Find distance of which wagon will travel before coming rest:
A train of wagons is first pulled on level track from A to B and then up a 5 percent upgrade as shown in the figure given below. At some point C, least wagon gets detached from the train, when it was traveling with the velocity of 36Km.p.h. If detached wagon is having mass of 5KN and the track resistance is 10N per KN, find distance through which wagon will travel before coming to rest. Take g = 9.8m/sec^{2}.
Sol: Given, Grade = 5% or sinθ = 5% = 0.05, u = 36Km.p.h. = 10m/sec, W = 5KN, V = 0, F_{r } = 10N/KN
Let s = Distance traveled by the wagon before coming to rest
Total track resistance F_{r} = 10 X 5 = 50N ...(i)
Resistance because of upgrade = msin θ = 5 X 0.05 = 0.25KN =250N ...(ii)
Total resistance to wagon = Net force = 50 + 250 = 300N
However, F = ma, 300 = (5000/9.81)a
a = 0.588m/sec^{2} ...(iii)
Apply equation, v^{2} = u^{2} - 2as
0 = (10)^{2} - 2 X 0.588 X s
s = 85 m .......ANS