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Papessa [141]
3 years ago
13

Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to

the horizontal. As the child slides down the slope, he has an acceleration that has a magnitude of 0.57 m/s2 and that is directed up the slope. What is the coefficient of kinetic friction between the child and the slope?
Physics
1 answer:
Debora [2.8K]3 years ago
5 0

Answer:

\mu_{k} \approx 0.719

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

\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

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a

g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a

\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}

\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )

\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)

\mu_{k} \approx 0.719

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