Answer
$$
H=30 \mathrm{cm}
$$
measured from the lowest plateau.
Work Step by Step
$$
K=m g d-\mu_{k} m g d=\frac{1}{2} m g d
$$
In its descent to the lowest plateau, it gains $m g d / 2$ more kinetic energy, but as it slides
across it "loses" $\mu_{k} m g d / 2$ of it. Therefore, as it starts its climb up the right ramp, it has
kinetic energy equal to
$$
K=\frac{1}{2} m g d+\frac{1}{2} m g d-\frac{1}{2} \mu_{k} m g d=\frac{3}{4} m g d
$$
Setting this equal to Eq. $8-9$ (to find the height to which it climbs) we get $H=3 / 4$ . Thus, block (for an instant) stops on the inclined ramp at right, at a height of
$$
H=0.75 d=0.75(40 \mathrm{cm})=30 \mathrm{cm}
$$
measured from the lowest plateau.
