The particle's initial mechanical energy is entirely potential, $E_i = mgh$. The particle will come to rest on the flat part, so its final mechanical energy is $E_f = 0$. The change in mechanical energy is due to the work done by friction, $W_f$.

Let $d_{total}$ be the total distance the particle slides on the flat part. The work done by friction is $W_f = -f_k d_{total} = -\mu_k N d_{total} = -\mu_k mg d_{total}$.

Using the work-energy theorem, $\Delta E = E_f - E_i = W_f$:

$$0 - mgh = -\mu_k mg d_{total}$$ $$d_{total} = \frac{h}{\mu_k}$$

Given $h = L/2$, the total distance traveled on the flat part is:

$$d_{total} = \frac{L/2}{\mu_k} = \frac{L}{2\mu_k}$$ $$d_{total} = \frac{40 \text{ cm}}{2(0.20)} = \frac{40 \text{ cm}}{0.40} = 100 \text{ cm}$$

The particle slides across the flat part of length $L = 40$ cm.

1. First pass (left to right): travels 40 cm.
2. Second pass (right to left): travels 40 cm. Total distance is $40 + 40 = 80$ cm.
3. Third pass (left to right): travels the remaining distance $100 - 80 = 20$ cm.

The particle stops 20 cm from the left edge of the flat part.