Work energy theorem by variable force

[rudransh verma]

Its Good to be Back!
From Resnik, Fundamentals of physics: Consider a particle of mass m, moving along an x-axis and acted on by a net force F(x) that is directed along that axis. The work done on the particle by this force as the particle moves from position $x_i$ to position $x_f$ is given by
$\int_{x_i}^{x_f} F(x) \, dx=\int_{x_i}^{x_f} ma \, dx$.
If $F$ is a variable force then how can we write $ma$?

 

[vanhees71]

The formula reads (after repairing the double-subscript problems)
$$\int_{x_i}^{x_f} F(x) \, dx=\int_{x_i}^{x_f} ma \, dx .$$
As you rightfully figured out yourself that simply doesn't make sense. As I wrote already some time ago, trying to clarify your questions concerning the work-energy theorem, the correct derivation of the work-energy theorem is an integration along the trajectory of the particle, i.e., along the solution of the equation of motion, which reads
$$m a=m\ddot{x} = F(x),$$
which is an ordinary differential equation, i.e., you look for a solution $x(t)$ of the equation, involving derivatives of this unknown function.

On the other hand, just from having the equation of motion you can derive some properties of this solution, even without explicitly knowing it. For the work-energy theorem, just multiply with $v=\dot{x}$ and integrate from $t_i$ to $t_f$,
$$ \int_{t_i}^{t_f} m \mathrm{d} t \dot{x} \ddot{x} = \int_{t_i}^{t_f} \mathrm{d t} \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{x}^2 \right) = \frac{m}{2} \dot{x}^2(t_f)-\frac{m}{2} \dot{x}^2(t_i) =\int_{t_i}^{t_f} \mathrm{d} t \dot{x}(t) F[x(t)]=W.$$
The integral is along the trajectory, i.e., in this very general case you cannot say much more than that the change of kinetic energy $E_{\text{kin}}=m v^2/2$ is the work done by the forces on the particle, $W$, along its trajectory.

 

[rudransh verma]

., along the solution of the equation of motion, which reads
ma=mx¨=F(x),
which is

When I did questions on laws of motion the force was always constant. $F_{net}=ma$. All those free body diagrams, the forces acted on the body were constant forces. I don't understand.
Your eqn of $F(x)$ is suggesting that force can never be constant.

 

[vanhees71]

Of course $F$ can be constant (e.g., the gravitational field close to Earth can be approximated as a constant force $F=mg$), but usualy it's not. The most simple example is a particle fixed at a spring. There the force is proportional to the elongation of the spring relative to its equilibrium length, i.e., if $x=0$ is the equilibrium position of the particle, then $F=-k x$ with $k=\text{const}$ (Hooke's law, valid for not too large elongations of the spring).

 

[rudransh verma]

Of course $F$ can be constant (e.g., the gravitational field close to Earth can be approximated as a constant force $F=mg$), but usualy it's not. The most simple example is a particle fixed at a spring. There the force is proportional to the elongation of the spring relative to its equilibrium length, i.e., if $x=0$ is the equilibrium position of the particle, then $F=-k x$ with $k=\text{const}$ (Hooke's law, valid for not too large elongations of the spring).

Ok! But $ma=m\frac{d^2x}{dt^2}=F(x)$ is not equation of motion. It’s the v=u+at, s=ut+1/2at^2, and v^2=u*2=2as.

 

[vanhees71]

No, if $F$ is not constant your solutions are not the solutions of the equation of motion. For my example of the force of a particle on a spring you get
$$m \ddot{x}=-k x.$$
It's easy to see that the general solution is
$$x(t)=A \cos(\omega t)+B \sin(\omega t) \quad \text{with} \quad \omega=\sqrt{\frac{k}{m}},$$
where $A$ and $B$ are integration constants, which you can fix with the initial conditions,
$$x(0)=x_0, \quad \dot{x}(0)=v_0,$$
which leads to
$$A=x_0, \quad B=\frac{v_0}{\omega}.$$

 

[rudransh verma]

No, if F is not constant your solutions are not the solutions of the equation of motion. For my example of the force of a particle on a spring you get

What solution?

 

[vanhees71]

The solution of the equation of motion of course!

 

[rudransh verma]

The solution of the equation of motion of course!

You mean the values of s,t,a,v,u?

 

[weirdoguy]

It’s the v=u+at, s=ut+1/2at^2, and v^2=u*2=2as.

These are true only if F is constant.

 

[vanhees71]

It depends on, how you define your quantities.

If $F=\text{const}$ in my notation, where $x$ is the coordinate of the particle along the direction of consideration you have
$$m \ddot{x}=F; \Rightarrow \; \ddot{x}=\frac{F}{m}$$
This you simply integrate with the initial conditions $x(0)=x_0$ and $\dot{x}(0)=v_0$:
$$\dot{x}(t)-\dot{x}(0)=\dot{x}(t)-v_0=\frac{F}{m} t \; \Rightarrow \; \dot{x}(t)=\frac{F}{m} t + v_0$$
and once more
$$x(t)-x(0)=x(t)-x_0=\frac{F}{2m} t^2 + v_0 t \; \Rightarrow \; x(t)=\frac{F}{2m} t^2 + v_0 t +x_0.$$
For other kinds of forces, which are not constant, these are obviously not the right solutions of the equations of motion.

