Consistent Approach for Calculating Work By Friction for Rigid Body in Planar Motion Question
How can the work done by friction on a rigid body be calculated?
Numerous questions and answers on this exchange deal with work done by friction. Some of the earlier questions/responses cause confusion in my opinion by how they refer to friction heating up an object without further elaboration.  For example, see the brief discussion of "heating" in the first answer for Tricky conceptual question: ball sliding and rolling down incline. In reality, friction does increase the internal energy ("heat" up") a body. However most treatments of friction in elementary mechanics assume a rigid body, and for a rigid body friction cannot change its internal energy.  A treatment of heating requires an application of the first law of thermodynamics; however, many of the questions on this site dealing with the work done by friction can be answered assuming a rigid body.  For example, the first answer to the above question for a ball sliding down an incline implicitly treats the ball as a rigid body; therefore, contrary to the answer there are no heating effects.  I address a ball sliding down an incline in detail later in example #2 in my detailed answer below.
I summarize a consistent approach provided by @user256872 in the following answer.
My Answer
Summary
Friction does result in “heating” of a body, or more accurately an increase in the internal energy of a body.  However, a great many mechanics problems treat the body as a “rigid body”. In a rigid body, by definition the particles always maintain a fixed position relative to one another and move only with the body as a whole. Hence, there can be no dissipation of energy within a truly rigid body.
For a system of particles, it is always true that $\vec F = M \vec a_{CM}$ and that $\vec \tau = { {d\vec J} \over {dt}}$ (for rotation about a fixed point or the center of mass for general motion).  $\vec F$ is the net external force, $M$ is the system mass, $\vec a_{CM}$ is the acceleration of the center of mass, $\vec \tau$ is the net external torque, and $\vec J$ is the angular momentum.  However if $\vec J$ as taken as $\tilde I \cdot \vec \omega$ (or $J = I \omega$ for rotation in a plane), where $\tilde I$ is the inertia tensor and $\vec \omega$ is the angular velocity, this is specific to a rigid body because the inertia tensor is developed assuming a rigid body. (See a physics mechanics text, such as Symon,  Mechanics or Goldstein, Classical Mechanics). Therefore, evaluations that use $J = I\omega$  implicitly assume a rigid body. Some evaluations that use $J = I\omega$, such as an answer in the earlier post, referenced in the Question section above, claim that friction causes heating which is incorrect; friction only causes a change in kinetic energy, as discussed in my detailed answer that follows.
I think an answer by @user256872 to an earlier question, provides the key for addressing the effects of friction for a rigid body.  This approach evaluates the work done by friction by segregating that work into two parts: work for translational motion and work for rotational motion; A detailed discussion this approach follows, elaborating on the approach used by @user256872, and some follow-up questions are posed.
Detailed discussion
Assumptions
For simplicity, the discussion is restricted to the following conditions. I consider a rigid body in planar motion. I consider motion where either there is no fixed point for rotation, or the fixed point of rotation is the center of mass (CM); I do not consider rotation about a fixed point that is not the CM.
For a rigid body, there is no internal dissipation of energy and therefore friction does not cause “heating” effects; all the work on the body due to friction changes the total kinetic energy of the body.  The total kinetic energy of the body is the kinetic energy of the CM plus the rotational energy about the CM: ${1 \over 2} mv^2 + {1 \over 2} I  \omega^2$, where $m$ is the total mass, $v$ is the speed of the CM, $I$ is the moment of inertia about an axis through the CM, and $\omega$ is the angular speed about the center of mass. The motion of the CM is obtained by the first law: $\vec F = m \vec a$ where $\vec F$ is the net external force, $m$ is the total mass, and $\vec a$ is the acceleration of the CM.  The rotation about the CM- even if the CM is accelerating- is obtained by $\vec \tau = I \vec \alpha$ where $\vec \tau$ is the net torque about the CM and $\vec \alpha$ is the angular acceleration about the CM.  The work done (by all forces) equals the change in the total kinetic energy.
Work by friction
Based on the earlier response by @user256872, for a friction force $\vec F_{fric}$, the work done by friction is $\int \vec F_{fric} \cdot\vec vdt$ +  $ \int \vec \tau_{fric} \cdot\vec \omega dt$ where $\vec v$ is the velocity of the CM, $\vec \tau_{fric}$ is the torque about the CM due to the force of friction, and $\vec \omega$ is the angular velocity about the CM.  The work done by friction has two terms: the work done by friction on the CM ,$\int \vec F_{fric} \cdot\vec vdt$, and the work done by friction with respect to the CM ,$ \int \vec \tau_{fric} \cdot\vec \omega dt$.  For certain situations the sum of these two terms is zero and friction does no work, while for other situations the sum of these two terms is not zero and friction does work.
The work done by gravity on a body is $-mg \Delta z$ where $g$ is the acceleration of gravity and $\Delta z$ is the increase (positive upward) in elevation of the CM. (This work is typically treated as the negative of the change in potential energy $mg \Delta z$.)  For the following examples the work done by friction on the CM simplifies to $\int  F_{fric} dx$ where $x$ is the displacement along the direction of the force of friction, and the work done by friction with respect to the CM simplifies to $\int \tau_{fric} d \theta$ where $\theta$ is angular distance with respect to the CM.
1. Rolling without Slipping
Consider a round body (sphere, cylinder, etc.) rolling down an incline, starting at rest, as shown in the following figure.

