26
Your conclusions are not correct.
Here is a simple counter-example.
Consider this force
$$\vec{F}=k(x\hat{y}-y\hat{x})$$
where $\hat{x}$ and $\hat{y}$ are the unit-vectors
in $x$ and $y$-direction, and $k$ is some constant.
From this definition we see, the magnitude
of the force is $F=k\sqrt{x^2+y^2}=kr$,
and its direction is at right angle to
$\vec{r}=x\hat{...
9
By "kinetic energy equation", I assume you mean the definition
$$KE = \frac{1}{2} mv^2$$
This does indeed arise from the work-energy theorem, which says that the net work done on an object of mass $m$ over some interval of time is given by
$$W_{net}=\frac{1}{2}mv_f^2- \frac{1}{2} mv_i^2$$
Looking at that equation, we simply notice that the quantity ...
6
The work over a short segment of path is the dot product between force $\vec F$ and small displacement $\delta r$.
$$\delta W=\vec F \cdot \delta\vec r$$
Now, the work over the entire path (i.e. not very small displacement) will be the sum of the work done over those infinitesimally small displacements. So what you end up doing is calculating the following ...
5
when I read about work it is defined as the ability to transfer energy, But when I read what energy is I find the definition is: the ability to do work
The issue that you are running into here is that physics is not one overall field of study, but rather has many smaller subdivisions. Very often one term will be defined differently in different branches of ...
4
You have hit upon the apparent paradox of the Oberth effect, and it is discussed in some detail here and here.
Tl;dr: No, the formula $W = F\cdot s$ is correct for a constant force.
What this really amounts to is the fact that kinetic energy is a non-linear function of the velocity. Let us look at your problem from the viewpoint of energy.
If the car ...
4
The general formulation of the first principle for a closed system says that $$L+Q= \Delta K + \Delta U + \Delta u$$
Where $L$ is the total non-conservative work done on the system. $Q$ is the heat entering the system. $K$ is the macroscopic kinetic energy, $U$ the macroscopic potential energy and $u$ the internal thermodynamic energy. Usually $U$ and $K$ ...
3
From a purely mathematical viewpoint, if you have a function f(x,y) then: $$df=f(x+dx, y+dy) -f(x,y)$$ and so $$\Delta f = f_f-f_i=\int_{f}^{i}df$$Since f is a well-defined, singular function, the integral can only depend upon the initial and final points. This would be true for any physical force defined by a singular function, such as gravity. Every ...
3
Welcome to this community!
A very short first answer is: no, work is not a function of state in an isobaric process. For several reasons:
1 – Suppose the state is given by the pair $(p,V)$. If the work between states $(p_0,V_0)$ and $(p_1, V_1)$, with $p_0=p_1$, were a function of state, it should be expressible in terms of $(p_0, V_0)$ alone, or of $(p_1,...
3
The work done on the object is greater than the PE if there is something other than gravity that opposes the object being pushed. The two typical examples would be friction and air resistance.
If you can neglect both of them then the work done on the object should be equal to the PE.
3
However, on the first disk, a force is applied tangentially and on the second disk, the same force is applied to its center of mass. The force is exerted over the same distance for both.
It is not possible for the force to be applied for both the same amount of time and over the same distance. The tangentially applied force will cover a larger distance over ...
3
OK, derivation of the work-energy theorem from F=ma
The qualification 'theorem' is indeed appropriate.
If we accept Newton's second law as axiom, and we accept as axiom that space is Euclidean, then the work-energy theorem follows logically.
First two standard kinematic relations, valid for the case of uniform acceleration. The derivation will capitalize on ...
3
If the line of action for the applied force does not pass through the center of mass, then the point of application of the force experiences a tangential acceleration. This means that for a given (short) time, the dR for the force is not the same as the dR for the center of mass. This difference is not considered in the given derivation. It is not valid if ...
3
If the force is applied at some $R$ above the CM, then $\mathrm{d}r=\mathrm{d}r_{\text{CM}}+R\omega \mathrm{d}t$, so you have an extra term. This extra term becomes:
$$ W_{\text{rot}}=\int FR\omega \mathrm{d}t=\int FR\omega \mathrm{d}\omega /\alpha=\tfrac{1}{2}I\omega^2$$
(where we replaced $\alpha=FR/I$ before integrating)
2
The central forces with spherical symmetry are also conservative forces.
You can show this proving that the work does not depend on the path.
HP: $$\vec{F}=F(r)\vec{r}$$ where $\vec{r}$ is the unitary vector to the position vector.
So you have: $$W_{AB}=\int_A^BF(r)\vec{r}\cdot ds$$ Since in polar coordinates elementar displacement in $d\vec{s}=d\vec{r}=dr\...
