53

Work is the dot product of a vector force and a vector displacement, hence a scalar. Knowing just the scalar distance isn’t enough to calculate work. That distance might be in the same direction as the force, but it might be perpendicular or even opposed. All of those would give different values for the work done.


46

Sound intensity is measured on the dB scale, which is a logarithmic scale of pressure. The "threshold of hearing" is given by the graph below: which tells you (approximately) that 0 dB is about "as low as you go" - the "threshold of hearing". Note that sound signal drops off with distance - we will have to take that into account in what follows. If you ...


36

No. No work would be done in this case, at least not at the macroscopic level. Work is the product of force and displacement in the direction of the force and in this case there is no displacement. I disagree with you that each person is "accomplishing a change in the book's movement". The book wasn't moving initially, or at the end, or at any time in ...


30

The general definition of work is $$W=\int\mathbf F\cdot\text d\mathbf x$$ Which essentially says, "Add up all of the dot products between the vector force $\mathbf F$ and the vector displacement $\text d\mathbf x$ along the path the object travels on." Since we are adding up dot products, which are scalar quantities, the work done by a force is also a ...


23

It depends on whether the force field is conservative or not. Example of a conservative force is gravity. Lifting, then lowering an object against gravity results in zero net work against gravity. Friction is non-conservative: the force is always in the direction opposite to the motion. Moving 10 m one way, you do work. Moving back 10 m, you do more work. ...


20

In physics we frequently leave off the limits of the integral when the limits can be figured out from the context. So, in the first case, the actual relation is: $$x(t) = x(0) + \int_0^t \dot{x} \operatorname{d}t'.$$ Most often, though, when the limits are left off the implied limits are over all possible values of the dummy variable. For example: $$ Q = \...


15

The distance referred to in the definition of work is, specifically, the distance that the object moves while the force is being applied. This is because the actual definition of work is a line integral, where the work $W$ along a path $C$ with tangent vector $d\vec{s}$ is defined as: $$W=\int_C \vec{F}\cdot d\vec{s}$$ In general, calculating the work done ...


14

Very nice question! You can see this from the second Newton's law: $$m\ddot{\mathbf{x}} = \mathbf{F}(\mathbf{x})$$ Now I would like to integrate this equation of motion with respect to time, to arrive at the energy conservation. To do so I multiply both sides with $\dot{\mathbf{x}}$: $$m\ddot{\mathbf{x}}\cdot\dot{\mathbf{x}} =\mathbf{F}(\mathbf{x})\cdot\...


12

The answer to your question depends on how we define work. Definition, A force is said to do work if, when acting, there is a displacement of the point of application in the direction of the force. In layman language, to do work you need displacement, not just force. In the equation, $dW=x.dF$, we are considering a change in force at a constant position ...


11

You are being confused by the shortcut that people took who wrote that expression. They mean for the integral to be taken between definite limits. It would be more proper to say $$x(t) = x(0) + \int_0^t \dot{x} dt$$ But that gets longwinded. Most people, when seeing the expression as you gave it, will understand it to mean what I wrote. But technically, ...


11

There are already several good answers. In this answer, we will just highlight a geometric argument. On one hand, work (within Newtonian mechanics) $$\mathrm{d}W=\vec{F}\cdot \mathrm{d}\vec{r}$$ is a scalar quantity, meaning that it is independent of coordinate system. On the other hand, the quantity $\vec{r}\cdot \mathrm{d}\vec{F}$ depends on coordinate ...


7

A coherent state is characterized by a complex number $\alpha \in \mathbb C$. Applying the displacement operator $D(\beta)$ to $|\alpha\rangle$ translates $\alpha$ in the complex plane by $\beta$, in the following sense: $$ D(\beta) |\alpha\rangle \sim |\alpha+\beta\rangle . \tag 1 $$ Here, $\sim$ means "up to a phase". The precise relation is: $$ {D}(\beta)...


7

A simple example is just an object starting at rest falling with a drag force proportional to the velocity of the object, $F_D=-bv$. Then the acceleration is given by $$a=\frac{dv}{dt}=g-\frac bmv$$ Therefore, the velocity over time is given by $$v(t)=\frac{mg}{b}(1-e^{-bt/m})$$ Typically you get $e$ popping up when the rate of change of something is ...


7

The first thing you need to understand: you are applying the creation of physics definitions backwards. You are asking, "why isn't work given by this equation?", but this question doesn't make sense if you think about it. It is not the case in physics where we think, "Hmm... I want to define something called "work". What should it's equation be?" This doesn'...


6

It's possible that the way in which these terms are used varies from person to person, even among professionals in the field. However, in the usage I'm familiar with, displacement is the change in position, period. Definition #2 is correct and #1 is wrong. (The length and direction of a line from a fixed reference point is just called position.) In this ...


