The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that
mgh = 1/2 mv²
where v is its speed at the bottom of the incline. It follows that
v = √(2gh)
If the incline is 20.4 m long, that means the block has a starting height of
sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m
and so the block attains a speed of
v = √(2gh) ≈ 10.1728 m/s
The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that
Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F
By Newton's second law, the net vertical force on the block as it slides is
∑ F [vertical] = n - mg = 0
where n is the magnitude of the normal force, so that
n = mg = (25 kg) g = 245 N
and thus the magnitude of friction is
F = -0.16 (245 N) = -39.2 N
(negative since it opposes the block's motion)
Then the block slides a distance of
d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m
