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Trimok is correct. Insofar as gravitation is a field theory with physical things represented by conserved quantities, there is no conserved energy tensor in GR. Another way to look at it is to simply assume energy must also exist in GR, and then deduce that because there is no tensor representation of it, then it must be non-local. Energy can disappear here ...


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$Ke = \frac{1}{2}m ( \int_0^{2\pi} (Rw_1-rw_1cos(\theta)+rw_2cos(\theta))² +(rw_2abs(sin(\theta))² d\theta) $ $Ke = \frac{1}{2}m \int_0^{2\pi} R²w_1²+r²w_1²cos²(\theta)+r²w_2²cos²(\theta)-2Rw_1²rcos(\theta)+2Rw_1rw_2cos(\theta)) -2rw1cos(\theta)rw_2cos(\theta)+r²w_2²sin²(\theta) d\theta$ $Ke = \frac{1}{2}m ...


2

As dmckee's comment says, your question doesn't really have an answer. However it's exactly the sort of question that fascinated me in my time as a teenage physics enthusiast, and I think it touches on some really interesting aspects of physics. To get a grip on this you need to understand how matter is described by quantum field theory. This is not ...


2

There are already many good answers. Besides the fact that the standard definition of work directly relates to the work-energy theorem and the notion of potential energy, here is a geometric argument. I) The force $F_i(x,v,t)$, $i\in\{1,2,3\},$ transforms as $(0,1)$ co-vector $$\tag{1} F_i ~=~\sum_{j=1}^3F^{\prime}_j \frac{\partial x^{\prime j}}{\partial ...


3

For starters, these are not the same thing. The integration by parts rule makes this fairly obvious: $$\int_i^f y\,\mathrm{d}x = y_f x_f - y_i x_i - \int_i^f x\,\mathrm{d}y$$ But then you might be wondering what makes $\int \vec{F}\cdot\mathrm{d}\vec{s}$ the "right" definition for work while $\int \vec{s}\cdot\mathrm{d}\vec{F}$ is the "wrong" one. In a ...


1

Your Hamiltonian has the "potential" $U = V + F$, just because $F$ is time-dependent doesn't make it not part of the potential. You are correct that the Hamiltonian is not conserved. The Hamiltonian is the total energy of the system if the Lagrangian does not contain terms linear in the velocity or when the generalized coordinates in the Lagrangian do not ...


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The reason the relationship $$ W=\int\mathbf s\cdot d\mathbf F $$ doesn't work is because Work is defined as the result of a force $\mathbf F$ on a point that moves along a distance. The point follows a curve $\mathbf s$ with a velocity $\mathbf v$. The small amount of work, $\delta W$, that occurs of the instant of time $dt$ is $$ \delta W=\mathbf F(\mathbf ...


4

Because, according to your definitions, if I strain a rubber bar with constant force until it rips apart, I haven't done one joule of work to it.


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Quantum-mechanicaly and relativisticaly the energy of a given object is never removed completely (Heisenberg uncertainty relations, relativistisc rest mass) Assuming we talk about kinetic energy, kinetic energy is defined with respect to a specific reference frame. This reference frame can be related to another object or not. The kinetic energy can be ...


0

Take the reference frame as centered in the fixed axis. The $R$ that connects the origin to the centre of the spinning disk forms an angle $\phi$ with the horizontal. Now, inside the disk of radius $r$, the angle of a certain point mass is given by the angle it forms inside the spinning circle, which we'll call $\theta$. Now take as generalised coordinates ...


5

Let me clarify some confusions in the notation that other answers have alluded to but not clearly mentioned. Historically, physicists liked to talk about two different definitions of mass The first is the rest mass of a particle $m_0$. This is the mass of the particle when it is at rest. For example, the rest mass of the electron is $(m_0)_{electron} = ...


3

The equation $$E^2=(mc^2)^2+(pc)^2$$ represents the correct energy-momentum relationship. It gives the total energy $E$ for an object of invariant mass (rest mass) $m$ that is observed to move with momentum $p$. This equation is applicable regardless whether the object is observed to be in motion ($p \ne 0$), or is observed to be at rest ($p = 0$). In the ...


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I agree with answer of ACuriousMind, but I think it might also help to think about it like this.... $E^2=m_0^2c^4+p^2c^2 =m^2c^4$ where $m_0$ is the rest mass and $m$ is the relativistic mass (or inertial mass), defined as $m = \gamma m_0 = m_0 / \sqrt{1 - v^2/c^2}$. The relatavistic mass increases as the momentum of the mass increases. At rest the two ...


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Both are correct, within the domains for which they are correct. More seriously, the general relation $$ E^2 = m^2c^4 + p^2c^2$$ holds for all objects, whether they have mass or not, whether they are moving or not. The special case $E = mc^2$ is for $p = 0$, i.e. objects which do not move, as you said. The special case $E = pc$ is for objects which have ...


1

You should be able to get the same result by using the electrical work formula - but note that you need to integrate since the force changes with position. That's really all the potential is - it is the integral of force for unit charge. That's why force has the $1/r^2$ relationship while potential has $1/r$ (with appropriate signs and constants...). ...


