Principle of Physics: Derivation of -dU/dx=F

[mark2142]

Hi, Everyone! This is the page(first image) from Principle of physics by resnik.
I want to ask the definition of work($W=F(x) \Delta x$) by variable force here is somewhat different from the usual integral version. I don't understand how is this valid definition?
Secondly, how did they reach to the derivative F(x)=-dU/dx by limit?
(There is another approach to proving this by taking potential energies and differentiating them which results in the field force involved. This is given in my another book which is to understand(second image)).

 

[vanhees71]

This is indeed a bit confusing. At best the conclusion is valid for 1D motions along a fixed "$x$ axis", because you can define
$$V(x)=-\int_{x_0}^x \mathrm{d} x' F(x').$$
Supposed $F(x)$ is continuous, then $F(x)=-V'(x)$, i.e., the force $F(x)$ always has a potential.

This doesn't hold true for 3D motion/forces. There the existence of a potential is constraining the kind of forces more. In this case you must necessarily have $\vec{\nabla} \times \vec{F}=0$ if there's a potential such that $\vec{F}=-\vec{\nabla} V$. If this constraint is fulfilled within a simply connected region of space, then there's a potential, and it's calculated by taking the line integral from any fixed position $\vec{x}_0$ to the point $\vec{x}$ within this region along an arbitrary path, and the potential is independent of the choice of this path.

 

[mark2142]

This is indeed a bit confusing.

Okay!

At best the conclusion is valid for 1D motions along a fixed " axis", because you can define

Supposed is continuous, then , i.e., the force always has a potential.

You mean we can define a potential as $U(x)=-\int F(x) \, dx + U(x_0)$ and then differentiate it we get $-\frac{dU}{dx}=F(x)$.
But how did you reach to the conclusion that $U(x)=-\int F(x) \, dx + U(x_0)$ where $U(x_0)$ is constant?

 

[vanhees71]

You just use the fundamental theorem of calculus, i.e., that integration is the "inverse" of differentiation.

 

[PeroK]

Okay!

You mean we can define a potential as $U(x)=-\int F(x) \, dx + U(x_0)$ and then differentiate it we get $-\frac{dU}{dx}=F(x)$.
But how did you reach to the conclusion that $U(x)=-\int F(x) \, dx + U(x_0)$ where $U(x_0)$ is constant?

If $U(x)$ represents potential energy and the work done by a position-dependent force is $F(x)dx$, then $$\int_a^b F(x)dx = KE(b) - KE(a)$$I.e. the integral represents the change in Kinetic Energy. Then, by conservation of total mechanical energy (KE+PE) that integral must also be the negative change in PE. So that $$\int_a^b F(x)dx = U(a) - U(b)$$ And now we can apply the fundamental theorem of Calculus to get $$F(x) = -\frac {dU}{dx}$$

 

[PeroK]

PS note that this only defines the potential energy up to a constant as the derivatives of $U(x)$ and $U(x)+ U_0$ are equal.

 

[mark2142]

by conservation of total mechanical energy (KE+PE) that integral must also be the negative change in PE. So that And

But how did you reach to the conclusion that $U(x) - U(x_0) =-\int F(x) \, dx$ or $\Delta U=-W$?( In book its given $$W=\int_{x_0}^x F(x) \, dx$$
We need to use a conservative force such as Gravity-
$$\int_{x_0}^x (-mg) \, dx$$
So, $$-mg \int_{x_0}^x \, dx$$
$$-(mgx-mg{x_0})$$
So we define mgx as potential energy )

Is this the proof of $-W=\Delta U$ ?

 

[PeroK]

If there is only PE and KE in a system and energy is conserved then $\Delta KE = -\Delta PE$.

 

[mark2142]

energy is conserved then $\Delta KE = -\Delta PE$.

Energy conservation comes after we have $W=\Delta Ke$ and $-W=\Delta U$ ?

 

[PeroK]

Energy conservation comes after we have $W=\Delta Ke$ and $-W=\Delta U$ ?

You just need:$$E = KE + PE$$ where $E$ is constant. That is conservation of total mechanical energy.

 

[mark2142]

That is conservation of total mechanical energy.

I know what it is. And that is derived in the book as $W=\Delta KE$ and $W=-\Delta U$. So, $\Delta KE + \Delta U=0$. So $KE+U=constant$.

But where has $W=-\Delta U$ come from?
Is it so obvious or done experimentally?

 

[vanhees71]

Just by definition you have
$$U(x)=-\int_{x_0}^x \mathrm{d} x' F(x') \; \Rightarrow \; F(x)=-U'(x).$$
Now you use it in the equations of motion
$$m \ddot{x}=-U'(x).$$
Multiply with $\dot{x}$ you get
$$m \dot{x} \ddot{x}=\frac{m}{2} \frac{\mathrm{d}}{\mathrm{d} t}(\dot{x}^2) =-\dot{x} U'(x)=-\frac{\mathrm{d}}{\mathrm{d} t} U(x).$$
Combining both sides of the equation gives
$$\frac{\mathrm{d}}{\mathrm{d} t} \left [\frac{m}{2} \dot{x}^2 + U(x) \right]$$
or, integrated,
$$\frac{m}{2} \dot{x}^2 + U(x)=E=\text{const}.$$

 

[mark2142]

If there is only PE and KE in a system and energy is conserved then $\Delta KE = -\Delta PE$.

