a proton, combine them and yes you get a smaller mass but what is the identity of the new creature. It cannot be an electron ... 2 There are three types of matter/energy we consider when calculating how the universe expands: Matter - both normal matter and dark matter Radiation Dark energy We measure the expansion of the universe using a scale factor that we normally denote by a. The scale factor increases with time as the universe expands, and if we look backwards in time we see ... 0 The expansion of space is like stretching a rubber sheet. (Don't take this analogy too seriously. It works for this explanation but fails elsewhere.) The mass of the rubber sheet stays the same as it gets bigger. Space expands, but mass does not increase with it. 0 The energy difference between two arbitrary speeds v and v' cannot be found by \frac{m(v'-v)^2}{2} except for the case where v=0. So your calculation about the energy change between 50m/s and 25m/s is incorrect. You instead have to calculate the initial and final energy states and subtract directly.$$E(50) = \frac{(20kg)(50m/s)^2}{2}$$... 1 The formula for kinetic energy is \frac12mv^2. If your initial velocity is v_i and your final velocity is v_f, then your initial kinetic energy is KE_i = \frac12 m v_i^2 and your final kinetic energy is KE_f = \frac12 mv_f^2. The difference is \Delta KE = KE_f - KE_i = \frac12 m(v_f^2 - v_i^2) It appears you're thinking that you can define ... 0 Here it explains what I think you are asking: We work with a formulation of Noether-symmetry analysis which uses the properties of infinitesimal point transformations in the space-time variables to establish the association between symmetries and conservation laws of a dynamical system. Here symmetries are expressed in the form of generators. We ... 1 I think I can remember the derivation for a conservative force field in classical mechanics, wich is a somewhat stronger assumption than pure time-translation invariance. Let \vec{F} be a conservative force field, that is$$ \nabla \times \vec{F} = 0 $$or alternatively$$ \phi := -\int_\gamma \vec{F} \cdot d\vec{a} $$does not depend on the path \gamma ... 0 I believe this is discussed in Nonlinear Dynamics and Chaos by Strogatz. I don't remember the details, but it's worth looking at his discussion of an energy function. 1 (After possibly introducing more variables) then OP is essentially considering an autonomous system of n coupled 1st order ODEs$$\tag{1} \frac{d\vec{z}(t)}{dt}~=\vec{f}(\vec{z}(t)), \qquad f: U \to \mathbb{R}^n , \qquad U\subseteq \mathbb{R}^n, $$i.e. without explicit time dependence, so that the system (1) possesses time translation symmetry. OP is ... 0 You have to think of what happens when object B is "converted to energy". Since there is no such thing as raw energy (dark energy excluded) B has to be transformed into other particles and or kinetic energy. As many of the other comment have mentioned the kinetic energy also has a gravitational pull, so initially after the "conversion" nothing has really ... 0 By "converting to energy," I'm assuming you mean converting part of the mass of the system to photons. These photons will quickly zoom off to large distances, making them irrelevant gravitationally. The original gravitational potential energy was caused by the original masses A and B warping spacetime. This warping is not instantaneously changed when mass B ... 0 I would need to look into this more, but I believe gravity couples to energy rather than the Lorentz invariant mass m. If this is true, then it doesn't matter if you view the mass of particle B as mass or energy. -3 Here you have to be clear how you are looking at the energy. Of course all energy can take the form mass times c2. In this particular case of object A and B we have a total energy equal to mc2 where m is mass of A plus Mc2 where M is mass of B. With this you subtract the gravitational potential energy since it is attractive. You then get the system total ... 2 It is better to think of the equation E =mc^2 as a true equality rather than a conversion. Mass is energy. If one has that mindset, then it is intuitive that energy has a gravitational field. A hot cup of tea weighs more than a cold one. 1 here is what I think he is trying to explain. You have two machines A and B. A is reversible B is not necessarily reversible. Both these machines are placed side by side. let both machines A and B be initially horizontal. (Left , right pans of machine A will hold 3 unit and 1 unit mass respectively. Similarly, the left and right pans of machine B will hold ... 0 Any aircraft has potential energy due to its altitude, and kinetic energy due to its velocity. The sum of these is its total energy. If the stick is pushed forward or back, the aircraft simply trades potential energy for kinetic energy or vice-versa, exactly like an earthbound roller-coaster. In order to descend to a landing, the aircraft must dissipate ... 0 You mention nothing about the drag on the airplane wing, and although I'm no aerodynamicist I'm pretty sure that an airfoil can't produce lift without producing drag. Add in drag, and your apparent perpetual flight machine fails. Ignore the propeller: think about a glider. Gliders always have a glide slope: they can never maintain speed at the same time as ... 