Newton's laws

  • Thread starter EndoBendo
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  • #1
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Homework Statement



A locomotive accelerates a 24-car train along a level track. Every car has a mass
of 50000 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 28 km/h, the magnitude of its acceleration is 0.3 m/s^2. (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 28 km/h?


The Attempt at a Solution



F-25(250v) = 25(50000)a
v=7.778
F = 24(50000)(0.3) + 24(250*7.778)
F= 360000 + 46,668
F = 406.7 kN

T = 406.7N x 10^3
T= mg*cos(theta)
406.7x10^3/24(50000)(9.81) = cos(theta)
theta = 1.98


am i correct?? second part was a mystery, i just tried so many things till the answer seemed right.
 

Answers and Replies

  • #2
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F-25(250v) = 25(50000)a
v=7.778
F = 24(50000)(0.3) + 24(250*7.778)
F= 360000 + 46,668
F = 406.7 kN
Not sure where or why you're getting your value for v, it gives you the speed of the train in the question.

Also, are you familiar with motion on an inclined plane? That would be the starting point of your second question I believe.
 
  • #3
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v must be in m/s to get to newtons
i included part 2 , i got 1.98. is that right??
 

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