New answers tagged

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The general relation between potential and a conservative force is $$\mathbf F(x) = - \nabla V(x) $$ Integrating along a path C, $$ V = -\int_C \mathbf F.d\mathbf s $$ The sign convention is natural as the direction of the force is opposite to the gradient of potential (c.f. gravity). This is the force applied by the field. If you think instead of the work ...


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There are charges that create the potential, which exert force $\mathbf{F}$ on the test charge and perform work $W$ when (slowly) moving this charge to infinity. And there are also external forces which do the opposite opposite, i.e. exert the opposite force $-\mathbf{F}$ and do the opposite work, $-W$. Distinguishing these two clarifies most of the ...


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The garage door is an example of constrained motion. End A is constrained to move in a horizontal line. Points B, D are constrained to move in a vertical line and to stay in a rigid position relative to A, C and C, E respectively. The object dropped from a height is not constrained while it is freely falling. A fairer comparison would be to examine the ...


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Let's consider one molecule that hits the piston. The kinetic energy of that molecule is supplied by breaking a chemical bond during combustion. Just before the molecule hits the piston the molecule's momentum is p = mv, and momentum of piston is p = 0. After the collision the law of conservation of momentum says that sum of momentum must be the same. Hence, ...


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Therefore we will conclude that whenever a negative charge moves from a lower potential to higher potential, work done by electric force is positive That is correct. The electron gains electrical potential but loses electrical potential energy. The energy for doing positive work on the electron comes from the electrical potential energy of the charge ...


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The increment of work done by a force is $dW = {\bf F} \cdot d\bf x$. The force on a charge is ${\bf F} = q {\bf E}$, where the electric field is ${\bf E} = -\nabla \phi$. That means that the increment of work done on a charge by the electric force is $dW = -q\nabla \phi \cdot d{\bf x} = -q d\phi$. So, by moving from a lower potential to a higher ...


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Say, you've got a negative charge and a plate near it. Now suppose that the plate has a positive charge. So, you can say that the "location" of the positively charged plate is at a higher potential than the "location" of the negative charge. Also, it is obvious that the positive plate will attract the electron towards it, and the electron would move from ...


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I do not know of any specific reference which shows this integral in the context of any work done, however one might come across this integral when dealing with ideal gases. Let's say we have an ideal gas whose volume is $V$, pressure is $p$, temperature is $T$ and it consists of $n$ moles. Thus by the ideal gas law, $$pV=nRT$$ Now let's assume it ...


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You require your system to be in equilibrium to be able to DEFINE quantities like pressure(how, for example, would you describe the pressure of a turbulent fluid-each point in it is at a different pressure!). By definition, then, the system is moving quasi-statically through these equilibrium steps. This is a reversible process. The equations of state hold ...


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You cannot. The quantity $PdV$ only makes sense when you are talking of a reversible process, as irreversible process may not even have a relevant $P-V$ plot to talk about the quantity $PdV$. This is because one can only draw a $P-V$ plot for states of equilibrium of a system. So, an irreversible process can have a $P-V$ plot consisting of just $2$ points - ...


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The way I see it, thermodynamical variables like $p$ and $T$ are only really defined in equilibrium. So when you draw curves in the $pV$ plane, you are implicitly saying that the process is quasistatic. Whether a quasistatic process is reversible or not is a confusing issue (to me); in the purest of situations it will be, but it depends on what happens in ...


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Since the weight lifter is lifting the weights normal to the ground there is no actual work being built up .. to justify : Work done = Force * displacement * cos(theta) work done = Force * displacement * 0 . . . . . . . . because theta here is 90 and cos 90 = 0 hence no work is done :)


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If the energy outputs is constant and linear with time, why isn't work also linear with the time of the net force (impulse)? You're making an assumption that the energy (power) output is linear with time, and it's not (or if it is, then the force isn't constant). As an example, a car engine can (for some regimes) be considered a constant power device. ...


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Does this mean that some curves 𝐿 have the property of being reversible paths and some don't? Yes. The work done by a reversible process between the same two equilibrium states will be greater than for the same process carried out irreversibly between the same two equilibrium states. Each has a different shape "curve" between the start and end points. ...


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I think there's a big conceptual issue here. 1- A field does not gain or lose energy, a material system gains or loses it. If a charge moves, it might gain energy, or lose energy. But we're talking about a particle. Energy belogns to systems, not to forces or fields. 2- A usual confusion: $W_{cons}=-\Delta E_p$ The work done by the electrostatic force ...


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When a charge is moved from infinity to $r$, the EPE is equal to the negative work done by the field. In this argument, is it also true that there is positive work done by the force to get from infinity to $r$. Or is it just that one exists. Its both. An external agent does positive work and the electric field simultaneously does an equal amount of ...


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definition of voltage as work done moving unit of electric charge between two points in electric field That's almost okay, but what's missing is the concept of electrical potential. And technically, you should say "work per unit charge" to get the units correct (volt = joule per coulomb). That's often not discussed in a lower level (high school) first ...


