Work done by friction when object is rolling WITH slipping So if I have an object that is rolling with slipping like a bowling ball, if I use W=F•dr definition, then work done by friction has to be negative. But if I use the W=τ•dθ definition, then work is positive because friction is causing the rolling motion. So does friction do positive or negative work? And how do I find the total distance or theta that friction works over?
More details on system: the original question is if I have an object on an inclined plane, plot the end kinetic energy as a function of theta of the plane if the objects starts a certain height above the ground each time it is rolled 
See: https://www.aapt.org/physicsteam/2015/upload/exam1-2015-1-8.pdf number 22
(I was also hoping just for a more generalized answer of W=F•dr vs. W=τ•d if possible)
 A: The power of the actions of a system of force on a solid is $P=\overrightarrow{R}\centerdot \overrightarrow{{{v}_{I}}}+\overrightarrow{{{M}_{I}}}\centerdot \overrightarrow{\omega }$ with $I$ a point of the solid.
If $I$ is the point of contact with the ground, $\overrightarrow{{{M}_{I}}}=\overrightarrow{0}$ and your result is correct.
On the other hand, if $I$ is the center of the wheel, you have to sum the two terms.
More precisely :
If $I$ is the center of mass of the wheel, we have $\overrightarrow{R}=T\overrightarrow{{{e}_{x}}}+N\overrightarrow{{{e}_{y}}}$ and $\overrightarrow{\omega }=\omega \overrightarrow{{{e}_{z}}}$. The power of the actions of contact is $P=T{{v}_{G}}-{{M}_{G}}\omega $. The moment in G is ${{M}_{G}}=-R\omega $.
So the power is $P=T({{v}_{G}}+R\omega )=T{{v}_{g}}<0$ with $({{v}_{G}}+R\omega )={{v}_{g}}$ the sliding speed.
In all cases, the power of the friction actions is negative: $P=T{{v}_{G}}-{{M}_{G}}\omega <0$ But the sign of each term depends on the initial situation.
If you put on the ground a wheel that turns with the center of gravity at rest. The friction will move the wheel forward and slow down the rotation. The reaction of the soil will be forward: $T>0$ We will have $T{{v}_{G}}>0$ and $-{{M}_{G}}\omega <0$ 
If you put on the ground a wheel that does not turn with the center of gravity thrown forward. The friction will slow down the wheel and accelerate the rotational movement. The reaction of the soil will be backward: $T<0$. We will have $T{{v}_{G}}<0$ and $-{{M}_{G}}\omega >0$
In any case, the slip will stop when the sliding speed is zero $({{v}_{G}}+R\omega )={{v}_{g}}=0$.
A: You are correct that if the bowling ball is skidding (slipping) down the lane, negative work is done because the friction force is in the opposite direction of the motion of the ball. The dot product between the friction force and displacement is -1.
If the ball is rolling without skidding (slipping) the friction force is static friction. Static friction allows the ball to rotate without slipping. Just like static friction between your shoes and the ground allows you to wall without slipping.  But no dissipative (heat producing) work is involved with static friction.  
Hope this helps.
