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17

"Curl" is a pretty well named mathematical term--it denotes the degree of "rotation" in the vector field. For this reason, if you go all the way around in a vector field, you'll find that the total integral along that path will depend on the curl of the field in question. If a force had a curl, you could go all the way around and have some net work done, and ...


3

The contact point between the wheel and the car is stationary - there is no "rubbing" there (well there is because the contact point is really a patch but let's keep it simple). To do work you need "force times distance" - and without relative motion there is no "distance". When you apply the brakes in a car you have sliding of the pads relative to the ...


3

I suppose you read this passage in the famous Feynman Lectures. I am fairly certain that what Feynman is referring to (and what you are looking for) is a proof that an electrostatic field is conservative. There are a number of equivalent ways of stating that a vector field is conservative, each of which can be taken as a definition. Let $\vec{F}(x)$ be a ...


3

One need not follow these steps. Indeed let $\gamma : I\subset \mathbb{R}\to \mathbb{R}^3$ be the trajectory of a particle. It's position at time $t$ is $\gamma(t)$, it's velocity is $\gamma'(t)$ and it's acceleration is $\gamma''(t)$. It's easy to see that $$(\gamma'\cdot \gamma')'(t) = 2\gamma'(t)\cdot \gamma''(t),$$ so the work done by the resultant ...


3

Conservation of energy, as you note, holds for "the system." For instance, if you push on a ball, that ball gains energy, but the energy of the ball is not conserved--only the energy of you and the ball. In this case, the system needs to include more than just "the sloth" because the sloth is not an isolated system--there are external forces at work. Here, ...


2

You are sloppy with units, but the result is correct. To go from 25C to 3C is 22 cal/g. When you multiply by 300 g you have cal and your conversion to kJ is correct. Converting to W-hr is silly, but that is the unit of energy, not W/hr. You have 8.3 W-hr you want to remove. That chills the water assuming no new heat is added, so insulate the water. ...


2

Q = mc(t1-t2), Now, m = (density)(volume), Specific heat of water, c(in joule/gramCelsius) = 4.186, Hence, you can find the energy it would require for this conversion. . And the work you do can be a bit more pertaining to your efficiency.


2

No the point of contact is not at rest. It moves with the block. You are probably confusing with rolling motion in which the point of contact is always at rest. There the point of contact is rest because the lowest point on the disk has two contributions, one due to forward motion of disk as a whole (v) and one in the backward direction due to rotation ...


2

I think you are confused about what $d$ is supposed to mean in the equation $W=F\cdot d.$ You seem to be under the impression that $d$ is the distance that the object being acted on moves relative to the object providing the force. But this is not the correct meaning of $d$ in the equation and you know it. Imagine if the car crate were in front of the ...


2

In terms of the Young's modulus the Hook's law (up to the overall sign) is written as $$F=ES \frac{\Delta L}{L_0}=kx, $$ the corresponding elastic energy (or the work that has be done to stretch a wire) is $$W=ES \frac{\Delta L^2}{2L_0}, $$ here you know the Young's modulus $E=1.3 \times 10^{10}$, the cross area $S=1.7cm^2$ and the initial length $L_0=0.89m$ ...


2

If the spring has been stretched then something must have stretched it. It's the work done by that something that you're calculating. That work has been done on the spring, so the potential energy of the spring has increased by an amount equal to the work done on it. To give a concrete example, suppose you're holding the end of the spring. Then your hand ...


2

$\vec F \cdot \mathrm{d} \vec r$ does not determine if a force is conservative or not. All normal forces (conservative or not) produce work equal to $\vec F \cdot \mathrm{d} \vec r$ but what determines if they are conservative is the integral of $\vec F \cdot \mathrm{d} \vec r$ in a closed loop. If that is equal to zero i.e. if $\oint \vec F \cdot \mathrm{d} ...


2

the heart of a force being conservative is that it is integrable, that, if we have a force ${\vec F}$, then it is possible to find a potential $\phi({\vec x})$ such that ${\vec F} = - {\vec \nabla}\phi$. The reason for this is that if we pick out two points $p$ and $q$, we want the difference in energy between the two points to be $\phi(p) - \phi(q)$, and ...


2

You are right. In Newtonian physics, work depends on reference frame. Force does not. Let's start with a book sitting on a table. The table exerts a normal force on the book, but it does no work because there's no motion. Next, imagine that the table is in an elevator and the elevator is going up at constant speed. The force between the book and table is ...


2

You are using the repulsive force as the force acting to move the charge from B to A(which is not actually moving the charge). We need an external force to move the charge from B to A, which will be taken into consideration(to calculate workdone).


1

No that's not right. a work can be calculated for each force individually. you are mixing it with the equation between work and energy that says: $$W = \Delta K \quad \text{K is kinetic energy}$$ but this work is actually the work of the overall force on the object. the work done by earth gravity force and your force is zero. But each is doing some work ...


1

It is zero; think of the conservation of energy. Actually, because of this law, all force's work on an object will be equal to only its change of kinetic energy (at least in the Newtonian physic), and in this case this is zero. We use the potential energies to ease up the calculations (so you don't have to calculate all times i.e. the Coulomb-force's work): ...


1

As you say, there is a downwards force on the plane of $Mg$ where $M$ is the mass of the plane and $g$ is the acceleration due to gravity. There is also a force due to aerodynamic drag, but let's ignore that for now. The key to understanding how the plane stays up is that force is equal to the rate of change of momentum. In this case the wings of the ...


1

The plane itself is exerting some force in order to overcome the effects of gravity. Consider that if you turn off the engine, the plane would go crashing down. The only reason you are able to observe a constant velocity is because the plane is exerting some power to work against gravity (and other phenomena such as air-resistance etc.) Now you may be ...


1

When a ball of mass m is brought with uniform velocity from infinity into the g field of the earth at a distance r from it, the potential energy of the ball earth system decreases from 0 to -GMm/r. What does this lost energy appear as? Kinetic energy, which is typically radiated away into space. Imagine your ball starts off a long long way from Earth, and ...


1

I think what may be referred to in the question is the determination of gravitational potential energy (GPE). The definition for GPE is the work that was done in bringing the two masses to a distance r apart from an initial separation that was infinite. (Tsokos, 396) What is important here is that this happens at a very slow uniform speed so the Kinetic ...


1

This might be better on the engineering SE site but here is some physics to consider: You are right the the heat you need to remove is the mass of water times the temperature difference times the specific heat capacity. Peltier devices and other heat pumps typically have a parameter called a COP - coefficient of performance. This compares their efficiency ...


1

Formally, we have the standard dot product identity $$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{A}\cdot\mathbf{B}=\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}\cdot \mathbf{B}+\mathbf{A}\cdot \frac{\mathrm{d}\mathbf{B}}{\mathrm{d}t}.$$ Inserting $\mathbf{A}=\mathbf{B}=\mathbf{v}$ gives $$\frac{\mathrm{d}\mathbf{v}^2}{\mathrm{d}t}=2\mathbf{v}\cdot ...


1

According to Newtonian mechanics, it is true that the table exerts an equal but opposite force against gravity that results in $\Delta y = 0$, where $y$ is the up/down dimension. However, sliding block is clearly moving in the $x$ dimension (i.e. horizontally across the table). And it is also acted on by a force, namely friction. The block does not generate ...



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