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Let's suppose I push my desk into the wall. It doesn't move at all, but I get tired. The sensation of getting tired has almost nothing to do with how much external work you do. A human is a very inefficient machine and tiredness is a very non-specific measurement. Whenever you are formulating a question in those terms it is likely not relevant to the ...


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Suppose you have a charge $q$ and an electric field $E(r)$ in space. By integration of the electric field, also an electrical potential $V(r)= \int E(r) dr$ can be found. The electrical potential energy of the charge at position $r$ is given by $\mathcal{E}(r_1) = qV(r_1)$. When the charge moves to another place by the electrical force $F(r)=qE(r)$, it will ...


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It doesn't move at all, but I get tired. The fact that you get tired is about you and not about the table. The work we are talking about is the work absorbed by/from the table. And there is no such work. Our human bodies just happen to expend energy within our muscles in order to generate a force. If you instead tie and rubber band around the objects so a ...


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While one can specify a voltage at a point, it must always be with respect to another point, sometimes referred to as common, ground or reference point where the voltage is arbitrarily assigned a value of zero. The zero voltage point is typically chosen at the negative terminal of the voltage source. For the simple circuit below, I have labeled points A, B,...


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Sometimes notation may induce misunderstandings. It is true that in many formulas voltage appears as $\Delta V$, making clear that it refers to the difference of voltage between two points. However, arrived at Ohnm's law, it is usual to find it expressed as $$ I=\frac{V}{R},$$ thus inducing to think that only voltage at one point is present. Actually in ...


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in Ohm's law we are measuring voltage at a point so work done should be very small since there is almost no distance This is the key mistake. In Ohm’s law the voltage is the voltage across the resistor. In other words, the voltage in Ohm’s law is measured at two points. One point is always the point where the current is entering the resistor and the other ...


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Wikipedia says In physics, work is the product of force and displacement. Or to put it mathematically $$\overrightarrow W =\int_{\overrightarrow x_{i}} ^{\overrightarrow x_{f}}\overrightarrow F\cdot \overrightarrow {ds}$$ here ${\overrightarrow x_{i}}$ and ${\overrightarrow x_{f}}$ are initial and final position vectors and that dot is dot product. As ...


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W = fdcos(theta). And clearly, there must be a force enacted to change the velocity of the object. Assume there is a distance present over which the force acts. Now, the work should be negative as cos(theta) is negative. Here is where you go wrong. $\cos(\theta)$ is negative in the beginning as the object slows from the initial positive velocity to 0. At ...


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(I) In first case the total force applied by man has two components ,that is , a horizontal force and a vertical force. Vertical force is doing no work as there is no displacement in vertical direction. But the vertical force is to be applied because of gravity. The horizontal force is doing work. So, W= F(horizontal) . S (ii) In this case the total force ...


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I believe there’s insufficient information to calculate the work done in (i). You need to know the velocity of the mass at 2 meters. Then, assuming it started at rest and its height didn’t change, you could calculate the net work done on the mass by the man using the work energy theorem which states the net work done on an object equals its change in kinetic ...


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Note that the question is about the work done on the weight, not the work done by the man. The work done by the man will be greater than the work done on the weight because the man's muscles are not 100% efficient and because energy is lost due to friction. In case (i), if the weight is carried at a constant height about the ground, then the force required ...


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Rather than thinking of the force applied by the man and the distance moved by the block, think of the energy difference. You will notice that in the first case, the body is just shifted parallel to the horizontal surface i.e. its P.E does not increase. But in the second case, the potential energy increases, which invariably means that he has done work. ...


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Short Answer: It is zero. Long Answer: Assume you are on a frame of reference of road (i.e you are standing on road) then the system is truck + block. You observe both of them moving now looking closely friction is holding block on truck by directing itself in forward direction, using Newton's third law equal and opposite forces are acting on the truck. ...


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Per the work energy theorem the net work done on an object equals its change in kinetic energy. Since the block begins and ends at rest its change in kinetic energy is zero. Therefore the net work done on the block by the static friction force is zero When the truck accelerated the work done by friction was positive giving it kinetic energy since the force ...


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The work $W_{\rm nc}(t)$ at instant $t$ of non-conservative (nc) forces in eq. (4.6) is ill-defined. Given a path $t\mapsto q^j(t)$, it is unclear what work $W_{\rm nc}(t)$ refers to. This dooms a variational principle of some action functional for nc systems, in agreement with standard lore. On the other hand, the infinitesimal virtual work $$\delta W_{\rm ...


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The notion required here is the concept of system. A system is a set of parts that interact. So, asking "If a force is applied on a body in an isolated system" is an ill-formed question. If the system is isolated, an external force cannot be applied to one of its parts. If an external force is applied to the system, the system is not isolated. As you ...


