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/sec²                                                                                                                                     ...(ii)

Apply equation, v² = u²  + 2as

v² = 0 + 2 X 0.0735 X 1000

v = 12.1 m/sec                                                                 .......ANS