I've given the next-most-simple example of a force law above ("harmonic motion").

 

[rudransh verma]

For other kinds of forces, which are not constant, these are obviously not the right solutions of the equations of motion.

I thought suvat eqns are called eqn of motion but it seems like you are calling Newton’s second law as eqn of motion.

 

[weirdoguy]

Sometimes Newton's second law equations are called dynamical equations of motion, and x(t), y(t), z(t) are called kinematical equations of motion.

 

[jack action]

If $F$ is a variable force then how can we write $ma$?

Because $a$ is also a variable. By definition:
$$F(x) = \frac{d(mv)}{dt}$$
Where $m$ is assumed to be constant such that:
$$F(x) = m\frac{dv}{dt} = ma$$
Therefore if $F(x)$ varies with distance, $a$ must also vary with distance since we already assumed $m$ doesn't. We can then conclude:
$$F(x) \propto a$$
And if $F(x)$ is constant, then $a$ is also constant.

Note that it is possible that your force is constant for a given distance, say from $x_1$ to $x_2$, and then is different, but still constant, from $x_2$ to $x_3$. Your equations of motion still stand, only with a different value for acceleration in both traveled paths.

 

[rudransh verma]

For other kinds of forces, which are not constant, these are obviously not the right solutions of the equations of motion.

Ok! I think I got you. I have another question. $\Delta U=-\int_{x_i}^{x_f} -kx \,dx$ for elastic potential energy of spring. We know that when integrating we take dx such that F becomes constant so that we can use the formula of $W=fd\cos\theta$. I don’t think $-kx$ is a constant for small element dx.

 

[vanhees71]

$$\newcommand{\dd}{\mathrm{d}}$$
That's an important point, where the work-energy theorem becomes much more profound. That's the case of forces that have a potential. Indeed for the linear-froce law of a spring you can write
$$F(x)=-k x=-V'(x) \; \Rightarrow \; V(x)=-\frac{k}{2} x^2 + V_0,$$
where $V_0$ is an arbitrary constant, which you can as well set to 0, which I'll do for simplicity, i.e., from know on we use the potential
$$V(x)=\frac{k}{2} x^2$$
to describe the force as
$$F(x)=-V'(x).$$
Now the right-hand side of the work-energy theorem becomes
$$W=\int_{t_i}^{t_f} \dd t \dot{x}(t) F[x(t)]=-\int_{t_i}^{t_f} \mathrm{d} t \dot{x}(t) V'[x(t)] = -\int_{t_1}^{t_2} \dd t \frac{\dd}{\dd t} V[x(t)]=-\{V[x(t_f)]-V[x(t_i)] \}.$$
Now you have
$$E_{\text{kin}}(t_f)-E_{\text{kin}}(t_i)=-\{V[x(t_f)] - V[x(t_i)]\}.$$
This you can now write in the form
$$E_{\text{kin}}(t_f)+V[x(t_f)]=E_{\text{kin}}(t_i) + V[x(t_i)].$$
Since now $t_i$ and $t_f$ are completely arbitrary that simply tells you that the total energy, defined as
$$E=E_{\text{kin}} + V(x)$$
is a constant for all solutions of the equation of motion. That's the energy-conservation law and much more useful than the very general work-energy theorem, because here you get useful additional information about the solution without knowing this solution before, because we could calculate the work in terms of the potential without having the concrete solution.

 

[rudransh verma]

Indeed for the linear-froce law of a spring you can write

So is -kx a constant force doing work for dx distance? Do we take a average value of x for dx distance making -kx a constant for dx distance?

 

[vanhees71]

No it's what I wrote $F=-k x$, where $x$ is the position of the particle counted from the equilibrium point of the spring, i.e., the force is proportional to the elongation of the spring relative to its equilibrium length and opposite to the direction (i.e., if you stretch or compress the spring it reacts such as to go back to its equilibrium position; for not too large stretching or compressing the force is proportional to the deviation of the length of the spring from its equilibrium length, which is Hooke's Law).

 

[russ_watters]

So is -kx a constant force doing work for dx distance? Do we take a average value of x for dx distance making -kx a constant for dx distance?

In calculus, dx distance is zero, so there is only one value and it is exact.

 

[ergospherical]

In calculus, dx distance is zero, so there is only one value and it is exact.

As you're no doubt aware - but to clarify for the OP - this description isn't correct. You partition the integration interval into a finite number of regions and choose a value $x_i^*$ of $x$ anywhere within the $i^{\mathrm{th}}$ region (e.g. the left, the centre, the right,...). When the limit is taken it turns out that the sum doesn't care where you chose $x_i^*$ in that region.

 

[rudransh verma]

No it's what I wrote F=−kx, where x is the position of the particle counted from the equilibrium point of the spring,

Ok! So that’s how the force is constant for dx distance and we can apply the definition of work.

 

[vanhees71]

There is no $\mathrm{d} x$ distance in the formula for the force. I gave a complete calculation of the work-energy theorem in terms of the potential of the force above in #16. There the corresponding integrals occur!

 

[rudransh verma]

There is no dx distance in the formula for the force. I

No I mean dx is small length in the formula of dW when finding the change in potential energy of gravitation and spring.

 

[vanhees71]

In some sense yes (see @ergospherical 's posting #20).