Answer:
a) [tex]d=236.280\,m[/tex], b) [tex]F_{coupling} = -8848\,N[/tex] The real force has the opposite direction.
Explanation:
a) Let assume that train moves on the horizontal ground. An equation for the distance travelled by the train is modelled after the Principle of Energy Conservation and Work-Energy Theorem:
[tex]K_{A} = W_{brake}[/tex]
[tex]\frac{1}{2}\cdot m_{train} \cdot v^{2} = F_{brakes}\cdot d[/tex]
[tex]d = \frac{m_{train}\cdot v^{2}}{2\cdot F_{brakes}}[/tex]
[tex]d = \frac{(51000\,kg)\cdot [(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (82000\,N)}[/tex]
[tex]d=194.360\,m[/tex]
b) The acceleration experimented by both trains are:
[tex]a = -\frac{v_{o}^{2}}{2\cdot d}[/tex]
[tex]a = -\frac{[(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s})]^{2}}{2\cdot (194.360\,m)}[/tex]
[tex]a = -1.608\,\frac{m}{s^{2}}[/tex]
The coupling force in the car A can derived of the following equation of equilibrium:
[tex]\Sigma F = F_{coupling} - F_{brakes} = m_{A}\cdot a[/tex]
The coupling force between cars is:
[tex]F_{coupling} = m_{A}\cdot a + F_{brakes}[/tex]
[tex]F_{coupling} = (31000\,kg)\cdot(-1.608\,\frac{m}{s^{2}} )+41000\,N[/tex]
[tex]F_{coupling} = -8848\,N[/tex]
The real force has the opposite direction.
