# Tag Info

5

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-----------...

3

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 ...

3

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 ...

3

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'...

2

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 ...

2

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 ...

2

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, ...

2

Mass is accounted for already in the force. $F = m \cdot a$ Then work formula could also be $W = m \cdot a \cdot d$ where the mass $m~ [kg]$ is accelerated at $a ~[ ^{m}/_{s^2}]$ over the distance $d ~[m]$ which is equal to $W ~[N \cdot m]$ or $[\frac{kg \cdot m}{s^2} \cdot m]$ Always look at units they are probably the most important thing to use to ...

1

In retrospect, I should have asked "Why is mass irrelevant when calculating work as in the Newton-Metre or Joule?" Although a larger mass will receive a force for a longer period of time while crossing the same distance, the resultant velocity is also proportionally slower, so the overall energy transferred to the object is not affected by the object's mass....

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In SI units: $1~\mathrm{N} ~\mathrm{m}=\mathrm{1~kg~ m^2 ~s^{-2}}$ and $1~\mathrm{J}=\mathrm{1~kg~ m^2 ~s^{-2}}$. So the units Newtonmeter and Joule are the same in SI-Units and there dimensions are actually equal too. This is a fact and just a matter of definition. Work along a curve $C$ is $W=\int_C \vec{F} \cdot d\vec{s}$. So in the simplest case where ...

1

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 ...

1

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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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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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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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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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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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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