About verification on Kinetic energy and work

[rudransh verma]

1. From resnik, Halliday “Kinetic energy K is energy associated with the state of motion of an object. The faster the object moves , the greater is the kinetic energy”
If I am right this means that greater the kinetic energy, greater is its speed.

2. Force transfers energy to the body due to which the velocity of the body increases and the body accelerates. We say the force has done some work.

3. “Pushing a wall does not cause an energy transfer to or from the wall and thus is not work done on the wall as defined here”. Well energy is transferred into something that’s why we get tired, like in the environment. It’s just not on the wall. So by definition we say no work is done on wall or from the wall.

4. $\frac12mv^2-\frac12mv_0^2=F_xd$
Left hand side is change in energy by the force and right hand side tells us the amount of change, ie the work. $W=F_xd$

5. In the formula $W=F.d$ ,F is a constant force. What does that mean? I mean when we apply a force to a object we push and remove the force or we continually apply the force all the way through its displacement. If the surface is frictionless we don’t have to apply continuous force. Does this formula not applicable for momentary push? Do we have to continually apply force on the body?

6. The force can transfer energy to the system or take out the energy from the system like the gravitational force take out the energy from a rising apple due to which it slows down and then give energy to the retuning apple increasing its speed. Work is done by the system and on the system respectively.

 

[vanhees71]

The greatest obstacle to understand the work-energy theorem is that books don't give a complete derivation athough it's not so complicated. You just start from the equation of motion for a point particle subject to some force
$$m \ddot{\vec{x}}=\vec{F}(\vec{x}).$$
Multiply this with $\vec{v}=\dot{\vec{x}}$ (dot product). Then you get
$$m \ddot{\vec{x}} \cdot \dot{\vec{x}} = \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right) = \vec{F}(\vec{x}) \cdot \dot{\vec{x}}.$$
Now integrate this from $t_0$ to $t_1$:
$$\int_{t_0}^{t_1} \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right) = \frac{m}{2} \dot{\vec{x}}^2(t_1)-\frac{m}{2} \dot{\vec{x}}^2(t_0) = \int_{t_0}^{t_1} \mathrm{d} t \dot{\vec{x}}(t) \cdot \vec{F}[\vec{x}(t)]=W.$$
That means that the change in kinetic energy
$$E_{\text{kin}}=\frac{m}{2} \dot{\vec{x}}^2$$
along the trajectory of the particle is equal to the work done due to the force acting of the particle along this path.

If the force is constant along the path, as e.g., for the motion of a particle in the gravitational field near the Earth, where $\vec{F}=m \vec{g}$ with $\vec{g}=\text{const}$, then the work is
$$W=\int_{t_0}^{t_1} \mathrm{d} t \dot{\vec{x}} \cdot m \vec{g} = m \vec{g} \cdot [\vec{x}(t_1)-\vec{x}(t_0)].$$
Since in this case the force has a potential,
$$\vec{F}=-\vec{\nabla} V \quad \text{with} \quad V=-m \vec{g} \cdot \vec{x},$$
you can formulate the energy-work theorem as the energy-conservation law:
$$E_{\text{kin}}(t_1)-E_{\text{kin}}(t_2)=m \vec{g} \cdot [\vec{x}(t_1)-\vec{x}(t_0)] \; \Rightarrow \; E_{\text{kin}}(t_1) + V[\vec{x}(t_1)]=E_{\text{kin}}(t_1)+V(\vec{x}(t_1)].$$
This tells you that the total energy
$$E=E_{\text{kin}}+V$$
is constant along the trajectory of the particle.

 

[A.T.]

5. In the formula $W=F.d$ ,F is a constant force. What does that mean? I mean when we apply a force to a object we push and remove the force or we continually apply the force all the way through its displacement. If the surface is frictionless we don’t have to apply continuous force. Does this formula not applicable for momentary push? Do we have to continually apply force on the body?

"Momentary push" is just an approximation. In reality, to do work you have to apply the force over a non-zero displacement. But if the push is just a short impact, and you don't know the displacement during it, then you cannot use the formula. It still applies, but you don't have sufficient information to use it.