2
If the function is (as you assumed) one of distance, then you are right. But there are many functions of position coordinates whose curl is not zero, hence non-conservative.
Edit: Try for example $$F = (xy,-xy,0)$$
2
Several answers have simply been affirming that $K := \frac{1}{2}mv^2$ is the appropriate definition and alluded to the Oberth effect (which this isn't quite), but I don't think they've addressed the crux of your issue, namely the apparent violation of energy conservation.
The issue you've laid out is that, since you have pressed the pedal at the same level ...
2
So doesn't the equation need to account for initial velocity as well?
No, the difference that you posted is in fact correct. We want the same force to give different values for work in the different cases.
The purpose of work is to quantify a transfer of energy from one system to another. Recall that kinetic energy is $KE=\frac{1}{2}mv^2$. So if you start ...
2
Your muscular energy used =mgh (this energy will used fully whether it goes to height h or h/2
When you throw want to stop it at h/2 some external force is required to stop it
$\int F_{ext}$+$\int F_{non conservative}$ = $\delta$ K.E + potential energy
On plunging values
$\int F_{ext}$+0= mgh
Note# (sum of potential and kinetic energy at every point in ...
2
What is equal to the change in PE is always the work done by gravity (with opposite sign). The work done by Any other force, like the one pushing up the body up the incline, has nothing to do with the PE, in principle. It may happen to be equal to the change in PE if the pushing force is so adjusted that the final KE is zero. And of course, if there is no ...
2
Good question.
Once you have specified a path, the thermodynamic differentials become exact differentials. The reason they are 'inexact' to begin with is that different paths end up in different changes in the state variables.
Note
: All statements made for reveriblse processes
2
It's pretty easy to see that if you take the inner product of the force four-vector with a displacement four-vector, you don't get a correct expression for the mechanical work done by the force. This is because the inner product of two four-vectors is a scalar, which is the same in all reference frames. But energy obviously depends on your frame of reference....
2
From what I understand, in one case you imagine a force with constant direction, orientation, and magnitude to be tangentially applied, at each instant, on the part of the disk that has highest vertical coordinate with respect to an observer in a reference frame in which the disk is initially at rest. This means that such a force (field) is applied to a ...
2
The Earth almost revolves around the sun in a circular orbit. If it were perfectly circular, then the Sun's gravitational field would do no work on the Earth, since the force is radial and would be at an an angle of 90 degrees to the Earth's motion.
With the small eccentricity, there will be some work done (the angle between F and d is no longer exactly 90 ...
2
So the work done by a force given by the vector field $\vec F$ is
$$W = \vec F \cdot \vec v \ dt$$
If the force varies from point to point, the displacement vector $\vec v dt$ may also change, as the object may follow a curved path in two or three dimensions.
Suppose that the path of an object is given by a vector function $r(t)$ at any point along the path, ...
2
I think you are simply forgetting the sideways speed of #2. Case #2 will indeed have a smaller vertical final velocity component of the reasons you describe. Your logic is fine. Now add it to the sideways component and the magnitude will turn out the same as the others.
2
Integrate Newton's second law with respect to distance along a path
$$ \int \mathbf F\cdot d\mathbf s = m \int \mathbf a\cdot d\mathbf s $$
We have $$\frac{d\dot x^2}{dx} = \frac{d\dot x^2}{d\dot x} \frac{d\dot x}{d t}\frac{dt}{d x} = 2 \dot x \ddot x\frac{1}{\dot x} = 2 \ddot x $$
Then, in Cartesian coordinates (for convenience) we may write the integral ...
2
For starters, energy is just a number that remains the same at any time in the universe, only its distribution changes. That is what energy is in a nutshell. Work is how that distribution can be changed in due course of time. What happened with energy is that we have been able to identify and calculate such a number that always gives us the same answer no ...
2
You should remember how we define energy, energy is defined as “ the capacity of an object to do work”, basically how much work is done on an object is equal to the gain/loss in the kinetic energy of the object, say I push an object how much work I did would be equal to the energy(in this case kinetic energy) of the object, I would also recommend reading ...
2
Work is the thing that makes kinetic energy Galilean invariant. Suppose you have a ball with mass $m$, on a spring ($k$), compressed a distance $L$.
That system has energy, in terms of kinetic ($T_i$) and potential ($U_i$):
$$ E_i = T_i + U_i = 0 + \frac 1 2 k L^2$$
Then you release the spring, and the balls rolls of with velocity $u$, so that:
$$ E_f = T_f +...
2
You've asked a bunch of questions, albeit all related, so I will just try to break them down.
From what I understand, energy is defined as the ability to do work
Essentially yes. But the term capacity is generally used instead of ability.
What does this actually mean and how is it different from work itself?
Work is the transfer of energy from one thing ...
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