1

2 is correct and so is 1. force is invariant under change in frame. However work varies under change of frame.This is because the displacement of a body changes with change in frame. For e.g., a body moves 5 meters with respect to a stationary body. However if there exists a frame that moves 2 meters at the same time then according to that frame the previous ...


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Your first question is answered quite easily - the force may be the same in both frames, but the distance traveled is not. In the frame where the change in KE is greater, the object will also cover a greater distance. So the change in KE is not frame invariant, and neither is the work done.


3

If the string is massless and taut, then the wave velocity is infinite - that is, a component of the force at one mass will immediately be felt at the other mass. But to answer your first question, not all the force of the impact will be transmitted along the string, as made clear by this diagram: As for your second question: momentum for the system is ...


2

No, if $S$ is really the only thing you know about your system then there is no way to know its energy. There is no relationship between the energy and the entropy that doesn't involve some other quantity such as temperature. ...but surely you know something about your system, other than its entropy? I mean, you must know something about what it's made of, ...


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The earth also gains the positive work the same way the rough floor does, only it's not apparent at first. To understand how, let's go to the microscopic level to see what actually happens when you rub an object against a rough surface. You have the molecules and atoms of the object above, and those of the surface below as in the image shown. The ...


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(Summary: In this post I argue that you need at least an energy of $m v_1^2(1-\cos(\theta))$ in the idealized kinematics situation to deflect an asteroid of mass $m$ and velocity $v_1$ by an angle $\theta$ using rockets, without changing the magnitude of $v_1$.) Energy isn't the be-all and end-all of motion. The problem is that momentum also has to be ...


2

Newton's 3rd law is nothing but the conservation of momentum. Recall that force can be expressed as the time rate change of momentum, $$\vec{F} = \frac{d\vec{p}}{dt}$$ In order for the conservation of energy to hold, when Object 1 pushes against Object 2, $$m_1d\vec{v}_1 = -m_2d\vec{v}_2$$ Since the mass does not change with time (conservation of mass), ...


2

Why do we say the centripetal force does not make any work on a circular uniform motion? In your case, the rocket does not do any work...on the asteroid. As others have noted, no such guarantee is provided for the accelerated propellant spewing out of the back end of the rocket. It is possible to apply a centripetal force to an object without ...


2

My Interpretation of your question : Why don't agents of the conservative force gain internal energy (i.e. heat) as in the case of friction, instead of gaining just potential energy. In any Conservative force, Mechanical Energy is conserved. $$\therefore~~ E_{k~(\text{initial})} + E_{p~(\text{initial})} = E_{k~(\text{final})} + ...


0

As long as the asteroid stays in a circle and has a constant speed, the centripetal forces does not do any work. But if you increase the centripetal force, the asteroid will no longer stays in the circle, in fact it will fall. This is equivalent to say that the asteroid develops a velocity in the direction of the centripetal force. So in this case, you do do ...


3

Fat can't just completely be changed to energy, there need to be chemical reactions in the body as the body metabolizes the fat. Some energy in released by the metabolism. By 1933, Tainter and Cutting discovered that dinitrophenol causes cells in the body to waste energy. So, yes, the energy of fat can be extracted without mechanical excercise. ...


1

Your finite square well potential looks like: where $V_0$ is the potential energy outside the well and $V_1$ is the potential inside the well. The depth of the well is $\Delta V = V_0 - V_1$. We normally take $V_0$ to be zero, in which case $V_1$ is negative (like your $-10$eV) and $\Delta V = V_1$. However you can add any constant value to the potential ...


2

No, gripping something with your hands will make you feel tired because that is how muscles work: blood is pumping, cells are working... But imagin the gripped object on a table, and on it a pile of heavy books. The effect on the object is the same (more or less) than when you grip it with your hand, but the books never get tired. You could harvers energy ...


5

Have a look at the answers to Why does holding something up cost energy while no work is being done?. Gripping things takes energy not because a constant, stationary force does any work but because of the way muscles work. A stationary force doesn't do any work so no energy can be harvested from it. The best you could do is capture the heat given off from ...


2

I reminded you elsewhere that you know very well when 3rd law is applicable, and that you even teach it to others for example here Newton's third law of motion is not applied on a single body. - user36790 but then, inexplicably, you forget to use it properly in your own posts: Let a body move and a conservative force oppose its motion. ...


1

The problem with your solution is that the inelastic collision and assumption that kinetic energy is conserved are mutually exclusive. You can see that in your math when you try to solve for $v_2$. Rewriting equation $(1)$ gives $v_1=\left(m+M\right)v_2/m$ which inserted into $(2)$ yields $$ m \left(\frac{m+M}mv_2\right)^2 = \left(m+M\right)v_2^2.$$ This ...


0

though momentum as well as energy is conserved but definitely the sum of individual momentum of particles is not equal to sum of individual K.E. of the particles. also there may be different value of K.E. for same momentum. So can not make any result by manipulating the equations.


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Both energy and momentum are conserved as always. But to understand why this statement is true you have to look at the system as you described it a little more closely: In order for blocks A and B to stick together after the collision, the force between them should be zero when the velocity difference is zero -otherwise that force would continue to ...