 

[ergospherical]

As with anything "momentary" it's a case of using delta functions. If you give the thing a kick at $x=0$ and regard the force $F = F(x)$ as a function of the position, which is fine so long as the velocity is of constant sign in a suitable domain $x \in \{ - \epsilon, \epsilon \}$, then put $F(x) = K\delta(x)$ and write the work done as
\begin{align*}
\Delta(\mathrm{kinetic \ energy}) = \int_{-\epsilon}^{\epsilon} K\delta(x) dx = K
\end{align*}

 

[rudransh verma]

The greatest obstacle to understand the work-energy theorem is that books don't give a complete derivation athough it's not so complicated. You just start from the equation of motion for a point particle subject to some force
$$m \ddot{\vec{x}}=\vec{F}(\vec{x}).$$
Multiply this with $\vec{v}=\dot{\vec{x}}$ (dot product). Then you get
$$m \ddot{\vec{x}} \cdot \dot{\vec{x}} = \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right) = \vec{F}(\vec{x}) \cdot \dot{\vec{x}}.$$
Now integrate this from $t_0$ to $t_1$:
$$\int_{t_0}^{t_1} \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right) = \frac{m}{2} \dot{\vec{x}}^2(t_1)-\frac{m}{2} \dot{\vec{x}}^2(t_0) = \int_{t_0}^{t_1} \mathrm{d} t \dot{\vec{x}}(t) \cdot \vec{F}[\vec{x}(t)]=W.$$
That means that the change in kinetic energy
$$E_{\text{kin}}=\frac{m}{2} \dot{\vec{x}}^2$$
along the trajectory of the particle is equal to the work done due to the force acting of the particle along this path.

If the force is constant along the path, as e.g., for the motion of a particle in the gravitational field near the Earth, where $\vec{F}=m \vec{g}$ with $\vec{g}=\text{const}$, then the work is
$$W=\int_{t_0}^{t_1} \mathrm{d} t \dot{\vec{x}} \cdot m \vec{g} = m \vec{g} \cdot [\vec{x}(t_1)-\vec{x}(t_0)].$$
Since in this case the force has a potential,
$$\vec{F}=-\vec{\nabla} V \quad \text{with} \quad V=-m \vec{g} \cdot \vec{x},$$
you can formulate the energy-work theorem as the energy-conservation law:
$$E_{\text{kin}}(t_1)-E_{\text{kin}}(t_2)=m \vec{g} \cdot [\vec{x}(t_1)-\vec{x}(t_0)] \; \Rightarrow \; E_{\text{kin}}(t_1) + V[\vec{x}(t_1)]=E_{\text{kin}}(t_1)+V(\vec{x}(t_1)].$$
This tells you that the total energy
$$E=E_{\text{kin}}+V$$
is constant along the trajectory of the particle.

Its given in my book very simply as
1.$F=ma$
2.$v^2=u^2+2as$
$$v^2-u^2=2as$$
$$\frac12mv^2-\frac12mu^2=Fs$$
$$K_f-K_i=Fs$$
$$\Delta K=Fs$$
$$\Delta K=W$$

 

[rudransh verma]

In reality, to do work you have t

In reality there is friction so that's why you have to continuously apply force but not in space. We can use the formula there for momentary push. We just need to measure displacement.

 

[vanhees71]

Its given in my book very simply as
1.$F=ma$
2.$v^2=u^2+2as$
$$v^2-u^2=2as$$
$$\frac12mv^2-\frac12mu^2=Fs$$
$$K_f-K_i=Fs$$
$$\Delta K=Fs$$
$$\Delta K=W$$

That's the problem! It's not given in explicit vector notation and it's using a very special case of a constant force. All to often one has students having memorized these "SUVAT equations" without a clou about their meaning!

 

[rudransh verma]

That's the problem! It's not given in explicit vector notation and it's using a very special case of a constant force. All to often one has students having memorized these "SUVAT equations" without a clou about their meaning!

I know the meaning. When the force F acts on the mass m it accelerates a amount. Due to this acceleration it covers a distance s and its velocity changes from u to v.During the acceleration a the force does work and this changes its kinetic energy by an amount Fs during the displacement s. This is work energy theorem and it states that work done changes kinetic energy.

 

[vanhees71]

Here you mix two different, important concepts, i.e., momentum and energy. You are not alone with this. Already Leibniz and Newton had one of their many fights about this ;-)).

