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The responses given so far are fairly accurate however, the question you should be asking goes to the experimental proof for the kinetic energy formula. Mathematically, the formulas for work and kinetic energy seem to function perfectly as taught. Unfortunately, there are at least 2 or 3 situations where it does not. No physics teacher ever looks at these ...


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To 'derive' conservation of Energy from $ \vec{F} = m \vec{a}$, we take a dot product ($\hat{i} \cdot \hat{i} = 1$) which means that we have one (scalar multi-variable non-linear differential) equation with potentially many unknowns. Energy is nice because it provides a common language with all the physical sciences, but in classical mechanics, it's mostly ...


2

You've already got some answers, but nobody mentioned Noether's Theorem yet. Noether's theorem maps a conserved quantity to each continuous symmetry. The relevant continuous symmetry needed to prove the conservation of energy is the one that leaves the laws of nature invariant, meaning the laws of physics don't change with time. Each continuous symmetry ...


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As you probably know, Newton thought that energy is linearly proportional to motion. The second law's original formulation reads: "Mutationem motus proportionalem esse vi motrici impressae" = "change of motion is proportional to the force impressed" It was Gottfried Leibniz, as early as ca. 1680, who first realized that KE (m = unitary mass) is ...


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I suspect, though I'm not sure, that a nineteenth century French mathematician and scientist, Gaspard-Gustave Coriolis, is your man. He was the first to define the notion of "work done" and even kinetic energy. His wiki reads: In 1829, Coriolis published a textbook, Calcul de l'Effet des Machines ("Calculation of the Effect of Machines"), which presented ...


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The master of the law of energy conservation was Hermann von Helmholtz (1821-94). See his classic 1847 paper "Über die Erhaltung der Kraft," translated into English as "On the Conservation of Force." (He called energy force.)


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Suppose you have a constant angle $\theta$ slope in the original frame (we suppose that the transitions from horizontal movements to the slope is quasi-instantaneous). Call $x$ and $x'$ the horizontal displacements in the original and moving frame. Call $T$ the total time for going to $z=h$ to $z=0$. Then you have $x'= x- v_0 T$ With $ x= h \cot \theta$, ...


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I believe the difference comes from the fact that forces can do different amounts of work in different reference frames. In particular, the normal force by the ramp does no work in the "lab" frame, but does do work in the moving frame (since there is a component of velocity that is now parallel to the normal force). I don't think you accounted for this work ...


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Some conservation laws are related to conservation of angular momentum. There is a famous example (from Feynman if I recall correctly), where you assume an infinite flat space and conservation of angular momentum about any point, and then you get conservation of linear momentum for free. Intuitively, to get say the $x$ component of linear momentum is ...


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the waves will obliterate each other but they will still exist, they just won't be moving they would just change form (energy cannot be destroyed it can only change form) so when the waves meet they will cancel each other so sound will change to potential and kinetic will change to sound or whatever


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In General Relativity, energy momentum flows from one region of spacetime to another. But there isn't necessarily a natural "total energy of the universe." It might help to contrast General Relativity with other theories. In Newtonian mechanics, a particle might gain kinetic energy while a corresponding gravitational potential energy decreases, thus you ...


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There is no extra energy - the blueshift is due to an increase in the observed frequency, not the actual frequency. The light is still being emitted at the same frequency so the same amount of energy is used.


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The energy is conserved but it becomes "lumpy"- more in some places (directions), less in others. Total over all directions is the same.


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Yes - the coiled spring has a certain amount of (potential) energy. When it gives up the energy to the ball, you could say the ball does negative work on the spring, so it loses (potential) energy.


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Write down the potential and kinetic energy as a function of position. When the spring is in the middle of the motion, all is kinetic. When it is at the extreme of the range, all is potential. Somewhere between these two extremes, the potential and kinetic energies will be the same; their sum should always be constant (when there is no loss). Recall that ...


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A simple picture here would be that dissipation leads loss of mechanical energy. But angular momentum has to be conserved. The lowest possible mechanical energy with a given value of the angular momentum of fluid inside a container is that of rigid body rotating around the axis of largest moment of inertia.


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You apparently mean "simulate" when you used the word "model". You'll need two things to accomplish this: A better rotational integrator than the one presented in this answer to your other question, and A physical model of a system that loses energy while conserving angular momentum. Regarding the first item, that integrator is not bad. It has the ...


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This formula is used for conservative forces like gravitational force. This formula is used when the force is not constant i.e. variable force. This formula conveys that conservative force is equal to negative potential gradient. This formula establishes relation between a vector quantity and a scalar quantity(PE).


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You should think of the formula the other way around, i.e. $$ \mathrm{d}W = F\mathrm{d}x$$ which means that the infinitesimal work done along an infinitesimal path is just the force $F$ times the length $\mathrm{d}x$ of the path along which the force was exerted. If we are now given a real path $\gamma : [a,b] \to \mathbb{R}^3$, the total work done along ...


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It is basically the same as two ordinary objects colliding and sticking together. The combined object's rest mass is the sum of the total energy of the original objects in the center-of-mass frame, which is their rest mass/energy plus their kinetic energy in that frame. Some of that would be carried away as gravitational radiation, but typically only a small ...


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The quantity $mv^2$ is not constant. You can argue this by an example. Consider a ball in a uniform gravitational field. If you drop the ball, the quantity $mv^2$ changes as it falls, so it's not constant. Or, more simply, $mv^2$ is twice the kinetic energy. Think of any system where the kinetic energy changes, and you've shown that the quantity can't be ...


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This is a very interesting question. The problem is simpler in kinematics. However, if we view it as a problem in dynamics, invoking forces and Newton's laws, then the question becomes a natural consequence and the answers become rather complicated. The question is one of reference frames, in kinematics. The collision is viewed from the laboratory frame of ...


1

Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out Ok, so sum of the forces is 0 and the acceleration is zero. and the book moves with a constant velocity. Spooky. Was the book moving initially? ...after the book has been lifted, and it has come to rest once again. According to ...


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There is a small error in your math by the way - the factor of $4$ should be a factor of $1$ - check the denominator of the quadratic equation. Why is this so unreasonable? At a glance, it looks like the momentum of the sail is a bit larger than what it started out with. If I make the reasonable assumptions that: The sail is non-relativistic ...


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You say: Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out and the book moves with a constant velocity. so I'm guessing your reasoning is that the net force on the book is zero so the amount of work done on the book is zero. And you are absolutely correct - no work is done on the book and ...


1

Let's say the book starts and stops from rest, as I believe you are assuming. The motion within this interval is unimportant, as you'll see. The increase in gravitational potential energy of the Earth-book system came from your body. You did positive work on the system since your hand force and displacement are in the same direction, resulting in an ...


0

If you lift the book with a force that is exactly equal (but opposite) to the force of gravity acting upon it the book won't go anywhere. After all, that's exactly what a table does when you leave the book on it. To get to book going you need to lift with a force greater in magnitude than that of gravity, resulting in a net force and an associated ...


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In inelastic collisions, kinetic energy is not conserved, so I'm going to assume you mean a totally elastic collision since you say energy is conserved. O.K, so when the ball hits the wall, the speed of the wall before and after is 0, so that means the kinetic energy of the ball is conserved and thus the magnitude of the velocity is the same before and after ...


1

You can zoom in and look at just the wire; if you do that E and J are parallel(given some resistivity, we can call it z-hat) and B is parallel to the surface of the wire(we can call that theta-hat), so by definition you do get a poynting vector into your wire(z cross theta = -r). This is a transfer of energy from the flowing charges to the heat in the wire. ...



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