A small block is released from rest from a height of 20 m on a...

By the principle of conservation of mechanical energy, the kinetic energy of the block at the lowest point before entering the rough patch is equal to its initial potential energy.

$K_1 = mgh_1$

The block loses $40\%$ of its kinetic energy on the rough horizontal patch. The remaining kinetic energy is:

$K_2 = K_1 - 0.40 K_1 = 0.60 K_1$

$K_2 = 0.60 mgh_1$

As the block moves up the second frictionless track, its kinetic energy is converted into potential energy. At the maximum height $h_2$, the potential energy is equal to $K_2$.

$mgh_2 = 0.60 mgh_1$

$h_2 = 0.60 h_1$

Substituting $h_1 = 20\text{ m}$:

$h_2 = 0.60 \times 20 = 12\text{ m}$

Answer: $12\text{ m}$