Missing questions in the text:
"determine (a) the distance traveledby the train before it comes to a stop (b) the coupling force between the cars as the is slowing down."
Note: The masses of the car are in Mg, not mg.
Solution:
(a) To find the solution of this part, we should write that the resultant of the forces acting on the x-axis (horizontal plane) is equal to the product between the total mass of the system (the mass of the two cars) and the acceleration (Second Newton law).
The only force acting on the light train system is the force of the brakes, Fb=40 kN = 40000 N, multiplied by 2 because it acts on both cars. The masses of the two cars are [tex]m_A=30 Mg=30000 Kg[/tex] and [tex]m_B = 20 Mg=20000 Kg[/tex], therefore Newton's law becomes
[tex](m_A+m_B) a = -2F_b[/tex]
where the negative sign in front of Fb means the force is acting against the direction of the motion. Solving for a, we find
[tex]a= \frac{-2 F_b}{m_A+m_B}= \frac{-2 \cdot 40000 N}{50000 Kg}=-1.6 m/s^2 [/tex]
where the negative sign means the light train is decelerating.
Once we find a, we can find the total distance covered by the train. We know the initial velocity, vi=90 km/h=25 m/s, the final velocity (0 m/s) and the acceleration, so we can use:
[tex]2aS=v_f^2-v_i^2=-v_i^2[/tex]
From which we find
[tex]S= \frac{v_i^2}{2a}=195.3 m [/tex]
(b) To solve this part of the problem, let's write Second Newton's law with the forces acting only on the car A. We have only two forces: the force of the brake, Fb, and the coupling force, Fc. So we can write
[tex]m_A a = F_c-F_b[/tex]
And so
[tex]F_c =m_A a+F_b=30000 Kg \cdot (-1.6 m/s^2)+40000 N = -8000 N[/tex]
Where once again, the negative sign means that Fc is against the direction of the motion.
