I'm given the following problem:
Now, the following is my attempt at a solution:
I have two problems: (1) With my equation for $v$, I end up having to take the log of $0$, which is obviously undefined, so something's gone wrong here.
(2) Secondly, I 'establish' that $ W \neq \Delta T $ (i.e. the work done is not the change in kinetic energy, contrary to the Work-Energy principle.
(Also, the reason I've printscreened my answer is that it would take a very long time for my to type in $\LaTeX$).
Could someone help me see the light? Thanks in advance.
It was interesting to follow your approach through to the end.
As a starting point, fix the first integral so it says $$\int_0^x{\left(-{\mu R\over m}\right)\,d\xi} = \int_{v_0}^v{\eta\,d\eta}\,.$$ Normally we'd save on the algebra by abbreviating ${\mu R\over m} \equiv a$. But since you're doing an unusual approach, I'll carry around the symbols so we can see how energy and kinematics relate.
Performing the integration, we have $$-\left({\mu R\over m}\right)x={1\over 2}(v^2-v_0^2)$$ or $$v = \sqrt{v_0^2-{2\mu R\over m}x}\,.$$ From this we find the stopping distance by setting $v = 0$ and solving for $x$; call the result $d$: $$d = {mv_0^2\over 2\mu R}\,.$$
This also shows that the work done by friction is equal to the change in kinetic energy; for the work done by friction is $W = \int{{\bf F}\cdot d{\bf s}} = -\mu R\int{ds} = -\mu R d$, and using the result for $d$ above we have $W = -{1\over 2} mv_0^2$.
Next let's find the trajectory $x(t)$. We have $dt = {dx\over v}$ so that $$\int_0^x{d\xi\over\sqrt{v_0^2-{2\mu R\over m}\xi}} = \int_0^t{dt^\prime}\,.$$
Substitute $u = v_0^2 - {2\mu R\over m}\xi$, turn the crank, and the equation becomes $${m\over 2\mu R}\int_{v_0^2-{2\mu R\over m}x}^{v_0^2}{du\over\sqrt{u}} = t\,.$$
The integration is trivial, so we find $$v_0 - \sqrt{v_0^2-{2\mu R\over m}x} = {\mu R\over m}t$$ which can be solved for $x$ to yield $$x = v_0 t - {1\over 2}\left({\mu R\over m}\right)t^2\,.$$ The horizontal velocity is $dx/dt$, or $$v_x = v_0 - \left({\mu R\over m}\right) t\,.$$ Solve for the time $t_\star$ when $v_x = 0$: $$t_\star = {mv_0\over \mu R}\,.$$
Note that $t_\star$ is the initial momentum divided by the magnitude of the stopping force.
This is by no means the way I would have solved the problem myself, but as I say it was interesting! I think that a person who is just learning energy methods would benefit from doing it this way so that each step could be compared to the familiar kinematical approach.