Given data, Mass m = 200KN, Frictional resistance F_{r} = 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 – F_{r } = ma

2.5 – 1 = (200/9.81)a

a = 0.0735m/sec^{2 }...(ii)

Apply the equation, v^{2} = u^{2} + 2as

v^{2} = 0 + 2 X 0.0735 X 1000

**v = 12.1 m/sec**