Answer:
[tex]\mu_{k} \approx 0.719[/tex]
Explanation:
The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)
[tex]\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0[/tex]
After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:
[tex]m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a[/tex]
[tex]g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a[/tex]
[tex]\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}[/tex]
[tex]\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )[/tex]
[tex]\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)[/tex]
[tex]\mu_{k} \approx 0.719[/tex]
