The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μ r = 0.002. Suppose a 180,000 kg locomotive is rolling at 10 m/s oust over 20 mph) on level rails. If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? How far will the locomotive move during this time?
The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μ r = 0.002. Suppose a 180,000 kg locomotive is rolling at 10 m/s oust over 20 mph) on level rails. If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? How far will the locomotive move during this time?
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 10 m/s oust over 20 mph) on level rails. If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? How far will the locomotive move during this time?
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 10 m/s (just over 20 mph) on level rails. If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? How far will the locomotive move during this time?
A drag chute must be designed to reduce the speed of a 1171 kg dragster from 342 km/hr to 80 km/hr in 4
seconds. Assume that the drag force is proportional to the velocity (kv).
a) What value of the drag coefficient k is needed to accomplish this?
kg · s?
(Leave an exact answer.)
m2
k =
b) How far (in meters) will the dragster travel in the 4-sec interval?
d =
meters. (Round to the nearest meter)
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A box rests on top of a flat bed truck. The box has a mass of m = 20 kg. The coefficient of static friction between the
box and truck is u, = 0.81 and the coefficient of kinetic friction between the box and truck is Hk = 0.62.
1) The truck accelerates from rest to v; = 16 m/s in t = 12 s (which is slow enough that the box will not slide). What is
the acceleration of the box?
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2) In the previous situation, what is the frictional force the truck exerts on the box?
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3) What is the maximum acceleration the truck can have before the box begins to slide?
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4) Now the…
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College Physics: A Strategic Approach (3rd Edition)
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