New answers tagged work
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Confusion in definition of Potential Energy
Without writing the explicit meaning of formulae, comments, and answers may only increase the original confusion.
The work done by a force ${\bf F}$ when a point-like body is displaced along any path ...
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Confusion in definition of Potential Energy
Your definition of potential energy is wrong. Yes, $ΔPE_{AB}=-W_{AB}$, but $W_{AB}$ is not the work done by the conservative force from $A$ to $B$. It is the work done by an external force (against ...
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How can power be defined as the derivative of work with respect to time if work is not a function?
My two cents. I think that's because sometimes it's possible to convert inexact differential to the exact one using integrating factors. For example, if the process is quasi-static adiabatic , then ...
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How can power be defined as the derivative of work with respect to time if work is not a function?
If so, why do I usually find power defined as the derivative of work
with respect to time if work is not a function
Just because work is shown as an inexact differential in the differential form of ...
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What if the point of application of force doesn't move but the force itself moves, does it count as work?
You should understand that Force is a vector quantity which has a property of 'translation', force aren't fixed in position, most of the problems that you've came across have forces that translate ...
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How to mathmematically direction of friction of spool?
We use the concepts of Impulse and Angular impulse.
If the force is applied below the center of gravity in forward direction then, frictional force will act in backward direction because the contact ...
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How to mathmematically direction of friction of spool?
With friction, the bottom doesn't slide. It matches the speed of the table, which is $0$. The spool rolls.
Suppose the spool was sitting on ice, so F was the only force. How would the spool accelerate?...
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Why isn't work $Fd \sec \theta$?
I assume you mean why work is not $Fd\sec\theta$. Only work done by a force perpendicular to displacement is zero. In your case the other component is not perpendicular and contributes to work
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How can the energy of a system increase even if net work done on it is zero?
This will result in increased kinetic energy of the blocks, thus result in an increase in the total energy of the system.
is a correct statement but must include an extra word,
$\dots$ increase in the ...
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How can the energy of a system increase even if net work done on it is zero?
In $W=\vec F \cdot \vec s$ the $\vec F$ is a force acting on the system and $\vec s$ is the displacement of the material of the system where the force is acting.
So in your example both forces are ...
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How can the energy of a system increase even if net work done on it is zero?
Either the pair of blocks are one thing, in which case the one thing's kinetic energy is the kinetic energy associated with the translation of the center of mass of the pair of blocks, which doesn't ...
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Can any irreversible work source be simulated by a reversible work source?
It briefly explains the reason; what the work source does is simply to
apply force to the piston, and therefore it does not matter how the
force is applied, whether or not it is applied by an ...
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Why is Power in an electric circuit equal to $VI$?
Very simply,
The voltage (potential difference) between two points measures how much energy is required to move a unit charge from one point to the other.
And current on a path between two points is ...
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Why is Power in an electric circuit equal to $VI$?
Heuristically (without diving into full microscopic theory):
Currents are made by moving charges, the power in classical dynamics is given by the $P= F v$ where $F$ is the force and $v$ is the speed ...
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Why is Power in an electric circuit equal to $VI$?
The level of bad teaching is beyond atrocious.
The question you have is really just down to definitions.
$$\begin{align}
\tag1\text{Power}&=\frac{\text{Energy}}{\text{time}}\qquad&\qquad P&...
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Would one be pushing as much as their weight with this machine?
His arms move a longer distance than he is lifted. Therefore the force he needs to push with is lower than his weight. There is a mechanical advantage. Much like to lift the rock in this picture, the ...
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Would one be pushing as much as their weight with this machine?
I was looking at the following machine in the video below, and was
wondering, would one be pushing as much as their weight?
His pushing force is less than his weight due to the mechanical advantage ...
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Why does the work done depend on the frame of reference?
A force perpendicular to velocity doesn't do any work. Work is the change in kinetic energy, so
$$\frac{d}{dt}\left(\frac{1}{2}m \vec{v}\cdot\vec{v}\right) = m\vec{v}\cdot\vec{a} = \vec{v}\cdot\vec{F}...
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Why does force perpendicular to the velocity change only its direction; not the speed?
Most answers rightly point out that a force, in order to achieve what you are stipulating (it changes the direction of the object, not its speed), must be constantly (at each instant) perpendicular. ...
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Why does force perpendicular to the velocity change only its direction; not the speed?
It's really very simple, because both force and Velocity are vectors, not scalars. That is, they have a direction and not just a value. When you exert a force on an object, the resultant acceleration (...
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Why does force perpendicular to the velocity change only its direction; not the speed?
The force perpendicular to velocity does not change the speed of an object because it only affects the direction of motion, not the magnitude of the velocity. This is a consequence of the fundamental ...
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Can any irreversible work source be simulated by a reversible work source?
It briefly explains the reason; what the work source does is simply to
apply force to the piston, and therefore it does not matter how the
force is applied, whether or not it is applied by an ...
- 68.8k
0
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Point forces doing work
This is intended to add one more perspective which is, I believe, consistent with the answers given by @BioPhysicist and @Farcher.
Clearly the energy that raises the boy up the ladder comes from the ...
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Newton's third law in thermodynamics
For question 1, when we do a force balance on a body, we include only the forces exerted by other bodies on that body, and not forces which it exerts on other bodies. So the two action-reaction ...
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Newton's third law in thermodynamics
Newton's 3rd law only describes what I consider to be the same force that acts on two objects, not two different forces. And forces not acting on the same object can never sum or cancel each other.
...
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Is Joule equivalent to Joule/radian?
A radian is not a unit in the same sense that a second or a metre is a unit. The last two are defined using special standardising items (e.g caesium light sources) whereas the radian is defined as the ...
