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0

If your friend's energy+ yours= F1, then you would see your own energy expenditure halved, which we know cannot be the case. If your friend helps you push the object, then you are no longer applying the same force, or (lazy answer) the force is no longer localised and motivates the part of the object most subject to friction. So, once the object is first ...


3

I'm sure everyone has had that concern when we encountered the definition for the first time, in school. There is a valid reason why this definition is still persisted with, despite the deficiency that you hit on. The most popular (and simple) forces in physics (also the ones with which we begin learning physics) are conservative forces, implying that the ...


32

If you're pushing a 10-ton truck and it's not moving, you are not doing any work on the truck because the distance $ds=0$ and the nonzero force $F$ isn't enough for the product $F\cdot ds$ to be nonzero. Your muscles may get tired so you feel that you're "doing something" and "spending energy" but it's not the work done on truck. You're just burning the ...


1

My assumption from hearing that problem is that you do not need to worry about fluid flow and the problem is only asking you to calculate how the mass of water is distributed. If the tank were cylindrical or a right prism, you could simply assume that all the mass was at the midpoint of the height of the tank, and you have to raise it in the gravitational ...


3

You may be imagining that if you push with constant force $F$, the spring will compress until the spring has such a resistive force. But since the spring was not counteracting that force, your constant force $F$ was accelerating the mass. Upon reaching the point where the spring has force $F$ as well, the mass does not stop but has a speed such that $KE = ...


0

The factor $\frac{1}{2}$ is due to the integral. The wrong sign of yours is due to the fact that you have to counter the force of the spring. So the Force if the Spring is $-kx$, but you have to pull in the direction it is extended, so apply the force $kx$, therefore the energy is positive $W=\frac 1 2 k^2 x$


5

Since the force is a function of distance, you need to integrate: $$F = kx\\ W = \int F\ dx\\ W = \int k\ x\ dx\\ W = \frac12kx^2$$ Add signs as needed... Your work considered the force to be constant - and that's not how springs work.


2

Work. Potential energy exists because of some force that exists, and moving an object relative to that force causes work to be done. And by the work-energy theorem, the work done on an object is equal to the change in kinetic energy of that object.


0

This is not specific to converting potential energy to kinetic, but the term you're looking for might be "transduction" or "to transduce"- the process of converting energy from one form to another.


0

Your intituition is totally different because ennumerous forces change the situation from the ideal situation predicted by work enrgy theorem, some of them are: Air Drag/Friction, Rotational Friction due to differences in size of tyres, different aerodynamic effects due to different body design.. etc.


1

I know which book you are referring to. It is the book "Finite elements in Engineering" by Chandrupatla and Balegundu.I also have the same question. We learnt in Physics that the Work done by the force is stored as Potential Energy. There was no mention of Work Potential.


4

Let $\mathbf x(t)$ be the path of a particle. Let $\mathbf F(t)$ be a force acting on the particle as a function of time, then the work done by the force from time $t_a$ to time $t_b$ is \begin{align} W(t_b, t_a) = \int_{t_a}^{t_b} \mathbf F(t)\cdot \frac{d\mathbf x}{dt}(t)\, dt. \end{align} where the center dot denotes dot product; \begin{align} ...


0

The definition of work, it is done on vectors, let's say $$ \vec F = F_x\hat i+F_y\hat j+F_x\hat k, $$ that would be the force and the displacement vector it is $$ \vec \ell=\ell_x\hat i+\ell_y \hat j + \ell_z\hat k $$ so you have the 3D. Then the definition is $$W=\int \vec F \cdot d\vec\ell$$ also it would be good to check anyways the dot product and the ...


0

You are mostly correct. Since you are only asked for a qualitative result, you can simplify what you said: If the boy pulls on rope with force $F$ for time $t$, it will move a certain distance $d$. If the other end of the rope is fixed, the work done is $F\cdot d$. If the end of the rope is moving towards him, the total length of rope he reels in is greater ...


2

While the total displacement as shown in your figure is zero, this does not mean that the work is zero! Work is force scalar displacement, $W = \vec F\cdot \vec s$. Diving the path $A \to B \to A$ into infinitesimal steps we are led to $$dW = \vec F \cdot d\vec s$$ and for the total work, adding up the contribution $$W = \int \vec F\cdot d\vec s.$$ I think ...


-4

The maths is all impressive but entirely redundant. It is not zero BY DEFINITION.


