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Firstly its the potential energy of the earth-object system as the two answers have said. "Potential energy of the object" is a loosely spoken phrase for the same for things happening on the surface of the earth where $g$ is taken to be a constant.Secondly work is not done by a person. Work is done by a force.There is an important thing you need to know. ...


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No it is still zero. Take $\theta$ to be the angle between the point at the bottom of the cylinder, C and P. Take the radius of the cylinder to be $R$ and max length of rope $L$. The vector from C to P is $$r_{CP} = R(\sin \theta,-\cos\theta).$$ The vector from P to the particle is $$r_{T}=(L-R\theta)(\cos\theta,\sin\theta).$$ So the position vector of ...


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Yes so-called pseudo forces do work and if they were to be describable as a conservative force, then yes the corresponding mechanical energy would be conserved. The best example I can find is the gravitational pull we feel at the surface of the Earth. It is in fact the sum of the "true" gravitational force owing to Newton's law of gravitation and the, ...


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A force does not require a constant input of energy to exist. Energy is only required to perform work, which is exerting a force over a distance. $$ W = \mathbf F \centerdot \Delta \mathbf x $$ That distance is key. In your example, if the size of the container does not change, no energy is expended no matter how long the force lasts. If the force is used ...


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Our equation for work follows from the conservation of energy. If we consider some object then we expect that if we do work $W$ on it then its kinetic energy must increase by $W$. So the requirement for the equation for work is that it must be equal to the change in kinetic energy. Proving this is usually done using integral calculus, but since you give the ...


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The notion of work in physics was first formulated by the French mathematician Gustave Coriolis in Calculation of the Effect of Machines, or Considerations on the Use of Engines and their Evaluation published in 1829. Coriolis defined work as "weight lifted through a height". He was concerned with developing a term that could measure the units of work ...


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There are some physical quantities that are usefull (and this is under statement), like energy. It is conserved, it is a function of some other very important quantities that can help you describe the motion of the body etc. If you can justify energy, there should be no problem in justifying work, which is energy transfered to a body by some force. Quantity ...


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Part of your problem comes from thinking that the potential energy is somehow located in or a property of the person alone. And the way the subject is usually introduced could easily lead you to think that, but it's not right. The potential energy is a property of the person-Earth system. In fact all potential energies are properties of systems of ...


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Entropy is fundamentally a subjective quantity, it is the amount of information needed to specify the exact physical state of a system given it's macroscopic specification. The fact that you don't know the exact physical state of a system, yet the laws of physics are such that information is never lost, implies the second law of thermodynamics (one also ...


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Introduction: Entropy Defined The popular literature is littered with articles, papers, books, and various & sundry other sources, filled to overflowing with prosaic explanations of entropy. But it should be remembered that entropy, an idea born from classical thermodynamics, is a quantitative entity, and not a qualitative one. That means that entropy ...


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Let's frame this question in terms of a heat engine (Carnot engine). Here is a diagram I made for a class when teaching this stuff. Heat flow $\dot{Q}$ has an associated entropy flow $\dot{Q}/T$. The job of a thermodynamic engine is extract/filter as much useful work as possible from a flow of energy and entropy. To answer your question in layman terms. ...


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There's no work done for a person climbing upstairs because the energy is converted to PE within system only. The person is the system. How true is the above statement? I think it's true enough. You do work on a brick when you lift it up. You add energy to it, and we call this energy gravitational potential energy. Then when you drop the brick this ...


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The above statement is not correct. First of all, you need to work against the force of friction while climbing stairs.So the energy is not entirely converted to PE.Rather a portion of it is dissipated. Secondly, even if we leave out friction, the basic flaw of the statement lies in the part: " the energy is converted to PE within system only. The person ...


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With regards to intuition, it might help to think about situations of mechanical advantage. For example, consider a simple pulley system. modified from "Pulley1a". Licensed under Public Domain via Commons - https://commons.wikimedia.org/wiki/File:Pulley1a.svg#/media/File:Pulley1a.svg You can work out using force that the weight $W/2$ balances the weight ...


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Work does depend on frame of reference, but so does change in kinetic energy. Work done and changes in kinetic energy should either both bother you or neither bother you. To know how much the kinetic energy changes from one location to another you need to know the force (if constant) and how far apart the locations are: ...


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As you point out, work done is a function of the frame of reference. More specifically, if you apply a force on an object, that force typically connects two different objects, and it's the relative velocity of these two objects that really concerns us. Example: you are walking in a train, and pull a suitcase behind you. The friction between the suitcase and ...


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All of this is all right, but the problem is that I'm taking a course on electrodynamics and the teacher said that the work $W_{\mathrm{ext}}$ done by one force external to the system is $$W_{\mathrm{ext}} = \Delta K + \Delta U,$$ that is the change in the total energy of the system. I don't know where this comes from It follows from the work-energy ...


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$$ m \ddot{\vec{r}} = \vec{F} $$ multiply by $\dot{\vec{r}}$ and integrat over t: $$ m \int_{t_0}^t \ddot{\vec{r}} \cdot \dot{\vec{r}}~ dt = \int_{t_0}^t \vec{F} \cdot \dot{\vec{r}}~ dt$$ With $\frac{1}{2} \frac{d}{dt} (\dot{\vec{r}}^2) = \ddot{\vec{r}} \cdot \dot{\vec{r}}$ it follows: $$ \frac{1}{2} m v^2 + \left( - \int_{t_0}^t \vec{F} \cdot d\vec{r} ...