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Is Joule equivalent to Joule/radian?
A radian has no physical dimension, it's a pure number. So, if you do dimensional analysis,
\begin{equation}
\dfrac{[torque]}{[angle]} = \dfrac{[torque]}{1} = [torque] \ .
\end{equation}
Recalling ...
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Why does force perpendicular to the velocity change only its direction; not the speed?
I want to come at this from a slightly different perspective.
If you were writing a computer program to simulate a ball on a string but you wanted to simulate it with forces in Cartesian coordinates, ...
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Sign convention of work
But, if work output is taken as positive, please explain why do we use
a negative sign to denote the work output in the equation above when
it should be positive?
If work is positive (work done by ...
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Why does force perpendicular to the velocity change only its direction; not the speed?
I want to highlight something you said:
Even if we assume that at the given instant , the change caused by the acceleration along it does not contribute significantly and magnitude effectively ...
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Why does force perpendicular to the velocity change only its direction; not the speed?
If we represent the movement in a 2D Cartesian plan, the velocity vector at any instant is $\mathbf v = (v_x,v_y)$. The velocity modulus is $\sqrt{v_x^2 + v_y^2}$. Now suppose that the derivative of ...
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Why does force perpendicular to the velocity change only its direction; not the speed?
The magnitude of a vector $\vec v$ can be expressed as the dot product $v = \sqrt{\vec v\cdot \vec v}$ so its derivative can be expanded as
$$\frac{\mathrm dv}{\mathrm dt} = \frac{\mathrm d}{\mathrm ...
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Why does force perpendicular to the velocity change only its direction; not the speed?
Note the following. Let $v$ be the modulus of the velocity and $\vec{F}$ a force perpendicular to it. Now, $$\frac{d}{dt}v^2=2\vec{v}\cdot \vec{a}=2\vec{v}\cdot \frac{\vec{F}}{m}$$
So you see clearly ...
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Why does force perpendicular to the velocity change only its direction; not the speed?
It is true that a force perpendicular to the velocity will change only the direction of the velocity. It's key to note that as the direction of the velocity changes, so too must the direction of the ...
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Why does force perpendicular to the velocity change only its direction; not the speed?
We can use a work/energy argument to deduce that the object's speed (the magnitude of its velocity) will remain constant, as follows.
A force $\vec F$ that acts on an object does work $\int \vec F . d\...
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A man moves on a straight horizontal road with a block of mass 2 kg in his hand. He covers a distance 40m with an acceleration of 0.5m/s^2
If you analyze this system from a frame moving with the man and block, then the work done is $0$ because the block is not moving in that frame. So, as you can see, the amount of work a force does ...
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Conservative force and change in the mechanical energy
Conservative forces can cause changes in kinetic energy. For example, the conservative force of gravity doing work on a falling object gives that object kinetic energy. It's just that conservative ...
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Question with work and Spring
I was able to get an answer a different way, finding the x value by
equating F = kx and solving for x, and then setting the equation up.
Don't understand what you mean by finding "x". It's ...
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What distance to use when calculating work?
For the calculation of work done by a Constant Force, we generally use $W = \vec{F} . \vec{S}$, Where $\vec{S}$ being the The amount of displacement of the object while the force is/was being applied, ...
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What distance to use when calculating work?
It is never the source (the "applier") that we consider. We only consider the properties of the object that is being influenced. In all laws and relationships (just think Newton's laws of ...
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What distance to use when calculating work?
In the equation $W = \vec{F}.\vec{s}$, the term $s$ refers to the displacement of the object on which the force is applied.
Then when climbing up stairs, I would be moving relative to the
stairs and ...
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Work of friction along a circular path using dot product
Yipee, Here we go, As a nice recall, force due to friction always acts in the opposing direction to motion.
So as we move around the circle, the force of friction will act in the opposite direction to ...
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I'd like to know how to calculate energy required to overcome static (or rest) friction
Yet I struggle with calculating how much energy would be wasted in
order to overcome the force of static friction.
It requires a force that exceeds the maximum possible static friction force in order ...
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A Conceptual Doubt in a Question on Work-Kinetic-Energy Theorem
In this answer I want to clarify the nature of the Work-Energy theorem. (The issue of the normal force exerted by the supporting surface has already been addressed in the answer by contributor Er ...
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A Conceptual Doubt in a Question on Work-Kinetic-Energy Theorem
The normal force corresponds to the defining constraint of a "solid surface" that motion can only exist along the surface and can't be through the surface. According to this definition, the ...
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A Conceptual Doubt in a Question on Work-Kinetic-Energy Theorem
You could indeed use the work-energy theorem to argue that the energy gained by the block is equal to the work done on it by the net external force acting on it, which is its weight $m\vec g$ plus the ...
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A Conceptual Doubt in a Question on Work-Kinetic-Energy Theorem
Indeed, since the net force is not calculated this question does not use the work energy theorem. The work energy theorem speaks only of the “net work”, which is the work done by the net force. You ...
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Conservative forces and Variation
mathematically rigorous:
If a differential form F is conservative, its differential vanisehs:
Be $U,\ V$ vector spaces and $F:U\rightarrow V$ a differential form. The vectorfield F is conservative is ...
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Escape velocity work done problem
to provide velocity to send to infinite distance from gravitational pull of earth so for that we have to apply external force
This is called an “initial velocity”. The problem begins at the “initial ...
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Definition of entropy as unavailable work
Entropy of a system is a measure of a part the internal energy that is not available for isothermal work. More specifically, the internal energy consists of several parts, thermal $TS$, volumetric $-...
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