91Ó°ÊÓ

The length of the flat part is L = 40 cm .

The coefficient of kinetic friction is${\mathrm{μ}}_k=0.20$ .

The height from which the particle is released is$h=\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ .

By using conservation of energy and equation for thermal energy generated $\mathbf{∆}{\mathit{E}}_{\mathbf{t}\mathbf{h}}$we can find the thermal energy at its first pass and second pass. After that by generalizing it for${\mathit{n}}^{\mathbf{t}\mathbf{h}}$pass and puttingheight h from which the particle is released, we can find thedistancedfrom the left edge of the flat part when the particle finally stops.

Formula:

The energy conversation is,$W=∆E_{mec}+∆E_{th}$

The thermal energy generated is, $∆E_{th}=f_{k}d$

The initial and final kinetic energies are zero, and we set up energy conversation in the form of

$$\begin{gathered} W=∆E_{mec}+∆E_{th} \\ W=0 \end{gathered}$$

If it occurs during its first pass through, then the thermal energy generated is

$∆E_{th}=f_{k}d$

Here,$d≤L$

$f_k={\mathrm{μ}}_{k}mg$.

If it occurs during its second pass through, then the total thermal energy is,

$∆E_{th}={\mathrm{μ}}_{k}mg\left(L+d\right)$

Generalizing to thepass through and solve:

$∆E_{th}={\mathrm{μ}}_{k}mg\left[\left(n-1\right)L+d\right]$

Therefore, we have,

$mgh={\mathrm{μ}}_{k}mg\left[\left(n-1\right)L+d\right]$

But,$h=\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ , we get,

role="math" localid="1661402112439" $\frac{d}{L}=1+\frac{1}{2{\mathrm{μ}}_k}-n$

The first two terms give

$\frac{d}{L}=1+\frac{1}{2{\mathrm{μ}}_k}=1+\frac{1}{2×0.20}=3.5$

So that,

The requirement$0≤\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$L≤1$}\right.$demands that n = 3 . We arrive at

$$\begin{gathered} \frac{d}{L}=\frac{1}{2} \\ d=\frac{1}{2}L \\ =\frac{1}{2}\left(40cm\right) \\ =20cm \end{gathered}$$

And that this occurs on its third pass through the flat region.

The distance d from the left edge of the flat part when the particle finally stops is

d = 20 cm

= 0.20 m