A locomotive accelerates a 25-car train a long a level track. Every car has a ma

Question

A locomotive accelerates a 25-car train a long a level track. Every car has a mass of 5 x10^4 kg and is subject to a frictional force f=250 v, where the speed v is in meters per second and the force f is in Newtons. At the instant when the speed of the train is 30 km/h, the magnitude of its acceleration is 0.2 m/s2. (a) What is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at 30 km/h?

Explanation / Answer

First off a)
Every car has a frictional force 250v. So the fictional force for all cars will be 25x250v=6250v. Newton says that every force must have to have an equal and opposite force if the objects are not in motion (they are not in motion w/r to each other, ex. the locomotive is not moving away from the cars, there attached). This equal and opposite force is tension, so tension is 6250v (could make this number negitive but why complicate it with sign notation?). So tension is 6250v=6250(30km/h)(1h/60min)(1min/60sec)(1000m/1km)=6250(8.33333)=52083Newtons (the extra bits are for conversion from km/h to m/s).

second off b)

You now the force of friction is 52083N, now i guess you could go the way of using the mass they gave you and find the force of gravity pushing something down a slope and soforth. However they give you an easy way out, they already give you the acceleration the locamotive can move the train at. So plug this in using arctan (or tan to the power of negitive 1) and us acceleration of gravity as the other acceleration and your done. arctan(0.2/9.8)=1.17degrees

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