Both laws are just integrals of the basic 2nd Law:
$$m \ddot{\vec{x}} = \vec{F}(\vec{x}).$$
Integrating this wrt. to $t$ from $t_0$ to $t_1$ along the trajectory, i.e., the solution of this differential equation of motion, leads to
$$m [\dot{\vec{x}}(t_1)-\dot{\vec{x}}(t_0)]=\vec{p}(t_1)-\vec{p}(t_0)=\int_{t_0}^{t_1} \mathrm{d} t \vec{F}[\vec{x}(t)].$$
The work-energy theorem, I've derived already in #2, i.e., you get in fact two different integral laws, one for momentum and one for (kinetic) energy.

 

[rudransh verma]

Integrating this wrt. to t from t0 to t1 along the trajectory, i.e., the solution of this differential equation of motion, leads to

How do you decide when to integrate? There must be something that integration does.

 

[vanhees71]

It's the operation that "inverts" differentiation, and as such it's used in the examples above. Of course in this form these integrals are not too useful, because you need the solutions of the equations of motion to do these integrals. The idea, however, becomes very useful for finding the solutions of the equations of motion in special cases, which often apply in specific problems (and that's not by chance, but because of deep mathematical properties concerning the geometry and symmetry of the Newtonian spacetime model):

In the case of the work-energy theorem you get the energy-conservation law if the force is derivable from a time-independent potential, i.e., if there is a scalar field $V(\vec{x})$ such that
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
If you now do the calculation as in #2, also the right-hand side becomes a total time derivative:
$$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right) = m \dot{\vec{x}} \cdot \ddot{\vec{x}} = m \dot{\vec{x}} \cdot \vec{F}(\vec{x})=-m \dot{\vec{x}} \cdot \vec{\nabla} V(\vec{x}) = -m \frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x}).$$
The difference to the general case is that now the line integral over the force is (at least locally) independent of the path along you integrate, i.e., you have the energy-conservation law
$$\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x})=E=\text{const}$$
which is formulated such that you don't need to know the solution of the equations of motion to make use of it to actually find the solutions of the equations of motion.

In the case of the momentum you can apply it to a closed system, i.e., a system where there are only interaction forces and not some "external forces" (which is anyway always an approximation to a real situation since all forces in fact are interaction forces). Take as the most simple example a closed system of two point particles interacting via some force (e.g., the gravitational force). Then you have
$$m_1 \ddot{\vec{x}}_1=\vec{F}_{12}, \quad m_2 \ddot{\vec{x}}_2 =\vec{F}_{21},$$
but according to the 3rd Law
$$\vec{F}_{21}=-\vec{F}_{12}.$$
So adding both equations gives
$$m_1 \ddot{\vec{x}}_1 + m_2 \ddot{\vec{x}}_2 = \vec{F}_{12} +\vec{F}_{21}=0,$$
and this you can integrate to get
$$\vec{P}=m_1 \dot{\vec{x}}_1 + m_2 \dot{\vec{x}}_2=\vec{p}_1+\vec{p}_2=\text{const}.$$
The total momentum is conserved!

 

[rudransh verma]

It's the operation that "inverts" differentiation, and as such it's used in the examples above. Of course in this form these integrals are not too useful, because you need the solutions of the equations of motion to do these integrals. The idea, however, becomes very useful for finding the solutions of the equations of motion in special cases, which often apply in specific problems (and that's not by chance, but because of deep mathematical properties concerning the geometry and symmetry of the Newtonian spacetime model):

In the case of the work-energy theorem you get the energy-conservation law if the force is derivable from a time-independent p

I mean when we differentiate we know we need to find the rate of something with respect to something . But why do we integrate. Sometimes people just multiply both sides of some eqn with dv or dx and then they integrate as if they know what they will get. Do you get me?
Do we know what we will get when we integrate?

 

[vanhees71]

Indeed, to solve the differential equations of motion is much harder than to get the derivative of a given function. Since integration is the inverse operation to differentiation one also says one integrates the equation of motion, and to solve differential equations by integration is an art, which you have to learn by solving a lot of problems. There are also some standard types of differential equations that can be solved in closed form pretty easily, and you get quite far with them.

