New answers tagged energy-conservation
1
Yes.
The 1st law of thermodynamics comes from basic energy conservation covering all involved energies:**
$$E_1+W+Q=E_2\qquad\text{or rewritten:}\qquad \Delta E=W+Q $$
where $E$ represents all present energy (the sum of kinetic, potential, chemical, thermal...) and $W$ and $Q$ represent all energy added (work or heat). We could, if we wanted to, expand the ...
0
No it is not the same. First of all, it is not the same to "minimize the energy" and "conserve the energy" as you seem to equate. In fact they are contradictory statements in some sense, since if the total energy is conserved its value is constant and can't be neither minimized nor increased without externel intervention. "...
2
Wrong. Consider a nonstationary external force (potentials). The energy is not conserved at all, but the equations of motion with the initial conditions describe the system behaviour.
0
Rubber is a bit of a special case, because it is "elastic", it heats up when stretched and cools when un-stretched.
see https://depts.washington.edu/chem/facilserv/lecturedemo/EntropyofRubber-UWDept.ofChemistry.html
As Niels answered, materials heat up when "plastically" deformed, because of "friction" when you drag sheets of ...
1
When you deform a piece of (for example) metal, on the atomic level you are forcing atoms to press against and squeeze past one another and assume new positions within the structure of the solid. And on the atomic level, there are friction forces that oppose that motion, like when you press your hands together and rub them against each other. This is called ...
4
Of course you can write $L=E-2V$, but you have to be careful with what is the total energy $E$ of the system. In the expression of the Lagrangian, the kinetic, potential, and total energy are considered to be defined instantaneously. The total energy is
$$E=\frac12m\dot x^2+mgx$$
You wrote instead that $E=mgh$. This is not the instantaneous total energy, but ...
1
In deriving the eqns of motion from the principle of least action, position and velocity are considered independent variables
19
OP is essentially asking:
Why can't we replace the Lagrangian $L=T-V$ with $L=E-2V$ by using energy conservation $T+V=E$, where $E$ is an integration constant?
Answer: Generically an action principle gets destroyed if we apply EOMs in the action.
Specifically, OP used energy conservation $T+V=E$, which were derived from EOMs. Here it is important to ...
3
The correct statement of the law of conservation of energy is that the total energy of an isolated system is conserved. This is valid both for classical and for relativistic mechanics.
In your example, the energy is not conserved as a consequence of the fact that the total mass of the system is not constant. In classical mechanics, the total mass can change ...
1
I try to explain the idea that I got following the computational study of a variable mass system.
Let us consider an isolated system in a certain state of motion. It will have a certain mass and geometry, i.e. a moment of inertia, which quantify its ability to oppose changes in its dynamic state.
Imagine that its mass varies over time for some reason.
The ...
0
Assuming the rod has negligible mass and that l is the distance to the center of mass of the block, I would think that h=l. Since the rod is rigid, there is no reason the velocity when the block is at the peak of the circle should not be essentially zero. So the initial potential energy mgh should be equal to the potential energy when the block is at the top ...
2
Recently, I learned however that it is unreasonable to talk about the
gravitational potential energy of a single object. So mgh actually
calculates the gravitational potential energy of the system made up of
the earth and ball.
That is absolutely correct. Potential energy (of any kind) is a property of a system and not a single object, because that energy ...
2
Assuming the hook does not come off the ring, if you start from height L, then you can arrive back at height L with no kinetic energy. That would be a position of unstable equilibrium.
5
The key point is that the bar is rigid, so the block can reach the top of the circle with zero velocity. So the block's kinetic energy at the point where it is at the same height as the ring about which it pivots (a distance $l$ below the top of the circle) must be at least $mgl$. But the block's kinetic energy at this point is equal to the potential energy ...
1
You must understand that velocity at the top is not zero but $ infinitesimally $ greater than zero and that zero here is a good approximation for calculating min velocity required.
If you get $v_{min}=5_{m/s}$ that means the velocity required for circular motion is something like 5.001or 5.0003 , anything greater than $5_{m/s}$
3
The simplest physics description of potential barrier tunneling can be found here:
According to classical physics, a particle of energy E less than the height U0 of a barrier could not penetrate - the region inside the barrier is classically forbidden. But the wavefunction associated with a free particle must be continuous at the barrier and will show an ...
2
The issue is that position eigenstate and energy eigenstates are incompatible. That is, if position is measured, the state then becomes a superposition of energy eigenstates. And vice versa.
So it doesn’t make sense to ask what the energy is in a localised region of space. The system doesn’t have those properties simultaneously.
