An extreme skier, starting from rest, coasts down a mountainthat makes an angle 25.0^\circ with the horizontal. The coefficient of kinetic friction be

Question

An extreme skier, starting from rest, coasts down a mountainthat makes an angle

{25.0}^{\circ}

with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 11.9 m before coming to the edge of a cliff. Without slowing down, she skis offthe cliff and lands down hill at a point whose vertical distance is 4.20 m below the edge. How fast is she going just before she lands?

Answer 1

First find the skier's final speed when she clears the cliff:
Before we can use the equation $v_{f}^{2} = v_{o}^{2} + 2 a d$ we need her acceleration.
If you drew a free body diagram, you should come up with thefollowing equation:
$m g \sin θ - F_{i c t i o n} = m a$
$F_{i c t i o n} = m g \cos θ$
$m g \sin θ - m g \cos θ = m a$
the masses cancel and you have:
$a = g (\sin θ - \cos θ) = 9.81 (\sin (25) - \cos (25) \cdot 0.2) = 2.37 \frac{m}{s^{2}}$
Now use acceleration to find her final velocity before leavingthe cliff:
$v_{o} = 0$
$a = 2.37 \frac{m}{s^{2}}$
$d = 11.9 m$
$v_{f} = \sqrt{(2) (9.81) (11.9)} = 7.51 \frac{m}{s}$
She leaves the cliff at this speed at 25 deg to the horizontal, so
$V_{\otimes} = 7.51 \cdot \cos (25) = 6.80 \frac{m}{s}$
$V_{o} y = 7.51 \cdot \sin (25) = 3.17 \frac{m}{s}$
Now we just need to find $V_{f} x$ and $V_{f} v$
, and then take their magnitude to get the final answer.
$V_{f} x = 6.8 \frac{m}{s}$
$V_{f} y^{2} = V_{o} y^{2} + 2 a d = {3.17}^{2} + 2 (9.81) (4.2) = 92.45 \frac{m}{s}$
$v_{f} = \sqrt{v_{f x}^{2} + v_{f y}^{2}} = \sqrt{{6.8}^{2} + 92.45} \approx 11.8 \frac{m}{s}$

An extreme skier, starting from rest, coasts down a mountainthat makes an angle 25.0^\circ with the horizontal. The coefficient of kinetic friction be

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