O yes! My mistake. I misunderstood it. The Energy is always conserved. And so there is a potential energy defined in the process. But potential energy is defined only if the force is conservative in nature. Why?

 

[PeroK]

O yes! My mistake. I misunderstood it. The Energy is always conserved. And so there is a potential energy defined in the process. But potential energy is defined only if the force is conservative in nature. Why?

The concept of potential energy is useful when the force is conservative. Otherwise, it cannot be a function of position but also depends on the path taken.

You could try to define a potential for the friction force and see what you get.

 

[mark2142]

Just by definition you have
$$U(x)=-\int_{x_0}^x \mathrm{d} x' F(x') \; \Rightarrow \; F(x)=-U'(x).$$
Now you use it in the equations of motion
$$m \ddot{x}=-U'(x).$$
Multiply with $\dot{x}$ you get
$$m \dot{x} \ddot{x}=\frac{m}{2} \frac{\mathrm{d}}{\mathrm{d} t}(\dot{x}^2) =-\dot{x} U'(x)=-\frac{\mathrm{d}}{\mathrm{d} t} U(x).$$
Combining both sides of the equation gives
$$\frac{\mathrm{d}}{\mathrm{d} t} \left [\frac{m}{2} \dot{x}^2 + U(x) \right]$$
or, integrated,
$$\frac{m}{2} \dot{x}^2 + U(x)=E=\text{const}.$$

As I understand you defined a potential energy and showed that the KE + PE=constant which it should be.

 

[vanhees71]

O yes! My mistake. I misunderstood it. The Energy is always conserved. And so there is a potential energy defined in the process. But potential energy is defined only if the force is conservative in nature. Why?

If the force cannot be written as the gradient of a potential, then there is no well-defined potential energy. Then all you have is the "work-energy relation",
$$T=\frac{m}{2} [\vec{v}(t_2)^2-\vec{v}^2(t_1)] = \int_{t_1}^{t_2} \mathrm{d}t \vec{v}(t) \cdot \vec{F},$$
where $\vec{x}(t)$ is the trajectory of the particle, i.e., the solution of the equations of motion.

 

[mark2142]

The concept of potential energy is useful when the force is conservative. Otherwise, it cannot be a function of position but also depends on the path taken.

You could try to define a potential for the friction force and see what you get.

If I understand it correctly then,
From $\Delta KE= W$
$\Delta KE= -\mu mg \Delta x$ where $\Delta x= x- x_0$
Since Change in kinetic energy $\Delta KE$ is -ve. So, it decreases.

Now $W=-\mu mg \Delta x$
$W=-(\mu mg x - \mu mg x_0)$
$-W=\mu mg x - \mu mg x_0$
We define $\mu mg x=U$ potential energy.
$-W=\Delta U$
-ve work causes increase in potential energy which depends on the $\mu$ of the path and x.

 

[PeroK]

If I understand it correctly then,
From $\Delta KE= W$
$\Delta KE= -\mu mg \Delta x$ where $\Delta x= x- x_0$
Since Change in kinetic energy $\Delta KE$ is -ve. So, it decreases.

Now $W=-\mu mg \Delta x$
$W=-(\mu mg x - \mu mg x_0)$
$-W=\mu mg x - \mu mg x_0$
We define $U=\mu mg x$ potential energy. -ve work causes increase in potential energy which depends on the $\mu$ of the path and x.

That's not potential energy in any shape or form.

 

[mark2142]

That's not potential energy in any shape or form.

Can you tell me why?

 

[vanhees71]

Indeed the potential of the force, $U$, usually is also called "potential energy", and
$$E=T+U=\text{const}.$$
For a constant force $F$ you indeed get for the potential
$$U(x)=-\int_0^x \mathrm{d} x' F=-F x.$$

 

[PeroK]

Can you tell me why?

An object at rest on a flat frictional surface has no potential energy dependent on its position. There is certainly not a linear energy profile.

 

[mark2142]

An object at rest on a flat frictional surface has no potential energy dependent on its position. There is certainly not a linear energy profile.

Ok! According to the $W=\mu mg \Delta x$ the work by the friction force is also dependent on the surface($\mu$) i.e. Non conservative in nature so we cannot define a potential energy function just depending on the position of the body. If we can't define the potential energy then the mechanical energy is not conserved for non conservative force. Yes?

 

[PeroK]

If we can't define the potential energy then the mechanical energy is not conserved for non conservative force. Yes?

Yes, that's the whole point.

 

[vela]

But where has $W=-\Delta U$ come from?
Is it so obvious or done experimentally?

In the work-energy theorem, you can separate the work done by conservative forces, which allow for potential energies, from the work done by non-conservative forces, which don't, i.e.,
$$\Delta KE = W_c + W_{nc}$$ Because the work done by conservative forces only depend on the endpoints, we move it over to the lefthand side of the equation and introduce the concept of potential energy.
$$\Delta KE + \underbrace{(-W_c)}_{\Delta U} = W_{nc}$$ So $W_c = -\Delta U$ is pretty much a matter of definition.

 

[mark2142]

Thank you guys...