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14

As you know, the Law of conservation of energy holds only for isolated systems. From Wikipedia, Law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. Now, what is an isolated system in physics? As stated in Wikipedia, it is a physical system so far removed from other ...


10

Krishnanand J's answer gives a good explanation, however, I assume (by how you asked) that you are considering yourself to be a part of an isolated system. Now, if you are applying a force on the system, and the system is accelerating, then you are doing work on it. And you cannot do the work without using your energy. So the kinetic energy of the system ...


8

Energy is defined as the ability to do work. Think about yourself floating in space and a high velocity projectile, say a meteor, coming your way. The only way that meteor can do work on you is via a collision (neglecting the mutual gravitational attraction between you and the meteor for the moment). The only relevant quantity in the calculation is not, ...


4

Relativistic energy is $E = \sqrt{p^2c^2 + m^2 c^4}$, so while stationary objects with no moving parts have an upper bound of $E=mc^2$ (by, for example, matter-antimatter interaction), the energy of a moving object is only capped by the momentum, $p$. Relativistic momentum, $p$, is $\frac{m v c}{\sqrt{c^2-v^2}}$. This has no upper bound. Once an object ...


3

Because both momentum and energy have to be conserved when the atom absorbs the photon. Suppose we have an atom with a mass $m$ and the energy difference between the initial and final levels is $E$. The photon energy is $hf$, so conservation of energy gives us: $$ hf = E + \tfrac12 m v^2 \tag{1} $$ where $v$ is the speed of the atom after absorbing the ...


3

It is confusing whats the difference between probabilities and probability densities. If you write the the probability distribution expressed with velocities you get $$ 1 = \int_0^\infty f(v)dv.$$ If you want to calculate the probability of a particle having velocity in some finite interval $a < v <b$ you get $$ P(a<v<b) = \int_a^b f(v)dv.$$ If ...


2

What puzzles me, is: 1 - why are some energies relative to reference frame and some are not? 2 - why the "absolute" energy from gasoline can be changed into kinetic energy of car, and therefore change into "relative" energy? The reason is there are two types of energy of an object or substance: Internal Energy and External Energy. The total energy of ...


2

Note that the question is about the work done on the weight, not the work done by the man. The work done by the man will be greater than the work done on the weight because the man's muscles are not 100% efficient and because energy is lost due to friction. In case (i), if the weight is carried at a constant height about the ground, then the force required ...


2

Many of the answers are right for the most part, but I think they have too much detail. The simple answer is that velocities are reference frame dependent, while lengths are not. Kinetic energy depends on velocity, so it is reference frame dependent. Potential energies depend on lengths between two objects (or from some reference point), so they are not ...


2

There are no free energy devices, even on a human time scale. In the real world, you would be skeptical about someone offering you free money. Physicists are like that about perpetual motion machines. There are good reasons to think they are impossible, and nobody has ever built a genuine one. Generally you don't even have to listen to details to know ...


2

Short Answer: It is zero. Long Answer: Assume you are on a frame of reference of road (i.e you are standing on road) then the system is truck + block. You observe both of them moving now looking closely friction is holding block on truck by directing itself in forward direction, using Newton's third law equal and opposite forces are acting on the truck. ...


1

Why must this transition be to another stationary state, which has definite energy? It does not! The OPs question is really an objection to the textbook presentation. Atomic transitions are a dynamical process When a photon or, more generally, a wave packet/driving pulse hit an atom, they set of dynamics of the electrons. In general, these dynamics are ...


1

There are two pressure effects on a gas that relate to the temperature and stored energy. The first, is that when applying pressure, one accelerates a few atoms as they bounce off the moving piston. This causes adiabatic heating (Diesel engines and clever little fire-starting devices put this effect to good use). If, however, you let the heated gas ...


1

If we assume a monatomic ideal gas then the internal energy of the gas is simply the kinetic energy of the gas particles, and the average kinetic energy of a single gas particle is dependent only on the temperature: $$ KE = \tfrac32 k_B T \tag{1}$$ If you have some fixed volume $V$ at a fixed temperature $T$ then the number of moles of gas in that volume ...


1

Per the work energy theorem the net work done on an object equals its change in kinetic energy. Since the block begins and ends at rest its change in kinetic energy is zero. Therefore the net work done on the block by the static friction force is zero When the truck accelerated the work done by friction was positive giving it kinetic energy since the force ...


1

The law of energy conservation means that energy cannot be created or destroyed. You can't get around that law merely by saying your device will wear out in the real world. Here's a crude analogy. If I give you a 10 litre bucket full of water, you may get less than the whole 10 litres from it, due to evaporation, spills, leaks, etc. But there's no way for ...


1

Rather than thinking of the force applied by the man and the distance moved by the block, think of the energy difference. You will notice that in the first case, the body is just shifted parallel to the horizontal surface i.e. its P.E does not increase. But in the second case, the potential energy increases, which invariably means that he has done work. ...


1

As for most questions, a more succinct answer is needed, rather than a more detailed one, so here it is: Stored energy is equivalent to mass. Thus we can never store more energy than mass, as the OP already suspected. This is just a very fundamental fact about nature. To understand and accept it, one can study some Special Relativity. To make some more ...


1

I will try to clarify. To simply the formulae, I will put the speed of light equal to unity, $c=1$, so that if time is seconds, distance is in light seconds and something traveling at half the speed of light has $v=1/2$. Energy-momentum can be written as a 4-vector $(E, \mathbf p)$. The magnitude of energy-momentum is mass, $m$, and obeys the relationship $$...


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