The most simple example is indeed the one with a constant force, e.g., the motion of a particle close to Earth. Then you can approximate the gravitational force on the particle as constant
$$\vec{F}=m \vec{g}.$$
Using a Cartesian basis with the basis vector $\vec{e}_z$ pointing up you have
$$m \begin{pmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -m g \end{pmatrix},$$
where $g \simeq 9.81 \text{m}/\text{s}^2$.

These equations of motion are easily integrated since the right-hand side does not depend on $(x,y,z)$ at all. So you indeed just do integrals. Cancelling the $m$ on both sides and then integrating once wrt. to $t$ you get
$$\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} = \begin{pmatrix} v_{0x} \\ v_{0y} \\ -g t +v_{0z} \end{pmatrix},$$
where $v_{0x}$, $v_{0y}$, $v_{0z}$ are arbitrary constants. They are given by the initial conditions, which you always need as input to get a unique solution of the equations of motion. Because they are of second order in time derivatives you need to know $x$, $y$, $z$ at some initial time (which I here choose to be $t=0$ for simplicity) and $\dot{x}$, $\dot{y}$, and $\dot{z}$ at the initial time $t=0$. These initial values of the velocity I've called $v_{0x}$, $v_{0y}$, and $v_{0z}$, respectively.

Now you can integrate once more to also get the positions
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} v_{0x} t + x_0 \\ v_{0y}t + y_0 \\ -g t^2/2 +v_{0z} t +z_0 \end{pmatrix},$$
where $x_0$, $y_0$, and $z_0$ are obviously the initial values for the position coordinates at the initial time $t_0=0$, and that's the well-known complete solution for the case of a constant force.

There are of course also more complicated cases, which however are still simple to solve. One of the most important examples is the harmonic oscillator. Here the force is proportional to the displacement of the particle from a certain equilibrium position. An example is a particle on a spring. For not too large elongations the force on the particle is proportional to the displacement from the equilibrium position (Hooke's Law). Here we have an example for 1D motion, and we just need to write done the one corresponding position component, which I'll call $x$. The equation of motion then reads
$$m \ddot{x}=-k x,$$
where $x=0$ is the equilibrium position (by choice of the coordinate system).

This is not as straight-forward to "integrate". Dividing by $m$ you get
$$\ddot{x}=-\frac{k}{m} x.$$
Now you have to remember which functions give twice integrated its negative (up to a factor). If you remember your differentiation rules, you find pretty quickly that these are cos and sin. So you make the ansatz
$$x(t)=A \cos(\omega t) + B \sin(\omega t)$$
with $A$, $B$, and $\omega$ some constants. Now indeed taking the derivatives you get
$$\dot{x}=-A \omega \sin(\omega t) + B \omega \cos(\omega t)$$
and once more
$$\ddot{x}=-A \omega^2 \cos(\omega t) - B \omega^2 \sin(\omega t),$$
but indeed this is of the right form, because for our ansatz
$$\ddot{x}=-\omega^2 x.$$
Comparing this with the equation of motion, you see that it solves indeed this differential equation if you set
$$\omega=\sqrt{\frac{k}{m}}.$$
Our ansatz is also the "complete" solution, because there is enough freedom to choose $A$ and $B$ to fulfill the initial conditions at $t_0=0$:
$$x(0)=A=x_0, \quad \dot{x}(0)=B \omega =v_0\; \Rightarrow \; B=\frac{v_0}{\omega}.$$
So the solution of the equation of motion with the given initial conditions finally is
$$x(t)=x_0 \cos(\omega t) + \frac{v_0}{\omega} \sin(\omega t) \quad \text{with} \quad \omega=\sqrt{\frac{k}{m}}.$$

 

[A.T.]

I mean when we differentiate we know we need to find the rate of something with respect to something. But why do we integrate.

To find something from the rate of change of that something.

 

[rudransh verma]

Indeed, to solve the differential equations of motion is much harder than to get the derivative of a given function. Since integration is the inverse operation to differentiation one also says one integrates the equation of motion, and to solve differential equations by integration is an art, which you have to learn by solving a lot of problems. There are also some standard types of differential equations that can be solved in closed form pretty easily, and you get quite far with them.

