# Finding force of train cars

1. Sep 9, 2009

### chenying

1. The problem statement, all variables and given/known data
A locomotive accelerates a 25-car train along a level track. Each car has a mass of 6.8e104 kg and is suject to a drag force f = 250v, where v is the speed in m/s and the force f is in N. At the instant when the speed of the train is 44 km/h, the acceleration is 0.20 m/s2.

(a) What is the force exerted by the locomotive on the first car?

(b) Suppose that the force found in part (a) is the greatest force the locomotive can exert of the train. What, then, is the steepest grade up which the locomotive can pull the train at 44 km/h?

2. Relevant equations

f=ma

3. The attempt at a solution

no attempt; no idea how to approach problem.

2. Sep 9, 2009

### rock.freak667

Okay. So ma is the resultant force on the train.

Force exerted by train (F) - friction (250v) = Resultant force

Now read the part in bold and see if you can't find F

EDIT: for the second part, they are asking for an angle θ for which the train can produce a constant velocity of 44km/h. Draw the train on an inclined plane at an angle θ. Don't forget to take the weight of train into account.

Last edited: Sep 9, 2009
3. Sep 9, 2009

### chenying

ok, but im confused with whether the force of the train is equal to the force of the train on one car?

4. Sep 10, 2009

### rock.freak667

You don't need to bother about the force on one car. The train is made up of 25 cars, the mass of the train is 25 * mass of one car.

5. Sep 10, 2009

### chenying

So then its basically this:

25 trains * 6.8e4 kg/train * .20 m/s^2 - 250 * 12.22222222m/s = ma

336944.444 = ma

is this the answer? I've entered this into webassign and it says its wrong.

6. Sep 10, 2009

### dvs0826

25 cars means 25 times the drag force too, right?