Assume the body rolls without slipping. Basics physics texts, evaluate this case using both the equations of motion and work/energy, and show the total energy at the bottom of the incline is the same using either approach.  For this case of pure rolling of a rigid body, friction does no work, as sometimes explained by the fact there is no relative motion (no sliding) between the point of contact of the body and the surface of the incline.  The only force that does work on the body is gravity, so the total energy at the bottom of the incline is $mgh = {1 \over 2}mv^2 + {1 \over 2}I \omega^2$. Since the body rolls without slipping, $\omega = v/R$, and $mgh = {1 \over 2}mv^2 + {1 \over 2}I v^2/R^2$.  The total kinetic energy is partitioned between the kinetic the kinetic energy of the CM and the rotational energy about the CM.
Since only gravity does work, the total kinetic energy is the same as for the case of a frictionless incline, $mgh$, except with no friction the total kinetic energy is the kinetic energy of the CM and the rotational energy about the CM is zero.  So, for pure rolling friction- contrary to the case with no friction- although friction does no work it causes the kinetic energy of the CM to decrease and the rotational energy about the CM to increase.  This is explained by considering the two terms for the work done by friction.  The work by friction on the CM is   $-\int F_{fric} dx$, negative since friction opposes the x motion; the work by friction with respect to the CM is $\int \tau_{fric} d \theta = \int F_{fric}R dx/R$, where $d \theta = dx/R$ with no slip.  The first term decreases the kinetic energy of the CM and the second term increases the rotational energy about the CM, but the sum of the terms is zero since the net work done by friction is zero.  Although friction does no work in total, the total work has two components; one component decreases the kinetic energy of the CM and the other component increases the rotational energy about the CM. For rolling without slipping, the sum of the two components is zero.
2. Rolling with Slipping
Next, consider the case where the body slips as it rolls down the incline.    Here the force of friction does work, as sometimes explained by the fact that there is relative motion between the point of contact of the body and the incline due to slip.  This case can be evaluated using the equations of motion; see Analytical Mechanics by Fowles, and also this post.  For slipping, the coefficient of friction, $\mu$, must not exceed ${\tan\theta} \over{1 + (RM/I)^2}$, otherwise the body rolls without slipping. Solving the equations of motion, the time to reach the bottom of the incline $t_1$ is $\sqrt{{2h} \over {gsin\theta (sin\theta - \mu cos\theta)}}$ where $\mu$ is the coefficient of kinetic friction. The velocity $v$ of the CM at the bottom is $\sqrt{{2gh(sin\theta - \mu \cos\theta)} \over {sin\theta}}$, and the angular velocity $\omega$ at the bottom is ${\mu mgR \over{I}}\sqrt{{2h} \over {gsin\theta(sin\theta - \mu sin\theta)}}$.  Using these results, the total kinetic energy at the bottom $1/2 mv^2 + 1/2 I  \omega^2  = mgh - mgh {\mu cos \theta \over sin \theta} + {5 \mu^2mghcos^2 \theta \over 2sin \theta(sin \theta - \mu cos \theta)}$.
The total kinetic energy at the bottom of the incline can also be evaluated considering the work done on the body, due to gravity and the two terms for the work done by friction. Using this approach, the total kinetic energy at the bottom for the case of sliding is $mgh -\int_{0}^{h \over sin \theta} F_{fric} dx +  \int_{0}^{\theta_f} \tau_{fric} d \theta$.  (Here due to slip $d \theta \ne dx/R$.)  $\theta_f = {1 \over 2} \alpha t_1^2$ where $\alpha$ is the angular acceleration. $F_{fric} = \mu mgcos(\theta)$ and $\tau_{fric} = RF_{fric}$ are both constant.  After evaluating the integrals, the total kinetic energy at the bottom is the same as previously evaluated using the equations of motion.  Here, friction does work due to slip, because the portion of negative work on the CM is greater in magnitude than the portion of the positive work causing rotation about the CM.  The motion just described agrees with the motion developed in an answer for this post, but contrary to a claim in the answer for this earlier post there are no heating effects involved in explaining this motion.
3. Pulling on a pulley
As a trivial example, consider a pulley rotating about an axle fixed in the inertial frame through its CM, with mass $m$ and moment of inertia $I$, as shown the following figure.

