Given data, Mass m = 200KN, Frictional resistance Fr = 5N/KN, sinθ = 1/80 = 0.0125,
Initial velocity u = 45Km/hr = 12.5m/sec, s = 1km = 1000m
Total frictional resistance = 5 X 200 = 1000N = 1KN ...(i)
Force responsible for sliding = Wsinθ = 200 X 0.0125 = 2.5KN
Now, Net force, F = F – Fr = ma
2.5 – 1 = (200/9.81)a
a = 0.0735m/sec2 ...(ii)
Apply the equation, v2 = u2 + 2as
v2 = 0 + 2 X 0.0735 X 1000
v = 12.1 m/sec