What is the work done by friction on a body which is rolling with slipping? 
A hollow sphere is released from the top of an inclined plane of inclination $\theta$.
(a) What should be the minimum
coefficient of friction between the sphere and the plane
to prevent sliding?
(b) Find the kinetic energy of the
ball as it moves down a length $L$ on the incline if
the friction coefficient is half the value calculated in part (a).

So I was solving this question and it was all good until I couldn't get to the solution to the second part using a work $-$ kinetic energy appraoch even though I thought that my method was correct. I got the first part easily, that coefficient of friction = $\frac{2}{5}\tan\theta$ and solved the second part correctly by treating the linear and rotational motions separately. I tried to calculate the kinetic energy and I tried to use the work $-$ energy relationship, thus giving me
$$mgL\sin\theta=\frac{mv^2}{2}+\frac{Iw^2}2+\frac{mgL\tan\theta \cos\theta}5,$$
but I didn't get the correct answer which was $\frac{7}{8}mgL\sin\theta$ and instead I obtained $\frac{4}{5}mgL\sin\theta$.
So can anyone please tell me what term am I missing in the work equation? I'm pretty sure it must be related somehow to work done by friction opposing rolling but I just can't get it.
 A: Thanks to the OP for this great question which requires a subtle understanding of the phenomenon of rolling without slipping. Since the question is tagged homework-like this answer will carefully avoid providing the direct solution but touch upon the subtle aspects involved in this phenomenon which enhance our understanding.

$$\underline{\textit{Analysis:}}$$
At the outset let us note that the static friction vanishes in the case that a uniformly dense circular body is rolling without slipping if the center of the body uniform velocity as calculated using the inertial ground reference frame on a flat surface to which the reference frame is fixed. This can be proved by contradiction, since the presence of static friction would cause the body to undergo rotational decelerate while linearly accelerating. Further, even if the motion of the center of the body is non-uniform and is instead linearly accelerating, the static friction performs no work (calculated using the inertial ground reference frame) on the body because the velocity of the material particle to which the  friction force is applied is identically zero.
The phenomenon considered is the planar motion of a hollow sphere on an inclined plane. In this discussion we use a reference frame with the preferred Cartesian coordinate system with the $X$ axis pointing down the plane and parallel to it's surface, the $Y$ axis perpendicular to the surface and upwards from the surface and the $Z$ axis pointing outwards from the screen.

Under the assumption of rolling without slipping, the situation depicted in the schematic is similar to a horizontal force being applied to the center of mass of a uniformly dense circular rigid body, so that the direction of friction is opposite to that of the applied force (direction is horizontally rightwards of the screen here). The same schematic applies to the motion of the sphere in the case of rolling with slipping as well. However, note that if instead, a torque was being applied to the axis of the body passing through it's center of mass, the direction of friction would be horizontally leftwards of the screen in that case.
There is a typographical error in the first enumerated point in the OP since it the required value of the coefficient of friction stated therein is erroneous.

*

*(addressing part a in the OP) If the sphere does not slip, then the velocity, $v$, and acceleration, $a$, components related to the angular velocity, $-\omega$, and acceleration, $-\alpha$, components by $v=-\omega r$ and $a=-\alpha r$ (observe that $\omega, \alpha<0$ in the situation being analyzed). The dynamical equations $ma=mg\sin\theta-f$ and $-I\alpha=fr$ where $f\leq\mu mg\cos\theta$ imply that $\frac{2}{5}\tan\theta\leq\mu$. Note that the apparent complexity of the negative signs in the previous equations could have been avoided by defining the $Z$ axis to point into the screen along with the necessarily downward point $Y$ axis, but the notation adopted here is so chosen because it emphasizes the subtlety of the kinematic analysis. In this notation, the vector quantities of interest are expressed as follows $\vec{\omega}=-\omega\hat{k}$, $\vec{\alpha}=-\alpha\hat{k}$ being the attitudinal velocity and acceleration of the sphere calculated using the inertial ground reference frame, $\vec{v}=v\hat{i}$, $\vec{a}=a\hat{i}$ being the linear velocity and acceleration of the center of the sphere calculated using the inertial ground reference frame and $\vec{F}_\text{gravity}=mg\sin\theta\hat{i}-mg\cos\theta\hat{y}$, $\vec{F}_\text{friction}=-f\hat{i}$ being the forces acting on the sphere due to gravity and friction.

$\underline{\text{Indeed, using vector quantities to obtain the scalar equations of motion is}}\\\underline{\text{ the best strategy to avoid analytical errors.}}$
In the following discussion, with convenience in mind, we abuse our notation and assume the variables $\omega, \alpha, v, a$ and $f$ to represent the absolute the absolute values of the physical quantities in the preceding discussion, unless otherwise stated.