0
Yes, the conservation of energy holds when there are net forces too, and the difference between the two cases that you proposed is the behaviour of kinetic energy:
if the net force is zero the speed is constant, as well as the kinetic energy; also consider that in reality -even when you try to lift a book at constant speed- the net force applied to the book ...
0
When you lift a book you are expending energy, you can accelerate it all the time you are lifting it, or you can lift it steadily at a constant speed. In both cases you are increasing the book's gravitational potential energy, also some of your energy may go to air friction and sound waves. So none of the energy you spend is lost.
0
To be clear, if you "lift a book at constant velocity" you are still having to provide a force against gravity and so are still "doing work". Perhaps what you mean is an object moving at constant velocity in a vacuum. If no external forces act on this object its energy doesn't change.
If you're asking if the conservation of energy holds ...
1
Energy is conserved in the whole universe *. Nevertheless, if you delimit a system and you focus only on that system, forces can vary the energy of the system. Total energy is conserved in the universe, but in your particular system migh not.
If you wide your system to include more objects, then you'll find energy conserved.
[*] Energy will be conserved as ...
1
As you say , potential and kinetic energy of large objects can be converted into thermal energy. For example, if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs $1kg$ and you drop it from about $2m$ , it will heat up by less than $.005$ degrees Celsius.
Considering ideal conditions, ...
4
In theories with dark energy, an energy density is just a property of space — even “empty” space devoid of matter and radiation. It’s closely related to Einstein’s old idea of a “cosmological constant”. Dark energy isn’t something material that comes from somewhere. As space expands, you simply get more dark energy because there is more volume; a dark energy ...
3
In the above types questions why we don't include the( initial and
final both) velocities of bullet while finding the initial kinetic
energy of the system in (COE) conservation of energy
In the example you gave, kinetic energy is not conserved. During the completely inelastic collision some kinetic energy is spent on permanently deforming the pendulum and ...
2
For an extended body in a uniform gravitational field, its potential energy is the same as that of a point mass located at the bodies' center of mass. This can be seen by writing out the total gravitational potential energy in terms of the gravitational potential energies of its constituent masses:
$$
U = \sum_{i} m_i g z_i = g \sum_i m_i z_i
$$
where $z$ ...
2
I would just add to Emilio Pisanty's comment by saying that the frequency remaining a constant as light travels across a boundary separating two media is required because it is required that the tangential component of E be continuous across the boundary.
In turn, that the tangential component of E must be continuous across the boundary separating two media ...
4
Light can be considered a particle, known as a photon, but the energy of each photon is not related to its velocity. Instead, it is given by the Planck formula,
$$
E=h\nu,
$$
where $h$ is Planck's constant and $\nu$ is the frequency of the light. Since the frequency of the light remains constant as it travels from one medium to another, the energy of those ...
0
I believe the confusion comes from the distinction between partial derivatives and total derivatives. I.e. there is a difference between potentials that have the form $V(x(t))$ and $V(x(t), t)$--the latter has an explicit time dependence not coming from the motion of the particle. For example, to show the conservation of energy with a system with only ...
0
The conservation of energy always holds for an entire system. If we have a particle in a changing potential the total energy of the entire system (particle + whatever's creating the potential) is constant.
If we take two containers of water, $A$ and $B$, the total amount of water in both containers is constant no matter how much we transfer the water between ...
1
Let look at this example one dimensional
Kinetic Energy is
$$T=\frac{m}{2}\,\dot{x}^2$$
and Potential Energy is
$$ U=U(x)$$
with Euler Lagrange you get
$$m\,\ddot{x}+\frac{\partial}{\partial x}\,U(x)=0$$
thus according to Newton second law
$$F=-\frac{\partial}{\partial x}\,U(x)$$
and the total energy
$$E=T+U(x)=~\text{constant}$$
but if the potential energy ...
2
You already have proved the "only if" part. What your derivation shows is that
$$\frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(t))\right)$$
equals zero only if
$$\mathbf{F} = -\nabla V.$$
In other words, your work proves
$$\mathbf{F} = -\nabla V \implies \frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(...
0
Your options are limited because the one thing you are choosing to exclude from your list of options is actually measuring the energy. Measuring the energy must be done by transferring the energy from one form to another. (Or some crazy relativistic stuff that will not work on any batter with less energy than, say, a large star)
The closest I think you can ...
0
This will provide an upper bound:
Measure the mass of the battery.
Count all the atoms in the battery (good luck!).
Compute the mass of all the atoms under the assumption of the atoms are non-interacting.
Since E=mc^2, the difference between the actual mass and the computed mass is an upper bound on the potential energy that could be discharged from the ...