5

Well, we can do a simple counter-example. Let $$ \vec{F}(\vec{x}) = F_0 \cdot \varrho(\vec x) $$ where $\varrho$ is the function that rotates vectors by 90° counter-clockwise (in matrix form $(\begin{smallmatrix}0 & -1\\1 & 0\end{smallmatrix})$ if you prefer that). Clearly, for the closed path $$ \vec{\gamma}\colon\quad [0, 2\pi]\ \to\ ...


5

For forces that change along the way, displacement is not the thing to calculate work with. Let $\gamma : [0,1] \rightarrow \mathbb{R}^3$ be the (closed or open) path that the particle the force is exerted on follows. Then, the work done along that path is $$ W[\gamma,F] = \oint_\gamma \vec{F}(\vec{x})\cdot \mathrm{d}\vec{x}$$ which is a line integral. If ...


1

When a force is applied over a certain distance, that force does mechanical work, $W$. If the force is constant $F$ and the object it is exerted on is moved by a distance $\Delta x$, then $W=F\Delta x$. If the force is not constant but a function of the position, this turns into an integral: $$W = \int_{x_1}^{x_2}F(x)\,\mathrm d x.$$ If you don't know ...


1

It's the same reason why simply holding the mass over your head uses energy, even though the mass is not moving. And the reason is the way your muscle tissue works, which is to continually contract and then relax - obviously with different parts of the same muscle firing off at different times.


0

Let's try to work this down. $W=F \cdot d$ Now for the sake of simplicity I will say we are dealing only in one direction, namely the x-axis, so $W=F \cdot x$ It may come in handy to derive the equation for force Let $F=ma=\frac {dp}{dt}$ Let's take this a step further though, since we know that $p=mv=m \frac {dx}{dt}$ So if follows that $F=\frac ...


6

If the displacement of the object is zero, then one can calculate the work done by each individual force, the work done by each force is zero. Why? Work is not defined in terms of what would have happened to the object in the absence of other forces; it is defined in terms of the motion that actually occurred. More concretely, if from time $t_a$ to time ...


0

Your definition of work needs a little...work, it only works for forces which are constant along the path the particle moves. The more general definition is: For any path $\gamma : [a,b] \rightarrow \mathbb{R}^3$ and any force field $\vec{F} : \mathbb{R}^3 \rightarrow \mathbb{R}^3$, the work the force does on a particle moving along $\gamma$ is ...


0

You specify a point particle, which means no internal structure, which means it's not deformable. If the object didn't move, then why choose 1 m? Why not 10 m, or 1,000 m ? No displacement, no work.


1

Here's my explanation (hopefully it's right but I'm no expert): Consider the simpler example of pushing a really heavy box; if we push on it there's a static friction force, but if we don't push on it there's no force. Similarly for a cylinder on a horizontal plane, if we push on it the static friction force causes it to rotate, but when we stop pushing ...


2

Here's a way to argue it isn't the displacement of the (center of mass) of the body. Take a spring that's at rest. Apply two forces on either end so it compresses. You can do this in such a way that the center of mass of the spring doesn't move. However, the energy of the spring system has changed (the potential energy increased). In order to satisfy the ...


1

First of all, I appreciate your doubt. It is conceptual and involves critical thinking regarding the topic. In order to understand work, you need to understand the 2 physical quantities namely - Force and Displacement . Note that using only one single word Displacement in order to define work is not appropriate. (Same is for Force ) . As, Displacement ...


2

It's the displacement of the point of application of the force. Elementary textbooks initially model objects as point particles. They have exactly two properties: position and mass. No others. So the displacement of the object is necessarily the same as the displacement of the point of application. Later, extended or compound objects may (or may not) be ...


4

It is the displacement of the point on which the force is applied. This is it. That is the real definition, period. The books which do it otherwise are wrong. I would like you to read Resnick Halliday Krane , physics Vol. 1, it has a separate chapter devoted to this problem with most of the books available. As an example to show that this is what the ...


0

Both scalar and vector products are definitions. $$ {\bf a} \cdot {\bf b} = a b \cos \theta\,\,$$ where $a$ and $b$ are the magnitudes of the two vectors and $\theta$ is the angle between them. This quantity only has a magnitude and hence no direction and is therefore a scalar. If you apply a (let's say constant for the sake of simplicity) vector force ...



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