5

To add to HDE 226868's correct answer "heat": Even the ram drag component, arising when the parachute losslessly exchanges momentum with the relatively moving air and thus feels a Ram Pressure (see Wikipedia article of this name, and also my footnote), ends up as heat because the air eddies and currents airising from ram effect (see my footnote) then ...


5

Heat Drag is the same thing as air resistance. It's a form of friction. Friction turns some of the kinetic energy of a moving object into heat; drag does the same thing, thus slowing a falling object down. When an object slows down due to friction, it heats up (and some of the heat dissipates to the surroundings). The same principle applies here: There will ...


0

There is energy in the magnetic field, $\mathcal U_B=B^2/(2\mu_0)$ in MKS units (as long as no magnetic materials are around). The act of turning on the magnetic field creates a changing $B$ field for a short time. A changing $B$ field creates an electric field, and this can cause current to flow in conductors. So there is some energy there too. ...


2

Let's make a concrete example with numbers: Suppose that $v_a = 6m/s$ and $v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$ . According to the conservation of energy and momentum: Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an ...


0

The thing to keep in mind is that for inelastic collisions, energy is not conserved within the system. Some energy is lost as thermal energy, and after all, it takes some energy to get the blocks to stick together. This doesn't mean that the Law of Conservation of Energy is false, though. It just means that energy has left the system that you're studying ...


1

In the 1d particle in the box the energy of the particle should be completely determined by the momentum of the particle that you observe correct? The Hamiltonian in the position basis is $$\hat H = \begin{cases}-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}, & 0\lt x \lt a\\\infty, & \text{otherwise} \end{cases}$$ and the energy ...


9

The source of gravity is not mass, but stress-energy-momentum, so you are correct that the energy converted in this process already has gravity and that that gravity is only rearranged The change in the gravitational field needs time to propagate, though, and this does indeed happen at the speed of light.


0

Why does torque, which is an analogy of force have the same units as energy, but force does not? The magnitude of (net) force is, in a sense, work done per unit displacement thus we can think of it as, e.g., Joules per meter. In a similar sense, the magnitude of (net) torque is work done per unit angular displacement, e.g., Joules per radian. But ...


1

This is a side-effect of treating angles as dimension-less. For translational systems, we have \begin{align*} [\text{linear momentum}] &= [\text{action}][\text{length}]^{-1} \\ [\text{force}] &= [\text{linear momentum}][\text{time}]^{-1} \\&= [\text{energy}][\text{length}]^{-1} \end{align*} Correspondingly, for rotational systems, we have ...


0

Torque is a cross product, and work is a dot product. So one big difference is that torque is a vector and work is a scalar. Another way to think about it is that work is a force being applied over a length interval, where only the force applied in the direction parallel to the displacement counts toward the work performed. On the other hand, torque is ...


6

The momentum operator $P$ in the infinite well can be defined as a self-adjoint operator by infinitely many ways with respect to the boundary conditions by: $$P_\theta=-i\hbar\frac{d}{dx}\\ \mathcal{D}(P_\theta)=\left\{\psi\in \mathcal{H}^1[0,a]:\psi(a)=e^{i\theta}\psi(0)\right\},$$ where $\mathcal{H}^1[0,a]$ is the Sobolev space, on the interval $[0,a]$. In ...


1

Is it possible for electrons to carry more than one charge? If by, one charge, you mean more electric charge than (the negative of) the elementary charge, the answer is no. More specifically, an 'electron' would not be an electron if its charge were not $-e$. However, electric charge is not the only type of charge electrons 'carry'. But this is a ...


2

Each electron has a fixed charge of $-1.602\times10^{-19}\,\mbox{C}$ (where (C stands for coulombs). If you gather $6.24\times10^{18}$ electrons, the total charge will be \begin{align*} \mbox{Total charge} &= \mbox{Charge per electron}\times\mbox{Number of electrons}\\ &= ...


5

A conservative force only returns the energy back when the object moves in a closed path, that is, it returns to the initial position (it doesn't matter if he returns due to other forces). This can be demonstrated as a theorem, but the intuitive explanation is that a conservative force depends only on the spatial coordinates, and not in the direction of ...


0

Almost all electrical machines can be run in both ways (generator or motor). If you're talking about the direction of rotation, it will work too. However, your message is so vague, we can't help you : we don't have any clue about the kind of machine, about the frequency, voltage, use...


1

There are several (equivalent) ways to look at this. One is to say that for any conservative force $\mathbf{F}$, one can define the potential energy Ep as an associated potential field such as $\mathbf{F}=-\frac{\partial Ep}{\partial r}$, or maybe more formally $\mathbf{F}=-\nabla(Ep)$. That's no more than a definition of the potential energy. ...


2

I think there is some confusion about the terms and the relationship between conservative forces and energy conserving systems. I will make a few general statements: 1) A conservative force can be described by a potential function $\Phi(x)$ with $\vec F=-\nabla{\Phi}$, a non-conservative force can not be reduced to such a function. 2) Friction is a ...



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