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1

The difference is that $dW$ is an infinitesimal ''quantity'', whilst $W$ is not. I assume the context here is thermodynamics, which make use of calculus. In calculus there is the concept of the infinitesimal. I suggest, for you, to concern yourself with the structure of calculus if you are to tackle thermodynamics.


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Work is done as long as the force is applied on the body, so in this case, the total work done would be the product of Force applied and the Displacement during the initial push only. If the object moves forever, it would do so with a constant velocity in this scenario, and consequentially its Kinetic Energy would be constant, implying that the work done ...


1

In this case it's because of who's doing the work. The derivation you cite regards the work as being done by the gravitational field. This means in bringing the object from infinity, the gravitational field has lost that energy to the object. Conversely, the work done on the object is the positive value. Everything else works out and when we take the ...


0

A varying magnetic field generates an electric field, and an electric field can do work on a particle. This is called Faraday's law of induction: $$\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$$ The full Lorentz force equation is $$\vec{F} = q(\vec{E} + q \vec{v} \times \vec{B})$$ So for example, if the magnetic field is increasing in ...


2

The force on a charged particle is called the Lorentz force, and it is give by: $$ {\bf F} = q({\bf v} \times {\bf B}) $$ where the $\times$ symbols means a cross product. This means the force ${\bf F}$ is always at right angles to the direction of motion ${\bf v}$, and therefore the work done on the charged particle is zero. The Lorentz force can ...


2

When you push something and it remains at rest your muscles transfer energy through isostatic muscle contraction/respiration. This means that even though the muscles don't move they convert the glucose into respiratory energy for muscle contraction that will be dissipated eventually by heating the surroundings. The only work done is that in contracting the ...


0

First of all, the conservation of energy states that energy is neither created nor destroyed. Therefore if the system was closed then: ME= ΔKE+ΔPE So when the body is first thrown, work is done on it (displacement and force). This work transfers an amount of K.E energy. from this point the kinetic and potential energy transfer to each other as the body ...


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I don't know the context of the original statement (in the textbook), and I have to guess what is meant by "the entire physics". But I can jump in with some thoughts that might help. Let's limit ourselves to classical (not quantum) physics. It's impossible to use the laws of classical mechanics to completely account for the behavior of any real system. ...


3

This is a common misconception. When you apply a force upward on the object, the "reaction" force in Newton's 3rd Law is NOT the force of gravity down on the object; they do not have to be equal, and as you said, cannot be equal if you are to accelerate the object upwards. It is just a confusing coincidence that the force of gravity kind of looks like a ...


3

That is nothing but a mechanical relation valid for every type of isotropic fluid (even viscous, in the quasi-static regime), so gases in particular, no matter any further state equation.


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By definition, work is the energy required by a force to displace something. So, you're not doing any work, you're just cancelling out the force being applied to you. If you wouldn't push back then the other force would be doing work by displacing you. So, your work done basically cancelled out the work done by the other force.


2

In the physics definition of "work done" energy is transferred from one object (the one doing the work) to another object or system. When you push against the stationary stone you apply effort but the energy transfer is all internal to you own body - glucose being metabolized, etc. You get tired but you do no work according to the definition above.


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Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another. Let us now prove that the above law holds good in the case of a freely falling body. Let a body of mass 'm' placed at a height 'h' above the ground, start falling down from rest. In this case we have to show that the ...


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so is that work being done on it due to the conversion of Potential into Kinetic Energy or Gravitational Force is doing the work? Yes to both statements. The gravitational force is causing the conversion from potential to kinetic energy. A force is required for an energy conversion.


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In your case, no work is done. Intuitively: If you want to move a wall, you could push on it and you might use a lot of force. The wall isn't moving, and you are simply "wasting" your energy in your muscles. You are not doing any work of any value. If you push a balloon you can push it far without any real effort. You might move it a long way but that ...


0

Yes. When you apply a net force to a mass (please note the word "net"), the object become accelerated . This acceleration means the body changes velocity, and a change in velocity means there is a change in the energy, because of the energy formula: $E=\frac{1}{2}mV^2$ The case of the circle is particularly interesting, because there is a force applied ...


1

No. In a uniform circular orbit the orbiting body maintains constant energy while a constant force, only changing in direction, operates on that body. Kinetic energy changes when a net force is applied in the direction that an object is moving. It will reduce if the net force opposes velocity, or increase if net force supports velocity.


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By the one who's exerting the force. It depends. When a system exerts some force on some other system, it does work on it and loses its energy. The lost energy is transferred to the second object, which in turn does work against the frictional force and loses its energy which is transferred into the molecules or atoms of the surface as heat.. There are a lot ...


1

$U$ in your equation in potential energy, and $W$ is internal work. That is, the work done by forces within the system. The system in question here comprises the object and the earth, and the internal force is gravity. The work that you do to lift an object is external to the system, and does not appear in your formula. In your scenario, lifting an ...


