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## New answers tagged work

2

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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Your approach is almost correct! To see what's wrong, consider the case $d = 0$. Then for a small change in volume, the change in potential energy of the water is zero (since you're just moving water from the left side to water at the same height on the right side). But the gas has definitely done $p \, dV$ work. The mistake is that you've neglected the ...

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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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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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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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The second one is correct. Work is defined by $$w=F \cdot dx$$ where x is travel distance, which can be calculated by $$dx \cdot S = dV$$ Here you can see dV is not $dV_B$. It should be total volume change.

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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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First, I should say that $\Delta W$ and $\Delta Q$ don't have meaning. $\Delta$ exists for state functions, but work and heat are path functions. A system doesn't have heat or work. Heat and work are recognized when they are transmitted through system's boundary. Second, the correct equation is $\Delta U=mC\Delta T$. So, I assume that you want to determine $... 1 It depends what work you refer to. If it is the work made by gas B on the piston at the right, then this work is$-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 ... 1 You can check your answer by figure below: $$\Delta V_B=S(x_{B,2}-x_{B,1})-S(x_{A,2}-x_{A,1})= \Delta V-\Delta V_A$$ 0 actually you got it wrong work energy theorem states that the net work done is equal to change in k.e here you have to concentrate on the word net .. because net work done work done by you and work done by frictional force is zero...... 1 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)·dsW=\int_{s1}^{s2} D·sds=\frac{D}{2}(s_{2}^2-s_{1}^2)$0 I have put the summary at the beginning. Suppose a mass$m$starting from rest at a height$h$from the Earth's surface and the kinetic energy of the mass$m$is to be found when the mass reaches the surface. The gravitational field strength is$g$and making the following assumptions: - the height$h$is much less than the radius of the Earth so the ... 0 There are two kinds of energy to keep track of: potential energy and kinetic energy. Work can be thought of as the change in energy (because of the "work-energy theorem"). And the key question to ask while the weightlifter keeps his weight in the air is: is there a change in energy? Kinetic energy is a function of movement. If there's no change in ... 0 When the weightlifter lifts the mass he does work and transfers some energy to the mass which is stored* in the mass as potential energy. Once the bob has been lifted the weightlifter applies a force to keep the bob lifted (to counter the force exerted by the mass downwards i.e. mg) but does no work as the force does not cause any displacement. Edit: *No ... 0 Both the forces are in opposite sense to the displacements of 1st and 2nd electrons respectively, i.e., they both are doing -ve work The force F in your diagram is the force on the left hand charge due to the right hand charge. Thus the force exerted by the left hand charge on the right hand charge is in the opposite direction ie to the right which is in ... 1 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 ... 1 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 ... 0 From wikipedia: Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity. So, it isn't matter how much work is done by other forces acting on a body. What is matter is work done by the gravitational field. 0 It is important to note, that you are totally skipping the kinetic Energy$T$part. The total energy of a system is given by$E=T+U$. There are several examples of systems which build up potential Energy over time. Some examples are: The mass of a pendulum is constantly cycling the total energy between kinetic and potential energy Objects orbiting a center ... 1 According to the definition of the work ($\delta W=F\mathrm dx$), it is better that we say friction doesn't do work. Because, at each point of contact area, friction force is fixed and doesn't move. Friction converts some portion of energy used to move the box, to heat. 3 Considering to your question and comments below it, I think that you have been confused about some assumptions. For example, when we assume that there is no friction, you should be able to imagine a situation that there is no friction in that. You shouldn’t imagine real life for that assumption, because in your common daily life, you cannot find a perfect ... 2 To put this simply, the work-energy theorem states that The work performed by a force$\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$is negative, the force is acting in the ... -1 No, we can't. I assume that you are talking about a moving object and therefore kinetic energy. You Can apply a resistive force, therefore slowing the object and reducing its energy. But as soon as the object reaches zero energy (read: kinetic energy), the body will be at rest. If you continue to apply the same force, from Newton's 2nd principle, the body ... 2 What is meant by a “change in volume of a system”? "Change in volume of a system" means "change in volume of a system", not anything else. System is a hypothetical concept. There is no specified system before we define it. We ourselves choose and define system. When someone talks about a system defined by himself/herself, he/she talks about that system not ... 0 Explanation of doubt 1: Consider a double pendulum made up of two massive rods. Here there are two degrees of freedom and hence two generalized coordinates. Assume that 5 separate non-constraint (applied) forces act on this system of double pendulum:${\vec {F_1}}$and${\vec {F_2}}$on the upper rod (rod 1) and${\vec {F_3}}, {\vec {F_4}}$and${\vec {...

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The system is supposed to consist of $n$ points at which the forces act. If no force acts at the $i$-th point, just set $F_i = 0$. The assertion that "this can only be true if external forces are the impressed forces" is false (of course all constraint forces are ultimately external). It is precisely the content of d'Alembert's principle that one may omit ...

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This result is consistent with the principle that says "the work is negative whenever the electric potential increases", and the one that states "the electric potential increases whenever a charge moves in the opposite direction of the force applied to it by the electric field of another charge". My question would be whether my answer is wrong and I am ...

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$U$ is the electrostatic potential energy of the system of the three protons. To assemble the three protons coming from infinity external work needs to be done so the electrostatic potential energy is positive. So the textbook answer has computed the external work which needs to be done to bring the protons together. Another approach is to compute the work ...

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I don't think it matters really, it's a matter of convention. In your picture if we start at infinity: $$W = \int_\infty^r dr' F(r')= C \int_\infty^r \frac{1}{r'^2} = \frac{-C}{r} \\ E_{pot} = -Q\int_\infty^r dr' E(r') = \frac{C}{r}$$ Energy conservation holds:$E_{pot}+W=0.$ Depending upon how you define work i.e. if the the system "carries out the work" ...

2

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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There are a lot of very physics-y answers here, so I'll give a more real-world example to help you wrap your head around the idea. Imagine (or better yet, try!) you have a weight on the end of a string and you are twirling it around your fingers. Once the weight is rotating, you let it wrap around your finger. You can see that as the object comes closer to ...

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The centripetal force is directed radially inwards . Work done due to the centripetal force is 0 (as, S(displacement in one rotation)=0). Work done = F.S.cos(\theta)=m . a . cos(\theta).The accln. is thus directed tangentially outwards. Thus, tangential velocity increases.

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When $\theta = 0$ the spring extension is zero and take this to be the zero of gravitational potential energy. With the rod at an angle $\theta$ from the geometry of the system find the extension $x$ of the spring and the vertical height $h$ through which the centre of mass of the rod has fallen. The potential energy of the system $V = \frac 12 k x^2 - mgh$...

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My guess: $\theta = 0$ might not be equilibrium cause if you stop holding the A box, it will fall. at $\theta = 90$ might be neutral equilibium (I assume the spring can pass through the wheel to the side of box A), since there is no tension in the spring, and the normal force on the B box balances the forces of gravity(assuming the B box is restricted to ...

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