The most simple example is indeed the one with a constant force, e.g., the motion of a particle close to Earth. Then you can approximate the gravitational force on the particle as constant
$$\vec{F}=m \vec{g}.$$
Using a Cartesian basis with the basis vector $\vec{e}_z$ pointing up you have
$$m \begin{pmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -m g \end{pmatrix},$$
where $g \simeq 9.81 \text{m}/\text{s}^2$.

These equations of motion are easily integrated since the right-hand side does not depend on $(x,y,z)$ at all. So you indeed just do integrals. Cancelling the $m$ on both sides and then integrating once wrt. to $t$ you get
$$\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} = \begin{pmatrix} v_{0x} \\ v_{0y} \\ -g t +v_{0z} \end{pmatrix},$$
where $v_{0x}$, $v_{0y}$, $v_{0z}$ are arbitrary constants. They are given by the initial conditions, which you always need as input to get a unique solution of the equations of motion. Because they are of second order in time derivatives you need to know $x$, $y$, $z$ at some initial time (which I here choose to be $t=0$ for simplicity) and $\dot{x}$, $\dot{y}$, and $\dot{z}$ at the initial time $t=0$. These initial values of the velocity I've called $v_{0x}$, $v_{0y}$, and $v_{0z}$, respectively.

Now you can integrate once more to also get the positions
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} v_{0x} t + x_0 \\ v_{0y}t + y_0 \\ -g t^2/2 +v_{0z} t +z_0 \end{pmatrix},$$
where $x_0$, $y_0$, and $z_0$ are obviously the initial values for the position coordinates at the initial time $t_0=0$, and that's the well-known complete solution for the case of a constant force.

There are of course also more complicated cases, which however are still simple to solve. One of the most important examples is the harmonic oscillator. Here the force is proportional to the displacement of the particle from a certain equilibrium position. An example is a particle on a spring. For not too large elongations the force on the particle is proportional to the displacement from the equilibrium position (Hooke's Law). Here we have an example for 1D motion, and we just need to write done the one corresponding position component, which I'll call $x$. The equation of motion then reads
$$m \ddot{x}=-k x,$$
where $x=0$ is the equilibrium position (by choice of the coordinate system).

This is not as straight-forward to "integrate". Dividing by $m$ you get
$$\ddot{x}=-\frac{k}{m} x.$$
Now you have to remember which functions give twice integrated its negative (up to a factor). If you remember your differentiation rules, you find pretty quickly that these are cos and sin. So you make the ansatz
$$x(t)=A \cos(\omega t) + B \sin(\omega t)$$
with $A$, $B$, and $\omega$ some constants. Now indeed taking the derivatives you get
$$\dot{x}=-A \omega \sin(\omega t) + B \omega \cos(\omega t)$$
and once more
$$\ddot{x}=-A \omega^2 \cos(\omega t) - B \omega^2 \sin(\omega t),$$
but indeed this is of the right form, because for our ansatz
$$\ddot{x}=-\omega^2 x.$$
Comparing this with the equation of motion, you see that it solves indeed this differential equation if you set
$$\omega=\sqrt{\frac{k}{m}}.$$
Our ansatz is also the "complete" solution, because there is enough freedom to choose $A$ and $B$ to fulfill the initial conditions at $t_0=0$:
$$x(0)=A=x_0, \quad \dot{x}(0)=B \omega =v_0\; \Rightarrow \; B=\frac{v_0}{\omega}.$$
So the solution of the equation of motion with the given initial conditions finally is
$$x(t)=x_0 \cos(\omega t) + \frac{v_0}{\omega} \sin(\omega t) \quad \text{with} \quad \omega=\sqrt{\frac{k}{m}}.$$

I didn’t catch everything. It’s foggy up here

 

[A.T.]

I didn’t catch everything. It’s foggy up here

Try this:

 

[rudransh verma]

Try this:

https://www.physicsforums.com/threads/a-truck-and-a-car-coming-to-a-stop.1012285/post-6599744
Here the poster has multiplied by dx on both sides with dx which is not wrong mathematically but why did he do it?
Then he took integral which is basically finding the area under the curve and the original function. How is area an original function and Why do you need the original function?
(The problem is to establish a relationship between mass and distance).