Explanation & answer:
Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.
Initially, the cube has a kinetic energy of
KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2
At the highest point when the block stops, the gain in potential energy is
PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2
By assumption, there was no loss in energies, we equate PE = KE
322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2
=>
h = 2000 /322 = 6.211 (ft)
distance up incline = h / sin(30) = 12.4 ft
