Answer:
[tex]v \approx 9.312\,\frac{m}{s}[/tex]
Explanation:
The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:
[tex]K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}[/tex]
[tex]K_{B} = K_{A} + U_{g,A}-U_{g,B} - W_{loss}[/tex]
[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot s\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cdot s \cos \theta[/tex]
[tex]\frac{1}{2}\cdot v^{2} = g\cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)[/tex]
[tex]v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}[/tex]
[tex]v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10\,m)\cdot (\sin 37^{\textdegree} - 0.2\cdot \cos 37^{\textdegree})}[/tex]
[tex]v \approx 9.312\,\frac{m}{s}[/tex]
