Work of a system: Tipler vs Halliday

[walking]

A strategy I use when learning is to consult multiple books. It is great most of the time, but occasionally I find that there seems to be a contradiction between different books. For example, when proving the conservation of mechanical energy theorem, Tipler & Mosca claim that the total work of a system is the total change in kinetic energy. However, in the Halliday & Resnick book, the authors claim that such a blanket statement does not hold for systems.

Tipler & Mosca:

Halliday & Resnick:

 

[jbriggs444]

A strategy I use when learning is to consult multiple books. It is great most of the time, but occasionally I find that there seems to be a contradiction between different books. For example, when proving the conservation of mechanical energy theorem, Tipler & Mosca claim that the total work of a system is the total change in kinetic energy. However, in the Halliday & Resnick book, the authors claim that such a blanket statement does not hold for systems.

Tipler appears to be using the term "total work" to refer to the sum of the work done across each system interface where a force is applied. That is the sum of the product of the force applied an interface and the parallel motion of that interface.

Halliday and Resnick are using the term "work" to refer to what I would call "center of mass work". That is, the product of the sum of the forces and the parallel motion of the center of mass.

Loosely speaking, Tipler is talking about the sum of the products and Halliday and Resnick are talking about the product of the sums.

A work-energy theorem works for both notions of work. But it works differently for each.

With "total work", you get the change in total energy. The sum of kinetic energy for the parts of the body individually. [Note that internal force pairs can do "total work"]

With "center of mass work", you get the change in bulk kinetic energy. The kinetic energy for the center of mass, ignoring the relative velocities of all the component pieces. [Note that internal force pairs cannot do "center-of-mass work"].

 

[vanhees71]

I'm not so sure what you are discussing here. What's behind the work-energy theorem is the following. Let's assume the most simple case of a closed system of point particles with pair-interactions, i.e., the equation of motion for the $i$-the particle read
$$m_i \ddot{\vec{x}}_i=\sum_{k \neq i} \vec{F}_{ik}.$$
Multiply by $\dot{\vec{x}}_i$ and sum over $i$, you get
$$\sum_{i} m_i \dot{\vec{x}}_i \cdot \ddot{\vec{x}}_i = \frac{\mathrm{d}}{\mathrm{d} t} \sum_{i} \frac{m_i}{2} \dot{\vec{x}}_i^2 =\dot{T}= \sum_{i,k;i \neq k} \dot{\vec{x}}_i \vec{F}_{ik}.$$
Integrating between times $t_1$ and $t_2$
$$T_2-T_1 = \int_{t_1}^{t_2} \mathrm{d} t \sum_{i,k;i \neq k} \dot{\vec{x}}_i \cdot \vec{F}_{ik}=\Delta W.$$
The change of total kinetic energy is the change of work calculated along the trajectory of the particle.

This becomes interesting if the pair forces can be written as the gradient of a potential. Since further according to Newton's 3rd Law $$\vec{F}_{ik}=-\vec{F}_{ki}$$ the potential should be of the form $V_{ik}=V_{ik}(|\vec{x}_i-\vec{x}_k|)$, where we also have used rotation invariance of Newtonian mechanics. Then you have
$$\vec{F}_{ik} = -\nabla_i V_{ik}(|\vec{x}_i-\vec{k}|).$$
Now writing
$$U=\frac{1}{2} \sum_{i,k;i \neq k} V_{ik}(|\vec{x}_i-\vec{x}_k|).$$
The $1/2$ is to avoid double counting, because each pair of particles appears only once in the total potential energy, $U$.

Now you have
$$\sum_i \dot{\vec{x}}_i \cdot \vec{\nabla}_i U=\frac{\mathrm{d}}{\mathrm{d} t} U,$$
and thus the energy-work theorem becomes
$$T_2-T_1=-\int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} U = -(U_2-U_1)$$
or
$$T_2+U_2=T_1 + U_1.$$
This is the energy-conservation theorem, and the great thing is that in this case you don't need to calculate the trajectory to get the work, because it's simply given as $$\Delta W=-\Delta U$$, and $U$ only depends on the positions $\vec{x}_i$ and is not dependent on the specific trajetories the particles go. Thus if the pair-interaction forces have a potential, the work-energy theorem is applicable without knowing the solution of the equations of motion and thus helps to solve them.

Independent of the work-energy theorem or the energy-conservation theorem the 3rd law implies momentum conservation. Because $\vec{F}_{ik}=-\vec{F}_{ki}$ (Newton's 3rd law) summing the equations of motion over $i$ leads to
$$\dot{\vec{P}}=0,$$
where
$$\vec{P}=\sum_{i} m_i \dot{\vec{x}}_i$$
is the total momentum of the particles. This is the law of total-momentum conservation, and integrating it with respect to time it says
$$\vec{P}=\text{const}$$
along the solution of the equations of motion.

Now you can write
$$\vec{P}=M \dot{\vec{X}}, \quad \vec{X}=\frac{1}{M} \sum_i m_i \vec{x}_i, \quad M=\sum_i m_i.$$
$\vec{X}$ is the center of mass, and the momentum conservation implies that the center of mass moves like a free particle, i.e., a particle on which no forces are acting,
$$\vec{X}=\vec{V} t + \vec{X}_0, \quad \vec{V}=\frac{1}{M} \vec{P}=\text{const}.$$