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P-V work is not the only kind of work that can be done on the contents of your system. In the case of your fan example, the fan is doing work on the gas within the container by exerting force on it through a displacement (of the fan blade). The kinetic energy imparted to the gas by the fan is then converted to internal energy by viscous dissipation (a ...


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Your intuition that the same amount of fluid goes down and then up by the same amount is incomplete, you are forgetting what happens inside the fluid. It is easier to see using solid blocks as in the figure below: Here you can see that the effect of moving block 1 down is to shift block 2 to the right, and moving block 3 back up the same amount that ...


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Considering to your question and comments below it, I think that you have been confused about some assumptions. For example, when we assume that there is no friction, you should be able to imagine a situation that there is no friction in that. You shouldn’t imagine real life for that assumption, because in your common daily life, you cannot find a perfect ...


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To put this simply, the work-energy theorem states that The work performed by a force $\mathbf F$ over a distance $\Delta \mathbf r$, $W=\mathbf F\cdot\Delta \mathbf r$, is equal to the change in kinetic energy $\Delta E_\mathrm{kin}$ of the relevant object. If the inner product $\mathbf F\cdot\Delta \mathbf r$ is negative, the force is acting in the ...


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For a conservative force the work done in going from position $A$ to position $B$ is independent of the path taken.When the pen hits the ground the ground is deformed. For a non conservative force the work done does depend on the path taken and the frictional force is an example of such a force. If you slide a block from position $A$ to position $B$ on a ...


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What is meant by a “change in volume of a system”? "Change in volume of a system" means "change in volume of a system", not anything else. System is a hypothetical concept. There is no specified system before we define it. We ourselves choose and define system. When someone talks about a system defined by himself/herself, he/she talks about that system not ...


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Work is always force times displacement in the direction of the force. The only place where the gas is doing work is at the bottom surface that is moving downward. The force it is exerting there is $PA$, where $P$ is the gas pressure and $A$ is the cross sectional area of the tube. If the lower surface moves downward a differential distance dx, the work ...


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Reversible work is done by conservative forces and so doesn't depend on the path. Non-conservative forces like friction, generate irreversibility and in presence of those forces, we cannot have a reversible process. Hence, if we want to determine reversible work, we should remove all irreversibilities i.e. all non-conservative forces. In thermodynamics (and ...


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This is a really deep question. My explanation will maybe be not so rigorous, but I hope it can help shed some light. Let's start by saying that reversible work is indeed path-dependent, so it is not a state function. Consider for example the two reversible transformations $A$ and $B$ in the picture: They both are composed by an isobaric and an ...


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Potential energy change of mass $m$ is equal to kinetic energy change of system ($m$ and $M$) plus wasted energy due to friction between spool and axle. $$mgy=\frac 12mv^2+\frac 12I\omega^2+W_f\tag 1$$ $\omega=\large{\frac{v^2}R}$ As the tension force of the string is constant, then the net force acting on mass $m$ will be constant ($F_{\textrm{net}}=mg-T$)....


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Start from the first principle of thermodynamics : $d U = \delta W + \delta Q$ where $\delta Q_{\text{rev}} = T dS$ so $\delta W_\text{rev} = d U - T d S$ hence at least for an isotherm, reversible work only depends on internal energy and entropy, both of which are state functions. So yes, in this particular case, reversible work between two states is a ...


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It depends what work you refer to. If it is the work made by gas B on the piston at the right, then this work is $-P\Delta V$ because the length the piston moves is given by the change in the total volume (regardless of the motion of the other piston). This will be also the net work done on the system A+B. To compute the net work by gas B you need to add ...


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You can check your answer by figure below: $$\Delta V_B=S(x_{B,2}-x_{B,1})-S(x_{A,2}-x_{A,1})= \Delta V-\Delta V_A$$


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It's not $W_{s}=\int_{0}^{s} Fs·ds$ But $W_{s}=\int_{0}^{s} F·ds$ No, it should be $W=\int_{s1}^{s2} F(s)·ds$ $W=\int_{s1}^{s2} D·sds=\frac{D}{2}(s_{2}^2-s_{1}^2)$


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First, let's pick a field to work with, because particles act differently in different vector fields. Let's say we're dealing with a charged particle in an electrostatic field. EM fields can be seen as a deformity in spacetime, the field is warping the space in which it is defined. In fact, for advanced EM we use tensors to describe electromagnetic ...


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First of all, we must be talking about a field that would affect (exert a force on) the object (like a charge in an electric field or an object with mass in a gravitational field). Now, what does potential energy mean? It is a measure of "stored energy" in the system. That means, if you released it, this energy would be released. Put a book on a shelf and ...


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According to the definition of the work ($\delta W=F\mathrm dx$), it is better that we say friction doesn't do work. Because, at each point of contact area, friction force is fixed and doesn't move. Friction converts some portion of energy used to move the box, to heat.



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