NLM BYU 31

The block moves on a rough horizontal floor with coefficient of friction $\mu$. The only horizontal force is kinetic friction $f_k = \mu mg$. Thus, the retardation (negative acceleration) is constant: $$a = -\frac{f_k}{m} = -\frac{\mu mg}{m} = -\mu g$$ The observation is that at time $t = \tau$, the block is at distance $s$. There are two distinct possibilities for this observation:

1. The block is still moving at time $\tau$.
2. The block stopped at distance $s$ at some time $t_{stop} < \tau$ and remained there.

Using the second equation of motion $s = ut + \frac{1}{2}at^2$: $$s = u\tau – \frac{1}{2}\mu g \tau^2$$ Solving for $\mu$: $$\frac{1}{2}\mu g \tau^2 = u\tau – s$$ $$\mu = \frac{2(u\tau – s)}{g\tau^2}$$ Condition: The final velocity must be non-negative ($v \ge 0$). $$v = u – \mu g \tau = u – g\tau \left( \frac{2(u\tau – s)}{g\tau^2} \right) = u – \frac{2u\tau – 2s}{\tau} = \frac{2s}{\tau} – u$$ For $v \ge 0$, we need $\frac{2s}{\tau} \ge u \Rightarrow s \ge \frac{u\tau}{2}$. So this solution is valid for $\frac{u\tau}{2} \le s < u\tau$.

If the block stops, it does so at distance $s$. The final velocity is 0. Using $v^2 – u^2 = 2as$: $$0^2 – u^2 = 2(-\mu g)s$$ $$u^2 = 2\mu g s \implies \mu = \frac{u^2}{2gs}$$ Condition: The time to stop $t_{stop}$ must be less than or equal to the observation time $\tau$. $$t_{stop} = \frac{u}{\mu g}$$ Substituting $\mu$: $t_{stop} = \frac{u}{ (\frac{u^2}{2gs}) g } = \frac{2s}{u}$ We need $t_{stop} \le \tau \Rightarrow \frac{2s}{u} \le \tau \Rightarrow s \le \frac{u\tau}{2}$.