*(addressing part b in the OP) We now analyze the rolling with slipping motion which necessarily occurs since the coefficient of friction is assumed to be half of that obtained in the previous part of the analysis and which is required to enable rolling without slipping. $\textit{Approach via kinematic analysis:}$ If the coefficient of friction is given by $\mu_\text{actual}=\frac{\mu}{2}=\frac{1}{5}\tan\theta$, the sphere will slip as it moves down the incline with the linear acceleration down the incline $a=g\sin\theta-\mu_\text{actual}g\cos\theta=\frac{4}{5}g\sin\theta$ and an angular acceleration component $\alpha=\frac{\mu_\text{actual}mg\cos\theta\cdot r}{I}=\frac{3}{10}\frac{g}{r}\sin\theta$ at all time instants. Applying the kinematic relationships $v(t) = a t$ and $\omega(t) = \alpha t$ arising from the constant derivatives of the linear and attitudinal velocity components and denoting the time at which the sphere's center reaches the distance $L$ by $t$, that is , $L = \frac{1}{2}at^2$, we can readily obtain the quantities required to calculate the change in kinetic energy $\frac{1}{2}mv^2(t)+\frac{1}{2}I\omega^2(t)=mL\cdot a + I \alpha^2 \cdot \frac{L}{a}$ at all time instants, which expresses the change in kinetic energy. Thus the change in kinetic energy $\frac{7}{8}mgL\sin\theta$ is accurately stated in the OP is obtained. $\textit{Approach via work$-$ kinetic energy analysis:}$ It is important to distinguish between the work $-$ kinetic energy theorem $W_{21}:=\sum \int_{t_1}^{t_2} F_i^{ext}(t) \cdot d\vec{r}_i(t)=\frac{1}{2}\sum m_i (v_i^2(t_2)-v_i^2(t_1)):=\Delta T$ which is identical to [the first law of thermodynamics] (in the case of mechanical systems which are devoid of thermal or other phenomenon) $\Delta U=Q+W$ and the application of the equation which is the integrated expression of the Newton's second law of motion applied to the center of mass of a system of particles (including, for instance, a rigid body). This particular example illustrates this distinction. Let us denote the absolute value of the external torque by $\tau:=fr=constant$, where for the motion being analyzed $f=\frac{1}{5}mg \sin\theta$. The integrated Newton's second law of motion for the translation dynamics of the center of mass of the rigid body being analyzed implies that $mg\sin\theta\cdot L - f\cdot L=\frac{4}{5}mgL\sin\theta=\int_{0}^{t}ma \;dx=\int_{0}^{t}mv \;dv=\frac{1}{2}mv^2(t)$. The integrated Newton's second law of motion for the attitude dynamics of the rigid body being analyzed implies that $\int_{0}^t\tau\;d\theta=fr\cdot \frac{1}{2}\alpha t^2=\frac{1}{2}I\omega^2(t)$. Indeed, the general work$-$kinetic energy theorem stated earlier for the rigid body dynamics can be formulated in the form $\int_{0}^t \vec{F}_{ext}\cdot d\vec{r}+\int_{0}^t\vec{\tau}_{ext}\cdot\;d\vec{\theta} =\frac{1}{2}(m\vec{v}^2(t)-\vec{v}^2(0))+\frac{1}{2}I(\vec{\omega}^2(t)-\vec{\omega}^2(0))$, which is directly applicable to this sub-problem. This more directly applicable form, can in fact be obtained by adding the two previous equations, which are time integrals of equations of motion. $\underline{\textit{Adressing the question in the title of the OP:}}$ The work done by the force of friction applied by the contact surface on which a circular rigid body is rolling with slipping is, in general, non-vanishing. However, in the case that the body is rolling without slipping on the surface, the work done by the friction applied to it vanishes because the velocity of the material particle on which the  friction force acts is identically zero.
$\underline{\text{Indeed, applying the first law of thermodynamics which is the most general }}\\\underline{\text{work $-$ kinetic energy relationship is the best strategy to avoid analytical errors.}}$
Please see this paper by Sherwood on distinguishing the similar but distinct equations related to the physical quantities of work and energy which were discussed above.

$$\underline{\textit{Learning value:}}$$

*

*Analysis of the phenomenon of rolling without slipping requires careful consideration of the direction of the friction force depending on the nature of the applied actuation, that is, a horizontal force at the center of the circular body, a torque at the center of the body or both.

*Using vector quantities, that is writing vector dynamical equations, to obtain the scalar equations of motion is the best strategy to avoid analytical errors when conducting dynamical analysis.

*Applying the first law of thermodynamics which is the most general work $-$ kinetic energy relationship is the best strategy to avoid analytical errors when conducting work $-$ energy analysis.

