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So my issue is best explained with an example.

Let us consider a ring of mass $m$ and radius $R$ sliding down an incline of angle $\theta$ ($I = mR^2$). It starts from a height h, and we wish to evaluate the energy of the block right before it reaches the bottom. Furthermore, let us assume that the coefficient of kinetic friction $\mu < \frac{tan\theta}{2}$, that is, the friction force is not large enough to cause the ring to roll without slipping.

First, from basic equations, $a = g\sin\theta - \mu g\cos\theta$, and $\alpha = \frac{\mu g\cos\theta}{R}$

Then, $v_f = \sqrt{2ah/\sin\theta}$, so $v_f = \sqrt{2gh \frac{\sin\theta - \mu \cos\theta}{\sin\theta}} = \sqrt{2gh(1 - x)}$ where $x = \frac{\mu}{\tan\theta}$

Thus $t_f = \frac{v_f}{a}$, and

$w_f = \alpha t_f = \frac{1}{R}\sqrt{\frac{2ghx^2}{1-x}}$

The final kinetic energy of the ball is $K = \frac{1}{2}mv_f^2 + \frac{1}{2}mR^2 w_f^2$

$= mgh(1 - x) + mgh\frac{x^2}{1 - x}$

However, if we look purely at work done, the work "wasted" due to friction would be $f\cdot \Delta x = \mu mg\,\cos\theta \frac{h}{\sin\theta} = xmgh$ (This step is probably where the discrepancy arises, but I'm wondering WHY this is "wrong")

If we looked at energy lost, then the final energy should just be $mgh(1-x)$, without the rotational energy term! In the calculation of $f \cdot \Delta x$, if we chose $\Delta x$ to be the distance "slid" instead, everything works out nicely, but WHY is this correct? Shouldn't $\Delta W = F \cdot \Delta x$?

I guess what I'm trying to ask here is: How do I know what to use as my $\Delta x$?

Thanks to whoever attempts to answer!!!

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  • $\begingroup$ Your equation for acceleration doesn't account for the rotational motion of the ring - it's just for a sliding object. $\endgroup$ – Señor O Sep 25 '17 at 3:15

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