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The answers are correct, the justification has some major problems. You have a ramp, so the slope is not 0. That means that the normal force (which is perpendicular to the ramp) and the gravity (vertical) are not collinear, so they cannot cancel each other. In a reference frame with an axis parallel to the ramp, the gravity has a component perpendicular to ...


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If we are dealing with the gravitational field of the earth, then because of the inverse square law, you should, in theory, need a tiny bit less force to raise an object from 50 m to 100 m, as you do lifting it from ground level to 50 m. But this difference in force is minute, I don't know offhand if we have instruments to measure it. But you can work it ...


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In a rigid body, according to Goldstein's definition, the distance between any two constituent particles does not change. Work done is force times distance moved in the direction of the force. There is no relative movement in the direction of any force. Therefore, regardless of the form of the internal forces, no work is done by or against them.


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As drawn the disc's angular speed $\omega$ is too fast as related to the velocity of the centre of mass of the disc $v$ for the no slipping condition ($v = R \omega$, with $R$ the radius of the disc) to be satisfied. You can think if the frictional force as trying to accelerate the centre of mass whilst at the same time the frictional force applies a torque ...


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Yes and I think this can be understood easily. When the same object is first lifted to 50m and then to 100 you can conclude that you need the same force but you need to apply it for a longer time(or distance). So, the force you need is same but there will be larger work done(consumption of energy) for larger distances


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Recall the difference between the weak and strong Newton's third law, cf. e.g. this Phys.SE post. If the internal forces satisfy the weak Newton's third law (but not the strong Newton's third law, i.e. without the collinarity assumption), then it is not guaranteed that the internal forces do no work, cf. e.g. Fig. 1. ^ F | | 2 x-----------...


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What do people actually mean by "rolling without slipping"? This question and the answer should give you better insight into your question. Think about the definition of rolling without slipping given and how friction is created and works for rolling objects maybe make a free-body diagram.


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To expand on the above answer by @PhysLab (which is very well written), the answer depends on how "in depth" you want to look at it. Before discussing energy transfer, let's cover a few fundamentals, because these are slippery concepts and often misunderstood. (Including by me, so fingers crossed!) Let's look at what "energy" and its "transmission" involves....


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Energy transfer can be thought to occur via the exchange of a 'virtual particle'. In nature, there are 4 fundamental forces, namely: 1. Electromagnetic force 2. Gravitational force 3. Strong force 4. Weak force Each of these forces have a different exchange particle: For instance, the exchange particle for EM is a photon whereas that for the strong force ...


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There is no acceleration in Y direction. If you consider the X and Y axis like this. Maybe this will help. Work done is P*d. Remember the net force will always be zero because there is no acceleration.


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Consider a piston cylinder arrangement. Pressure * Area equals force and this force moves by a distance ds (consider piston moving upward by a distance ds) then work done = Fdx = PAreads =Pdv. Volume of cylinder equals Area *height


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Q is heat and W is work. Q is considered positive when it is added to the system and work is considered positive when work is done by the system. Also remember that Q and W are energies in transit I.e energies associated when system moves from one state of equilibrium to another state of equilibrium. One cannot say that "Q" amount of heat is present in the ...


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On the other hand this does not makes sense, because Carnot theorem requires that in case 1 (in wich all processes are reversible) the efficiency is the same of a Carnot engine working betweeen highest and lowest temperature (in this case $T_{D}$ and $T_{B}$), that is, in case 1., efficiency should be $$\eta_{\mathrm{reversible}}=1-\frac{T_D}{T_B}\tag{II}$$ ...


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First, for an isochoric process, $Q=nc_v\Delta T$ is not true in general, because you can increase internal energy of a system by other means (paddle work, for example). Let us assume that there is no such process involved in your case, so indeed $Q=nc_v\Delta T$ for isochoric process. You are right in saying that this relation holds whether or not the ...


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There are basically two possible answers, depending on the character of the irreversibility. So let's think: What would an irreversible process look like? Here are some examples... The hot reservoir is much hotter than the ideal gas, and/or the cold reservoir is much colder than the ideal gas. This does not change the engine efficiency. It merely ...


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Infact during any isochoric and isobaric processes (reversible or not) $$Q_{\mathrm{isochoric}}=n c_v \Delta T$$ $$Q_{\mathrm{isobaric}}=n c_p \Delta T$$ I think this is wrong. Because $Q$ is a path function. True formula (I think) is as below: $$\Delta U=n c_v \Delta T$$ And it is same for reversible and irreversible processes (if $\Delta T$ is same). ...


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if there is a fan inside the tank and if the tank's volume is kept constant, what will be happen? The air will circulate and viscosity will convert kinetic energy to thermal energy, i.e. temperature will increase. The electrical energy input will be eventually converted to thermal energy which is $\delta Q$. So $W=pdV-Pt$ is not correct. The equations are, $$...


