# A train of wagons is first pulled on a level track from A to B and then up a 5% upgrade as shown in fig.

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A train of wagons is first pulled on a level track from A to B and then up a 5% upgrade as shown in fig. At some point C, the least wagon gets detached from the train, when it was traveling with a velocity of 36Km.p.h. If the detached wagon has a mass of 5KN and the track resistance is 10N per KN, find the distance through which the wagon will travel before coming to rest. Take g = 9.8m/sec2. +1 vote
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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.