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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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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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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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Energy is dissipated in the form of "internal energy", which means that all of the objects kinectic energy is transfered to internal movement of atoms and mollecules of both the object and the surface. When there is a large deformation and no restitution you can argue that some of the energy is stored in some kind of ellastic energy of the mollecular bonds ...


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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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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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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 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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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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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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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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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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$p$ 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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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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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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$$ m \ddot{\vec{r}} = \vec{F} $$ 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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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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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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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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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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For turbulent flow, the friction goes up as the square of the velocity. So the force (pressure) goes as $n^2$. And you are moving $n$ times more liquid. The velocity (or volume flow) increases by $n$. The two factors combine to give the cube law since power = force times velocity, or pressure times flow rate.



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