Evaluate the speed of train:
A train of mass 200KN has frictional resistance of 5N per KN. Speed of train, at the top of an inclined of 1 in 80 is 45 Km/hr. Find speed of train after running down the incline for 1Km.
Sol: Given,
Mass m = 200KN, Frictional resistance F_{r} = 5N/KN, sin = 1/80 = 0.0125,
Initial velocity u = 45Km/hr = 12.5m/sec, s = 1km = 1000m
The total frictional resistance = 5 X 200 = 1000N = 1KN ...(i)
Force responsible for sliding = Wsinθ = 200 X 0.0125 = 2.5KN
Net force, F = F - F_{r} = ma
2.5 - 1 = (200/9.81)a
a = 0.0735 m/sec^{2} ...(ii)
Apply equation, v^{2} = u^{2} + 2as
v^{2} = 0 + 2 X 0.0735 X 1000
v = 12.1 m/sec .......ANS