The frictional force will be:
Ff = 160,000 x 9.8 x 0.002 = 3136 N
acceleration would be:
a = Ff/m = 3136/160000 = 0.0196 m/s^2
we know that, a = v/t => t = v/a
t = 12/0.0196 = 612.2 sec
The distance travelled will be:
D = Vavg x t = 12/2 x 612.2 = 3673.5 m
Hence, t = 612.2 s and D = 3673.5 m.
The rolling resistance of steel is typically mu_r = .002. If a train weighing 160,000 kg...
The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically ? r =0.002 . Suppose a 180,000 kg locomotive is rolling at 25 m/s on level rails. Part A If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? Express your answer using two significant figures. Part B How far will the locomotive move during this time? Express your answer to two significant figures...
B) A 1300 -kg car is pushing an out-of-gear 2200 -kgtruck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4500 N . The rolling friction of the car can be neglected, but the heavier truck has a rolling friction of 760 N , including the "friction" of turning the truck's drivetrain. What is the acceleration aT of the truck? C) An...
The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μr=0.002. Suppose a 180,000 kg locomotive is rolling at 12 m/son level rails. You may want to review (Pages 138 - 142) . Part A If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? Express your answer using two significant figures. Δt = s SubmitRequest Answer Part B How far will the locomotive...