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Recall how work is defined via the non-conservative force: $W_{nc}[x, \dot{x}, t]=\int_{\vec{x}(t_1)}^{\vec{x}(t)}\vec{F}_{nc}(x', \dot{x}', t)\cdot d\vec{x'}$ Requiring $\int_{t_1}^{t_2}(\delta L + \delta W_{nc})dt=0$ gives, by a completely analogous argument to ordinary Euler-Lagrange, $(-\frac{\partial}{\partial\vec{x}} +\frac{d}{dt}\frac{\partial}{\...


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Krishnanand J's answer gives a good explanation, however, I assume (by how you asked) that you are considering yourself to be a part of an isolated system. Now, if you are applying a force on the system, and the system is accelerating, then you are doing work on it. And you cannot do the work without using your energy. So the kinetic energy of the system ...


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As you know, the Law of conservation of energy holds only for isolated systems. From Wikipedia, Law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. Now, what is an isolated system in physics? As stated in Wikipedia, it is a physical system so far removed from other ...


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The work-energy principle, when applied properly, says that the net work done by all forces acting on the system matches the change in the kinetic energy. That means that instead of accounting potential energy, you let that count for work. If the change in PE is $\Delta U$, then the work done by the conservative force is $-\Delta U$ (by definition). If some ...


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It might be helpful to first get the result, then ask why it is not zero. The spring initially had no potential energy, but upon being compressed, stores a potential energy $$U=\frac{1}{2}kx^2$$ $x$ being 0.1m here. By energy conservation, this has to be supplied from outside, and is equal to the work done on the spring. So, why is the work done not zero? ...


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What came first: the work-energy theorem OR work and energy individually which were later reconciled in a theorem? The work-energy theorem states that the net work done on an object equals its change in kinetic energy. It can be derived from Newton's second law. See the following for a derivation: https://courses.lumenlearning.com/boundless-physics/...


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The work is the dot product of the force and the displacement, and displacement is a vector; we have to take into account what direction it is pointing. If an object is traveling in a circle, then it has to have a centripetal force, so it doesn't have a constant force. It could have a constant magnitude force, though. The centripetal force is perpendicular ...


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If, by $p_{gas}$, the author means the force per unit area that the gas exerts on the piston face, then this discussion is incorrect. That is because, for a massless, frictionless piston, by Newton's 2nd law, the force the gas exerts on the inner piston face must match the external force on the piston, $p_{piston}A$. On the other hand, if, by $p_{gas}$, ...


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The difference between the two pressures integrates to a residual force whose work is invested in accelerating the piston. However, at the end of the process the piston is static, so there must exist additional resisting forces acting during the process in the opposite sense, and it must hold that work done by the resisting forces must be equal to (minus) ...


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The reason for $p_{gas}\ne p_{piston}$ in equilibrium or during a quasi-static process is that you have some friction between the walls and the piston or viscous effects within the gas, both processes being dissipative are irreversible.


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The reason for this is that your understanding of the definition of work contains an error: "product of ... distance (scalar)" Work is not defined using a product of force and a "distance", but a displacement. A displacement is the difference of two positions, which are points, and displacements are vectors. So it is actually the dot product of two ...


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The general definition of work is $$W=\int\mathbf F\cdot\text d\mathbf x$$ Which essentially says, "Add up all of the dot products between the vector force $\mathbf F$ and the vector displacement $\text d\mathbf x$ along the path the object travels on." Since we are adding up dot products, which are scalar quantities, the work done by a force is also a ...


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Work is the dot product of a vector force and a vector displacement, hence a scalar. Knowing just the scalar distance isn’t enough to calculate work. That distance might be in the same direction as the force, but it might be perpendicular or even opposed. All of those would give different values for the work done.


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Without all the context I assume that the book is trying to make a distinction between a "real" battery, and an "ideal" battery. An ideal battery would deliver the same voltage regardless of the current draw. A real battery essentially has an internal resistance. Thus as the current increases the voltage decreases.


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The work energy theorem is indeed a tricky beast and it is easy to misuse it. Unfortunately, most derivations are rather lax about identifying the assumptions in the derivation. Most derivations assume that the system being analyzed is a point particle with no internal degrees of freedom. Your system here has internal degrees of freedom and an internal ...


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It's because the work depends both on force and displacement. Let's say that the first body was displaced by $\vec d_1$ and the second body by $\vec d_2$, then the work: $$ A=\vec F_1\cdot \vec d_1+\vec F_2\cdot \vec d_2=\vec F_1\cdot \vec d_1-\vec F_1\cdot \vec d_2=\vec F_1\cdot(\vec d_1-\vec d_2). $$ Thus, unless the displacement of one body relative to ...


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$F_{ab}+F_{ba}=0$ tells us that there is no net force on the system consisting of the two particles. That is, there is no external force on the system, and no work done on the system. The total mechanical energy of the system is constant. But there are internal forces in play in this system, and work is done internally on both particles. But the total ...


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