A 30-ton engine exerts a constant horizontal force of 40(10^3)-lb on a train hav

Question

A 30-ton engine exerts a constant horizontal force of 40(10^3)-lb on a train having three cars that have a total weight of 250-tons. If the rolling resistance is 10-lb per ton for both the engine and cars, Determine how long it takes to increase the speed of the train from 20 ft/s to 30 ft/s. What is the driving force exerted by the engine wheels on the tracks? A 30-ton engine exerts a constant horizontal force of 40(10^3)-lb on a train having three cars that have a total weight of 250-tons. If the rolling resistance is 10-lb per ton for both the engine and cars, Determine how long it takes to increase the speed of the train from 20 ft/s to 30 ft/s. What is the driving force exerted by the engine wheels on the tracks?

Explanation / Answer

Force exerted by engine = F_E = 40000 lb

Total mass = 250+30 = 280 ton = 280*2000= 627200 lb

Friction Force = f = 10*(250+30) = 28000 lb

Net force = 40000-28000=12000lb

acceleration = 12000*32.2/627200 ft/s^2= 0.62 ft/s^2

time taken, t = (30-20)/0.62= 16.12 sec

Drving force on tracks = rolling friction = 28000 lb

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