0
I don't think it's experimentally feasible, but if one could measure its change in mass precisely enough, it will change by $1/𝑐^2$ times its change in potential energy, which will will be the energy produced during discharge.
If you discharged it through a resistor, then the total energy as heat generated by the resistor (plus any other small effects like ...
0
You have to understand how the rope affects the motion of the particle.
Imagine to cut the rope at an instant, e.g. when the angle is obtuse. What would the particle do? It would no more move along the circle. Rather, it would follow a parabolic trajectory (because of gravity alone) which is tangent to the circle at the point in which the particle was when ...
1
Let me first discuss a reduced case, for the purpose of focus.
(The concluding statement - all the way down - explains the non-equivalence.)
I will first discuss the case of motion in a space with 1 spatial dimension. In classroom demonstrations this is of course the demonstration of two or more carts moving over an air track
In this reduced environment, ...
1
$F=BqV$ is the force experienced by a moving charge in a magnetic field.so, as charge moves in circular path we equate this to the required centripetal force(=$\frac{mv^2}{r}$)now you can proceed with the solution
1
The electrons are undergoing uniform circular motion due to the magnetic force. Therefore, the centripetal force component is equal to the magnetic force. I'll leave the calculation (if you can call it that?) to you.
1
The other answer is very good, but not useful if we don't understand Noether's theorem (or Lagrangians). Instead, we can reason about this in a less difficult but more roundabout way.
You ask if Conservation of Energy is equivalent to Conservation of Momentum, i.e.
$$CoE \iff CoP$$
First, we will disprove
$$CoE \rightarrow CoP$$
We can do this by finding a ...
4
is conservation of energy not equivalent to conservation of momentum?
The conservation of energy is not equivalent to the conservation of momentum.
Per Noether’s theorem momentum is conserved whenever the Lagrangian is symmetric under spatial translations and energy is conserved whenever the Lagrangian is symmetric under time translations. Since it is ...
2
When charging a capacitor the energy from the battery is transferred to the capacitor. If the wires have resistance, some of this energy is lost, i.e. dissipated. If the resistance is zero, there are no losses - but there is still the energy transfer from the battery to the capacitor.
0
If $v$ is treated as the velocity instead of the speed of the object then $m(dv/dt)$ has direction and would be simply the rate of change of momentum, rather than the rate of change of the magnitude of the momentum.
The reason for this imprecision probably stems from the two definitions of kinetic energy: it can be calculated as $\frac{1}{2}mv^2$, where $v$ ...
4
You are incorrect in saying that, “the total force being applied in both directions is $1.366$ N.” Forces are vectors, not scalars. This means they conform to vector addition and not scalar addition.
The magnitude of the total force, $F$, is given by Pythagoras:
$$F^2 = (1\sin 30^{\circ})^2 + (1\cos 30^{\circ})^2$$
which gives $F = 1$ N, as you would expect.
...
1
Energy (including energy transport caused by a temperature difference, known as heat) is an extensive variable. In other words, if you duplicate a system and its input and output conditions, then all the energy terms will double. Linear superposition of energy fluxes of $\dot{Q}_A$ and $\dot{Q}_B$, if valid, will produce a flux of $\dot{Q}_A+\dot{Q}_B$.
In ...
0
Potential energy doesn't have any physical meaning. Only potential energy difference has physical meaning. You can add a constant to the potential energy at all points, and it will not make any difference. Hence, we have the liberty to choose any point in space as the point with potential energy as 0. Choosing potential energy at infinite as 0 is most ...
2
I agree with the answer @Dale provided.
To put things into perspective, the energy stored in the magnetic field of a straight conductor is minuscule compared to the energy dissipated in the resistance of the conductor.
The inductance of a straight copper wire 1 mm in dia and 10 cm long is about 105 nH. The energy stored in the magnetic field of the wire ...
3
1.Is energy needed to create magnetic field in general?
Yes. When you are using circuit theory the mechanism for dealing with magnetism is called inductance and is usually represented by the variable $L$. The inductance gives the total magnetic field so that you can still use a lumped element approach and do not need to know the details of the magnetic ...
1
Yes, energy is needed to create the magnetic field. Once the field has been created, no further energy is needed. The electric current in the circuit preserves the magnetic field. 2. The chemical energy in the cell is converted to electrical energy that moves the charges in the current, and as the charges start to move, they produce the magnetic field and ...
1
In quantum mechanics you have to be careful what you mean by 'the energy'. In QM the energy is calculated using the Hamiltonian operator $\hat H=-\frac{\hbar^2}{2m}\nabla^2+U(x)$. Naively you would expect you can just apply it to wavefunctions $\hat H\psi(x)$ and wherever this gives you a larger number the energy is high. But the only way to calculate the ...
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