The rotation is due to unequal tensions $T_2$ greater than $T_1$ on a rope that causes rotation.  The axle provides a force of constraint $\vec F_c$  upwards equal in magnitude to $T_2 – T_1 + mg$ to keep the CM fixed.  Friction between the rope and the pully- assuming no slip- causes rotation, therefore the tension forces can be regarded as frictional forces on the pulley.  Assuming no initial rotational motion, using the equation of motion for the pulley after time $t_1$, the total angular distance traveled is $\theta_f = {{(T_2 – T_1)Rt^2} \over{2I}}$, and  the rotational energy is ${(T_2 – T_1)^2R^2t_1^2}\over {2I}$.  Using the two-term relationship for the work by friction, the first term- for the kinetic energy of the CM- is zero since there is no motion of the CM due to the force of constraint.  The second term is the kinetic energy for rotation about the CM up to time $t_1$ is  $ \int_{0}^{\theta(t_1)} (T_2 – T_1)R d \theta = (T_2 – T_1) R \theta_f$.  Using the earlier result for $\theta_f$, this is the same result as obtained from the equation of motion for rotation.
Summary
In conclusion, the work done by friction can be evaluated by segregating that work into two parts; this approach provides insight into the effects of friction on translational and rotational motion.  Again, this approach is based on the earlier response by  @user256872. Comments on this approach are appreciated.
Pseudowork and Work
Other questions/answers on this exchange and in the literature discuss the difference between work as defined in mechanics (sometimes called pseudowork) as different from work as defined in thermodynamics.  See Pseudowork and real work
Sherwood, Bruce Arne
The American Journal of Physics , Volume 51 (7) – Jul 1, 1983.  The concept of pseudowork seems useful to me.
Some Questions to Consider

*

*For the first example with rolling without slipping, one explanation for friction doing no (net) work is that there is slip (no relative motion) of the point of contact of the body on the incline.  However, for the last example where friction drives the pully there is also no relative motion, yet here friction does work.  Does this mean that no slip does not always imply no work by friction?


*How can this approach for evaluating the work by friction be extended to general three dimensional motion of a rigid body where the inertia tensor is needed?


*For fluid mechanics, if frictional heating does not cause an appreciable change in the temperature of a fluid, the Bernoulli energy equation can be modified to include a frictional “head” loss to account for friction; for an appreciable change in temperature, the evaluation is complex, requiring the first law, and momentum relationships.  For the case of solids experiencing friction, is there an approximate approach (similar to the modified Bernoulli for fluids) that can be used?


*Where can we find discussions of how to evaluate frictional heating for solids using both an approximate and a detailed approach?
 A: 
How can the work done by friction on a rigid body be calculated?

The work done by friction on a body is calculated the same way the work done by any other force is calculated: $$W_F = \Delta E_F = \int \vec F \cdot \vec v \ dt$$ where $\vec v$ is the velocity of the body’s material at the point of application of the force.
This formula applies for all mechanical situations: static friction or dynamic friction, rigid or non-rigid, etc. Note importantly that the velocity is the velocity of the material at the point of application of the force, not the velocity of the center of mass.

Does this mean that no slip does not always imply no work by friction?

It is easy to find no-slip situations where friction does work. For example, static friction does work on a box on the bed of an accelerating truck, or when a person plants their feed and pulls an object attached to a rope, the static friction from their hands does work. The key is that in all cases where static friction does work the object’s material is moving at the point where the friction force is acting.

How can this approach for evaluating the work by friction be extended to general three dimensional motion of a rigid body where the inertia tensor is needed?

No extension is needed. It applies as-is.

For the case of solids experiencing friction, is there an approximate approach (similar to the modified Bernoulli for fluids) that can be used?

In kinetic friction the velocity of one surface is different than the other. The frictional forces are equal and opposite, but since the velocities are not the same, the work is not equal and opposite. There is always more mechanical energy that leaves one object than enters the other. The difference is the thermal energy produced at the interface.
Once the amount of thermal energy produced at the interface is determined, then the distribution and flow of heat follows normal thermodynamic rules.
A: Long post. I'll offer a small clarifying point.

For the first example with rolling without slipping, one explanation for friction doing no (net) work is that there is slip (no relative motion) of the point of contact of the body on the incline. However, for the last example where friction drives the pully there is also no relative motion, yet here friction does work. Does this mean that no slip does not always imply no work by friction?

No slip does not mean that friction does no work. Here is a block resting on a slab resting on an ice-skating rink:

When I push the slab along the ice, the top block does not slip on the slab. Static friction does work to change the block's kinetic energy.
Here is a cylinder in a truck bed:

When the truck accelerates forward, static friction drags the cylinder forward and rotates it counter-clockwise, the cylinder rolls without slip off the end of the truck. Static friction did work during roll without slip.
