Question

A train of mass 200KN has a frictional resistance of 5N per KN. Speed of the train, at the top of an inclined of 1 in 80 is 45 Km/hr. Find the speed of the train after running down the incline for 1Km.

Answer 1

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/sec² ...(ii)

Apply the equation, v² = u² + 2as

v² = 0 + 2 X 0.0735 X 1000

v = 12.1 m/sec