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It doesn't seem possible for the original premise to be correct. If you start at a certain state, and carry out a reversible Carnot cycle ending up at the same original state, you can design a great big Carnot cycle and you can design a tiny little Carnot cycle. Certainly the reversible work for these two cycles will not be the same.


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The area enclosed within the loop is the net work and for the complete cycle you've illustrated, work (energy) flows both into and out of the system at different points within the cycle. But there is a net loss. From $A$ to $B$ and $B$ to $C$, the direction is positive (by right hand convention) and so energy flows into the system. From $C$ to $A$ the ...


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You have to be careful when drawing irreversible transformations on the PV plane. Apart from the fact that drawing them is meaningless because they are not a set of equilibrium points, you cannot just assign some arbitrary property to them. You drew the blue curve and called it an "irreversible adiabatic", giving it some arbitrary slope. Fine, since we don'...


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Remember than the definition of the efficiency is the ratio between the work "extracted" or transferred to the machine and the heat taken from the hot source. $\eta=\frac{W}{Q_h}$ For a reversible process this expression is equal to $\eta=1+\frac{Q_c}{Q_h}$ But for an irreversible process it is not. The reason is that the work done by the gas, is ...


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First of all note that human muscles require energy even when doing no work, so be a bit cautious about analysing any situation involving humans manipulating objects. See Why does holding something up cost energy while no work is being done? for more details. If we replace the human by some form of mechanical cantilever then you are quite correct that the ...


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Image object $m$ is at some point $a$ and you were to supply a force opposite to the gravitational force caused by $M$. This force is equal to $F_{stop}=-F_g$ so that $m$ hovers completely still at point $a$. M ------------------- a ↑ m Obviously this force can't do any work because the object doesn't move and ...


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Saying that work done is +ve or -ve is a mathematical convention used for calculating energy transfers. It is +ve when it is done by us on the system, and -ve when it is done by the system on us. Positive work is done in pushing against a force to reach a configuration - eg pushing a car uphill. Negative work is done if a force pulls in the direction we ...


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The object would experience a net negative force, and be moved a negative displacement. Potential is defined in terms of the work done by an external force. The object has a negative force acting on it due to the gravitational attraction so the external force acting on the object must be in the positive direction to have a net zero force on the object. It ...


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The work done by an external agency in bringing the body to a point is needed. Note that the process must be quasistatic, otherwise kinetic energy terms will be needed. Now gravity will tend to attract a mass. Thus to keep the process quasistatic, one must oppose this gravitational attraction. In other words, F is directed oposite to the gravitational ...


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according to me you have erred in assuming that change in KE is 0 as the block will have some velocity when it is compressed.since total mechanical energy remains constant loss in P.E will result in gain of K.E


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You can compute the integrals $W=\int\vec F\cdot d\vec r=q\int E\cdot d\vec r$ for each charge $q$ you bring from infinity, where $\vec E$ is the electric field generated by the other charges present and then add them all together. On the other hand, since electrostatic fields are conservative the work done is independent of the path, so it is conservative ...


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Find the potential of the system once the charges have been arranged. Can you relate that to the work done to assemble the system?


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Work is defined by formula below for any kind of processes not only for quasi static process. $$\delta w=P\mathrm dv$$ But, what is the $P$? $P$ is the pressure that resists against the system boundary movement not pressure of the system itself. So, you don't need to have a quasi static process for calculating the work because you don't need pressure of the ...


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When you write "energy given out" or "energy gained" you are expressing a choice of sign for the energy transfer that you are specifying, and each of them is different. Notice that in your scenario A loses energy, so it's "energy given out" is positive, but it's "energy gained" would be negative. Employing a sign convention means always stating changes of ...


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The work is not done by a conservative field. If the field were conservative the force made by the piston should be the same at a given volume, that is for a given position of the piston, regardless of the state of the gas. But this is not true. For instance, you can add heat at constant volume to the gas, with results in an increase in pressure. The force ...


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Consider the case of an ideal gas satisfying the relation $$PV=nRT$$ where $P$ is the pressure, $V$ is the volume, $T$ the temperature and $n$ the number of moles of gas particles. The work done is (assuming an isothermal process) $$W=\int_{V_1}^{V_2} PdV=nrT\int_{V_1}^{V_2} \frac{dV}{V}=nRT\ln\left(\frac{V_2}{V_1}\right)$$ At first sight one may see ...


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You have added the negative sign in front of your integral and then put in the cos(180) as well. Pick one. Since your external force is opposite in direction to the force of the sphere, and you've already put the cos(180) in there, there is no need for the extra negative sign in front.


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The force exerted on a unit positive charge is $ \dfrac {kq}{r^2} (+\hat r)$ and so an external force $ \dfrac {kq}{r^2} (-\hat r)$ must be applied to move the charge. If the step is $d\vec r$ then the work done by the external force in moving that step is $ \dfrac {kq}{r^2} (-\hat r) \cdot d\vec r = -\dfrac {kq}{r^2} dr$. You have moved from $\vec r$ to $\...


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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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