1

It would help me, but even more importantly you, if you defined what your symbols are supposed to mean. You did that with $U_i$ and $U_j$. So let's be precise together, as an exercise: Let's explicitly state that we consider a point mass $m$ in a gravity field caused by a much larger mass $M_E$ at the origin of the coordinate system, so that we can assume ...


0

For your case work is done by the one who applies force and displaces an object in any direction. You take this example "if an object falls from a height then the work is done by gravity (look who is applying force here )" I hope u understood it . Work cannot be done by a body who does not have it's own energy source like an hammer can never do work.


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For your case work is done by the one who applies force and displaces an object in any direction. You take this example "if an object falls from a height then the work is done by gravity (look who is applying force here )" I hope u understood it .


1

So why have we suddenly jumped from a line integral to not a line integral For a point charge, the electric field has a radial component only $$\mathbf E = \frac{kQ}{r^2}\hat{\mathbf r}$$ Thus, the dot product of the electric field and the infinitesimal displacement vector is $$\mathbf E \cdot d\mathbf l = \frac{kQ}{r^2}\hat{\mathbf r}\cdot \left( ...


0

Strictly speaking, if a body is moved from a point to another by some force, it is said that the force does some work on the body.


0

Probably not the correct way to think about it but this is my reasoning if you applied force on the body you had to spend the energy to do that work therefore you did the work the body gained the energy in some form say you lifted it of the ground the body has gained potential energy while you lost some bio-mechanical energy.


0

If you exert a force on a body and it becomes displaced, then it is said that work is done by you or, by the force on that body.


1

Sally is correct. Of course they both do the same amount of work lifting the bag, but Sally does additional work moving the bag to the truck. This is because the bag had an initial horizontal velocity of zero. It is not possible to move something at a constant velocity that has in initial velocity of zero. First it will be necessary to accellerate it for ...


1

I suspect that the variable-density case isn't a common exercise because a non-uniform chain/rope isn't very common! But the definition of work, $$ W=\int F(x)\,dx=\int g\mu(x)\,dx $$ will work just fine here for the non-uniform chain. As an aside, my calculus textbook, Thomas' Calculus (10th edition), covers non-constant forces of the form $F(x)=a-bx$. A ...


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It seems like a pretty straightforward problem, and there is no reason why you couldn't solve it with standard calculus methods. Go ahead and create your problem (and the solution). There's nothing tricky here.


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You are supposing that the acceleration $a$, which is the second derivative of the distance $s$ (I don't use $d$ because $\mathrm{d}d$ looks awful), $$ a = \frac{\mathrm{d}^2s}{\mathrm{d}t^2}$$ can be written as $$ a = \frac{\Delta v}{\Delta t} = \frac{\Delta(\frac{\Delta s}{\Delta t})}{\Delta t} = \frac{\Delta^2 s}{\Delta t^2}$$ which simply doesn't ...


1

The overall potential energy change is zero. But as you will have noticed, you get tired doing this. That's because you are doing work on the escalator. If a person stands on a perfectly efficient friction-free escalator, work will be done on it as they go down, and the esclator would generate electricity. This is the conversion of potential energy into ...


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Essentially, what you are missing is that you need to look at your situation from the point of view of the escalator steps. The escalator "sees" you "climbing" it because each step you take places you at a later escalator step. You don't keep landing on the same step. A force is applied by your foot on each escalator step. Therefore, each escalator step ...


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The force is applied at the step. You apply that force downward and the step moves downward at the same time; work is force times distance (with a directional factor that is roughly 1 in this case). So the work is non-zero (and is similar to the work you would do climbing the same number of steps if the elevator was stopped.1 Now, Rahul analyzed the problem ...


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yeah,you are right. since you are at the same distance above the ground so you have your potential energy constant so you are not doing any work though if you were on the frame of escalator then you are moving so work is done in form of kinetic energy.


0

Gravitational potential is defined such that it is $0$ at infinity, and has negative values at all points other than the one at infinity. So, an object technically has the highest gravitational potential at infinity. (an extremely extremely far distance.) The incline can be considered as path from an area of lower gravitational potential to a higher one. ...


0

The sun gets its energy from the nuclear fusion, mainly the fusion of hydrogen atoms to form helium atoms. This energy passes through the vacuum of space via Electromagnetic radiation. Plants collect some of this radiation and use it to perform photosyntheses, which in turn is used to make carbohydrates and such. You eat that plant or some other animal ...


1

When you walk up a hill, pushing a bicycle or not, you increase your potential energy by spending chemical energy. One of the reasons you need to eat is to ingest fuel, so to speak, that allows you to spend energy on doing your daily tasks. For example, your body can metabolize sugar (most notably glucose) by oxidizing it, which frees energy that you then ...


0

We normally define work something along the lines of "force applied over a distance," but for the case of work done on a rotating object, it would be force applied over an arc-length. So, for some small amount of work $dW$, need to have a force at some distance form the center of rotation (actually a torque), $\tau$, which goes over some small angle, ...



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