6

Of course, $e$ is ubiquitous in kinematics. For example, consider a repulsive force proportional to $x$, $$F = kx.$$ Then the acceleration is $$a = \frac{k}{m} x = \omega^2 x, \quad \omega = \sqrt{\frac{k}{m}}.$$ This differential equation has solutions of the form $e^{\omega t}$ and $e^{- \omega t}$. In particular, suppose that $x(0) = 1$ and $v(0) = \omega$...


5

what exactly does the area under a displacement-time graph denote? I think it just represents what you said: the area under a displacement-time graph. I can't think of any other use for it. There are two main reasons for this: Your quantity, let's call it $f(t)$, retains a memory of where the object has been. That's because the area under the graph depends ...


5

I suggest a totally different approach. But it's only a partial approach with much guesswork, too. The ear is able to perceive 20 µPa. (at 2 kHz). Of course you could calculate some pressure changes at the closing void, but these actually have nothing to do with the sound pressure at your ear drum. Let's do some energy calculations. 20 µPa at 1 cm² area at ...


5

Its just a matter of what you use to call as displacement and as distance . I have seen the usage of: dx ds dr as the displacement too. Wikipedia says : The work done by a constant force of magnitude F on a point that moves a displacement (not distance) s in the direction of the force is the product, W = Fs. Note the usage of s as ...


5

If you 'carry' an object 10 meters in one direction then return it back 10 meters from where you started the work done on the object is not the force you expended times distance walked. The formula you write is often misunderstood and misused. In your example, when you lift the object in a gravitational field, the work being done on the object is its weight ...


5

Well, say that you are looking at a man lifting boxes. Each box weighs 10kg. At first you look at him standing, but since just looking at him made you tired, you decide to lie down. Now, from your horizontal position, the scene looks different, but is the man doing more, less, or the same amount of work per box? The word “scalar” is often used to just mean “...


5

Position is a single point. Usually in space we indicate positions with coordinates like $(x,y,z)$ in Cartesian coordinates, $(r,\phi,\theta)$ in spherical coordinates, etc. We can also define the position as a vector, i.e. the position vector, that is a vector that points from the origin (subjectively defined) to the position of the particle in question. It ...


5

$\color{blue}{\text{Position}}$: $$\color{blue}{\vec p(t)}$$ $\color{red}{\text{Displacement}}$†: $$\color{red}{\vec p(t_2) - \vec p(t_1)}$$ $\color{green}{\text{Distance Traveled}}$: $$\color{green}{\int_{t_1}^{t_2} \left\|\frac{d\vec p(t)}{dt}\right\|dt}$$ †Displacement is sometimes defined as the scalar $\|\vec p(t_2) - \vec p(t_1)\|$.


5

In affine spaces the derivative of a curve $P=P(t)$ is a vector since couples of points uniquely define vectors: $$P(t + h)-P(t)$$ as a consequence of the affine structure. As it is a well-defined vector it can be multiplied with a number and the velocity is defined as the limit of the vector valued function $$\lim_{h\to 0} h^{-1}(P(t + h)-P(t))\:.$$ ...


5

Their lesson today says that "work" is done only when a change in position is accomplished by application of force. The key word here is change. If an object is not moving, then no work is being done to it$^*$. Therefore in your scenario, no work is done on the book by any of the forces acting on it. More formally, work done by a constant force on an ...


4

If you travel with constant speed $V$ for a time $T$, you will travel for $V\times T$ distance. Example: if your speed were $2$ m/s, and you were walking for $3$ seconds, you'd walk $2\times3 = 6$ meters. Now, when calculating distance traveled while accelerating (or decelerating), we can approximate it if we split total time of travel into sub-intervals, ...


4

Angular displacement is an example of what's generally called a pseudovector. This is a quantity that is similar to a regular vector, except for the fact that it behaves differently under improper rotations such as reflections (it gains an additional sign flip). Any quantity which is the cross-product of two polar vectors will generally be a pseudovector. ...


4

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 ...


4

Kudos to the question-asker for thinking about everything they read! :-) I was pleased to note that the author of the previous answer mentioned "audible" means "audible" to the human ear. Note also that "audible" also depends on the frequency a bit...generally-speaking, as humans, for high-frequency sounds we need them a little more intense if we're going ...


4

If you drive along the road on the side of a mountain) there are (at least) two kinds of forces on your car: friction and gravity. If you drive to the top of the mountain and back, the net work done against gravity is zero. This corresponds to the fact that your displacement is zero. On the same trip, you had to overcome friction. That depended on your ...


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