The total mechanical energy of the block-Earth system changes due to the work done by friction. We use the work-energy theorem, setting the gravitational potential energy reference ($U=0$) at the level of the lower plateau.

The block starts from rest at an initial height of $d + d/2$ above the reference level. The initial mechanical energy is entirely potential:

$$E_i = mg\left(d + \frac{d}{2}\right) = \frac{3}{2}mgd$$

As the block traverses the two plateaus, friction does negative work, dissipating energy. Work done by friction on the first plateau: $W_{f1} = -f_k d = -\mu_k mgd$. Work done by friction on the second plateau: $W_{f2} = -f_k \frac{d}{2} = -\frac{1}{2}\mu_k mgd$. The total work done by non-conservative forces is $W_{nc} = W_{f1} + W_{f2} = -\frac{3}{2}\mu_k mgd$.

The block's final state is momentarily at rest ($K_f = 0$) at a height $H$ on the last ramp. The final mechanical energy is:

$$E_f = mgH$$

Applying the work-energy theorem, $\Delta E = E_f - E_i = W_{nc}$:

$$mgH - \frac{3}{2}mgd = -\frac{3}{2}\mu_k mgd$$ $$mgH = \frac{3}{2}mgd - \frac{3}{2}\mu_k mgd = \frac{3}{2}mgd(1 - \mu_k)$$ $$H = \frac{3}{2}d(1 - \mu_k)$$

Substituting the given values:

$$H = \frac{3}{2}(0.40 \text{ m})(1 - 0.50) = \frac{3}{2}(0.40 \text{ m})(0.50) = 0.30 \text{ m}$$

Since $H>0$, the block does not stop on either plateau and reaches the final ramp.