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What about using Galilean free-fall? From $S= \frac 12 g t^2$ and $v = g t$ you get that velocity after a fall $h$ follows $$h= \frac 1{2g} v^2$$ We conclude that if the ball is consuming some essence to get velocity from the line of fall, this essence must be "stored" in space as the square of the velocity. The idea works because if we have already a body ...


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When you write: Derivations (or at least, convincing arguments) of the kinetic energy formula that didn't require the work formula required relativity to make sense, which is unbelievable considering that Newtonian mechanics were established well before relativity. I assume you are referring to arguments like Ron's argument. Although such ...


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I think your approach isn't wrong; however in your calculations you're making the assumption that the potential difference between plates, $V$, is constant: What remains constant is the charge on each plate. So the equation becomes: $$W=\int_0^d {{q^2} \over {2\epsilon_0A}} \;\mathrm{dx}={{q^2d} \over {2\epsilon_0A}}$$ Since $C=\epsilon_0A/d$ we obtain ...


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From what I understand, this energy equals the work of the electrostatic forces needed to get the plates from a zero separation (when they touch) to a separation d. That's not the ordinary understanding and, from the electrical circuit perspective, the energy stored equals the work done by the external circuit separating electric charge in the ...


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The calculation is not right, but the result is true. You need infinite work to pull apart two oppositely charged infinite plates, they have each infinite charge! The electric field due to one infinite plate is $\frac{\sigma}{2 \epsilon_0}$. The force on an area $A$ of the second plate will thus be $\frac{- \sigma^2 A}{2 \epsilon_0}$, which is infinite for ...


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As John Rennie marked, the result should be $\tfrac{1}{2}CV^2$. Let me deduce this for you; Let's start with an uncharged condenser & by some means you remove one an electron from one plate & transfer it to the other plate. You have to do hardly any work to transfer the first electron but as you gradually continue the process, the field that is ...


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The answers you get refer the phenomenon of deflection of charged particles in a magnetic field to a quantitative description, derived from experiments. The mechanism, how this happens, stays hidden. The electron has three well known properties, its electric charge, its magnetic dipole moment and its intrinsic spin. All three are constant quantities. And to ...


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Energy is conserved, and then you can treat it like a budget and say it is accounting. But then you just ask what is conserved and what is the budget making an accounting of and it gets rather circular. So let's start with the basics. You have configurations. These are possible ways thing can be, for instance having two objects and one object here and ...


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If you do not know calculus or vectors, I have a review of both that I can insert here in this answer. I firmly believe that trying to understand Newtonian mechanics without some understanding of calculus and vectors is a Bad Idea, in the sense that you can turn the first few weeks of such a physics course into a vectors-plus-calculus course and cover ...


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$p$ does not necessarily mean ‘the pressure of the surrounding’, its meaning depends on the problem. In an archetypical problem where a piston is moved down a cylinder filler with an Ideal Gas, an infinitesimal amount of work $dW=pdV$ is performed when the piston is moved an infinitesimal amount, with $p$ the pressure inside the cylinder. To obtain a ...


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Let's say that you are standing on the ground and you observe two horizontal forces of equal magnitude, but acting in opposite directions on the block, because of which the net vector sum of all horizontal forces is zero, i.e. $$\sum F_x=0 \implies \left(a_{x}\right)_{net}=0 $$ Thus the block moves with constant velocity. Now, you wan't to calculate the net ...


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Work is a scalar and does not have a direction. If your box was moving, because of the forces you apply than your forces may be equal in value but surely not opposite in direction or spot of application or both. In this case you would count the work of a sum of two forces, which cant be zero in any case if the box is moving. If the box is moving at constant ...


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By the way, it's customary to use primes or some other indicator inside the integral to be slightly less confusing. I.e. $$W(t) = \int_{t_0}^{t} P(t')Q(t')dt'$$ Be careful about how you think about Q(t). You've described it as a "flow rate", but really, you are saying $Q(t) = \frac{dV}{dt}$, or the rate of change of the volume with time. Now, the term ...


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In addition to the other reply, it can be added that by definition, in an ideal gas, there is no interaction between molecules, and therefore no potential energy associated with the average distance. This is why in a Joule-Thomson expansion, there is no change in the temperature of the gas: only the volume changes, no work is extracted, and the average speed ...


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The internal energy of an ideal gas is independent of volume when considered as a function of volume and temperature. If we choose to consider internal energy as a function of volume and some other thermodynamic variable we will find that the dependence of the energy on volume will change because we are keeping a different variable constant as volume is ...


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General Question: Why should I use just the friction force rather than the net force to integrate over distance when conserving energy? Answer: In energy conservation problems each way of storing energy generally gets it's own term. In the example problem there is a gravitational potential energy term (GPE), a kinetic energy (KE) term, and a friction ...


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Is it correctly understood that energy is continuously put into the system, in order to maintain the orbit? And that gravity is thus an infinite source of energy? Consider the general case of one particle orbiting another in an ellipse (this is general because we can a always reduce the two body problem to an effective one body problem, and the general ...


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I think you need to read up a bit on vectors and scalars, how we define things like acceleration. Vectors are physical quantities that have a magnitude and a direction. Velocity is a vector (speed is its magnitude). Acceleration is also a vector. Acceleration is defined as the rate of change of velocity, basically it measures how velocity changes with ...


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Because in the second scenario you described, the exhaust gases from the rocket fly off at high speed into space. This is where the extra energy goes. A better alternative to get circular motion of the satellite with the earth removed, would be a second satellite, connected to the first satellite with a long cable. If, for example, both satelittes have the ...



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