Given that,
Grade = 5% or sinθ = 5% = 0.05, u = 36Km.p.h. = 10m/sec,
W = 5KN, V = 0, Fr = 10N/KN
Let s = Distance traveled by wagon before coming to rest
Total track resistance Fr = 10 X 5 = 50N ...(i)
Resistance due to upgrade = msinθ = 5 X 0.05
= 0.25KN = 250N ...(ii)
Total resistance to wagon = Net force = 50 + 250 = 300N But,
F = ma, 300 = (5000/9.81)a a = 0.588m/sec2 ...(iii)
Apply the equation, v2 = u2 - 2as 0 = (10)2 – 2 X 0.588 X s
s = 85 m.