**Given that, **

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 wagon before coming to rest

Total track resistance F_{r} = 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/sec^{2} ...(iii)

Apply the equation, v^{2} = u^{2} - 2as 0 = (10)^{2} – 2 X 0.588 X s

**s = 85 m.**