Answer:
[tex]m_{c} = 6768\,kg[/tex]
Explanation:
According to the Principle of Energy Conservation and the Work-Energy Theorem, the system is modelled as follows:
[tex]K_{o} = K_{f} + W_{loss}[/tex], where [tex]\frac{K_{f}}{K_{o}} = 0.78[/tex].
Then,
[tex]K_{f} = 0.78\cdot K_{o}[/tex]
[tex]0.5\cdot (m_{f}+m_{c})\cdot v_{f}^{2} = 0.39\cdot m_{f}\cdot v_{o}^{2}[/tex]
Besides, the Principle of Momentum Conservation describes the following model:
[tex]m_{f}\cdot v_{o} = (m_{f}+m_{c})\cdot v_{f}[/tex]
The final velocity of the system is:
[tex]v_{f} = \frac{m_{f}}{m_{f}+m_{c}}\cdot v_{o}[/tex]
After substituting in the energy expression:
[tex]0.5\cdot \frac{m_{f}^{2}}{m_{f}+m_{c}}\cdot v_{o}^{2} = 0.39\cdot m_{f}\cdot v_{o}^{2}[/tex]
[tex]0.5\cdot m_{f} = 0.39\cdot (m_{f}+m_{c})[/tex]
The mass of the caboose is:
[tex]0.39\cdot m_{c} = 0.11\cdot m_{f}[/tex]
[tex]m_{c} = 0.282\cdot m_{f}[/tex]
[tex]m_{c} = 0.282\cdot (24000\,kg)[/tex]
[tex]m_{c} = 6768\,kg[/tex]
