Question

A light train made up of two cars is traveling at 90 km/h when the brakes are applied to both cars. Know that car A has a mass of 31 Mg and car B has a mass of 20 Mg, and the braking force is 41 kN on each car. 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.

Answer 1

Answer:

a) $d=236.280\,m$, b) $F_{coupling} = -8848\,N$ 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:

$K_{A} = W_{brake}$

$\frac{1}{2}\cdot m_{train} \cdot v^{2} = F_{brakes}\cdot d$

$d = \frac{m_{train}\cdot v^{2}}{2\cdot F_{brakes}}$

$d = \frac{(51000\,kg)\cdot [(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (82000\,N)}$

$d=194.360\,m$

b) The acceleration experimented by both trains are:

$a = -\frac{v_{o}^{2}}{2\cdot d}$

$a = -\frac{[(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s})]^{2}}{2\cdot (194.360\,m)}$

$a = -1.608\,\frac{m}{s^{2}}$

The coupling force in the car A can derived of the following equation of equilibrium:

$\Sigma F = F_{coupling} - F_{brakes} = m_{A}\cdot a$

The coupling force between cars is:

$F_{coupling} = m_{A}\cdot a + F_{brakes}$

$F_{coupling} = (31000\,kg)\cdot(-1.608\,\frac{m}{s^{2}} )+41000\,N$

$F_{coupling} = -8848\,N$

The real force has the opposite direction.