0 When the lift produced by the wing is greater than its weight, the wing begins to accelerate vertically. I can see where you're coming from - essentially questioning the physicality of wing lift on the basis of thermodynamics. In a simple conception of the situation, this does seem like a problem. But let's look more closely at the gravitational ... 1 When you push the block the block 'pushes' you with the same force and you both gain equal and opposite direction momentums. Both block, and you have now some momentum and hence kinetic energy. The work done on both you and the block is:$$ W=\int F \,dx$$where F is a force applied. It must be equal to the total kinetic energy of you and the block (if ... 0 In the pure hypothetical situation you pose, you, the block, and no forces from gravity, friction or wind resistance - the moment you 'push' the block you will impart a 'packet' of energy for only as long as you can extend your arms and maintain contact with the block. For at that moment the block will move away from you, and you will move away in the ... 0 In a coordinate system rotating at constant angular rate \omega, neither energy nor angular momentum are conserved and one has coriolis and centrifugal forces. The bead is forced outwards by the centrifugal force: the energy increases by the work done on the bead by the rotating system. In fact, since the Hamiltonian$$H=\frac{p^2}{2m} - ... 1 An elastic collision is defined as one which conserves energy. When you jump against a wall, most of your kinetic energy is dissipated as heat into your tissue as your legs and muscles absorb the impact. Therefore, energy is not conserved so by definition this is an inelastic collision. 2 A significant source of power loss in a transformer is the induced eddy currents in the core. Just as the varying magnetic field induces current in the secondary coil, it can also induce currents in the core itself. These currents do nothing but dissipate energy, and so are to be avoided. To reduce eddy currents, you either build your transformer out of a ... 1 I suspect most of the loss is simply resistive heating in the coils and possibly some heating due to hysteresis in the iron core rather than coupling of the magnetic field to any external power leakage 1 A magnetic field is determined by the current and a changing electric field. And it has energy just for existing. It takes energy to make the magnetic field, for instance to increase the current, and you get energy back when magnetic fields decrease in strength. For a common inductor the magnetic field and associated stored energy are due solely to the ... 1 Energy, in the context of the universe and the models used in describing it , is in the stress energy tensor, which contains the energy/mass transformations. The accepted at the moment model for the universe is the Big Bang model, as summarized in this plot , is a good fit to the available observations using all known physics to date. Known physics is ... -2 This looks like another Chicken or the egg problem. I think I have the answer. The amount of energy and matter are constant. Remember there is only energy. Matter has been called "energy at rest." The two often change places but the quantity in the infinite universe is the same. The evidence for the Big Bang was recently presented as a residual background ... 0 My interpretation: The definition of a perpetual motion machine is one from which more energy is produced than consumed i.e. getting something from nothing. Classic examples of perpetual motion machines are those which involve some sort of repeating cycle, and there is an expectation of excess energy in some form at the conclusion of each cycle. Among the ... 2 The energy is not entirely from from the Big Bang since a lot of the material had to come from supernovas. This would mean that some of the energy came from the supernovas separating the dust that makes up the objects. Yes, you can harvest energy from the falling objects, if they fall onto the targets. 2 Total energy is conserved. Let$E$stand for total energy. Let$T$stand for kinetic energy. Let$U% stand for potential energy. $E = T + U$ If $T$ increases and $E$ is constant (which it is in a closed system due to conservation of energy), then $U$ must decrease. Simply put, the kinetic comes from the potential energy. Potential energy comes from ...

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E is the initial mgh, a constant. V is mgx, a function of x.

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The kinetic energy you calculated is in joules, not kilojoules. So x is 1.0m. You can calculate the peak acceleration using the elastic force: F=kx=ma; a=kx/m

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Half of the energy is lost to the battery's internal resistance (or other resistances in the circuit).if you try to consider an ideal battery with 0 internal resistance, the notion of charging the capacitor breaks down.since the capacitor and the battery are connected by a (0 resistance) wire, their voltages are the same the instant they are connected, no ...

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At the moment the circuit is completed, the capacitor has zero voltage, while the supply has $V$. This voltage difference creates an electric field that accelerates charges. This acceleration sets up a current. As the current flows, the capacitor charges until the voltage reaches $V$ as well. At this point there is no voltage difference. But the ...

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