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Inessa [10]
3 years ago
13

If santa slides 5.00 m before reaching the edge, what is his speed as he leaves the roof?

Physics
1 answer:
AveGali [126]3 years ago
8 0

Even with no friction, it depends on the slope of the roof. That is, it depends on how much elevation (altitude) he loses during the slide.

Whatever that number is ... call it 'h' ... Santa's speed when he reaches the edge is

Square root of (19.6h) meters per second.

It doesn't matter how much he weighs, or how far he has slud. Only how much altitude he lost on the slope while sliding.

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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
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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)

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