mountain angle and kinetic energy theorem

**rcmango** · November 9th 2007, 09:27 AM

An extreme skier, starting from rest, coasts down a mountain that makes an angle 25.0° with the horizontal.

The coefficient of kinetic friction between her skis and the snow is 0.200.
She coasts for a distance of 12.3 m before coming to the edge of a cliff.

Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.70 m below the edge.

How fast is she going just before she lands?
in m/s

I believe I need to use the kinetic energy theorem.
i'm not good with the friction formulas.

**earboth** · November 10th 2007, 07:48 AM

Originally Posted by rcmango

An extreme skier, starting from rest, coasts down a mountain that makes an angle 25.0° with the horizontal.

The coefficient of kinetic friction between her skis and the snow is 0.200.
She coasts for a distance of 12.3 m before coming to the edge of a cliff.

Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.70 m below the edge.

How fast is she going just before she lands?
in m/s

I believe I need to use the kinetic energy theorem.
i'm not good with the friction formulas.

Hello,

the friction is a force which has an effect opposite to the force which causes a motion (the motion can be zero!). The value of the friction depends on the normal force to the surface on which a solid is moving.
With your problem the weight of the skier causes a normal force to the snow and an accelerating force. Now the friction decreases this accelerating force. (see attachment)

Let $w$ be the weight of the skier
$F_N$ be the normal force
$F_a$ the accelrating force and
$F_f$ the friction

Then you have:

$w = 9.81 \cdot m$
$F_n = w \cdot \cos(25^\circ)$
$\frac{F_f}{F_N} = 0.2~\implies~F_f=0.2 \cdot f_N = 0.2 \cdot w \cdot \cos(25^\circ)$
$F_a=w \cdot \sin(25^\circ) - 0.2 \cdot w \cdot \cos(25^\circ) = w(\sin(25^\circ)-0.2 \cdot \cos(25^\circ)$

$F=m \cdot a$ . That means: $w(\sin(25^\circ)-0.2 \cdot \cos(25^\circ) = m \cdot a$

$9.81 \cdot m \cdot (\sin(25^\circ)-0.2 \cdot \cos(25^\circ)) = m \cdot a~\implies~$ $a = 9.81 \cdot (\sin(25^\circ)-0.2 \cdot \cos(25^\circ)) \approx 2.3677\ \frac m{s^2}$

With a uniformly accelrated movement you have:

$d = \frac12 \cdot a \cdot t^2$ . You know the distance already and the accelration:

$12.3 = \frac12 \cdot 2.3677 \cdot t^2 ~\implies~ t \approx 3.2233\ s$

and $v = a \cdot t$ . That means $v = 2.3677 \cdot 3.2233 \approx 7.3168\ \frac ms$ when the skier arrived at the edge.

Now the horizontal speed $v_h = v \cdot \cos(25^\circ)$ doesn't change but the vertical speed increases because the skier is falling down:

Use $d = v \cdot t$ to calculate the time the skier is flying through the air:

$d = v \cdot t \cdot (-\sin(25^\circ)-\frac12 \cdot 9.81 \cdot t^2$ . Plug in the values you know:

$-3.7 = 2.3677 \cdot t \cdot (-\sin(25^\circ)-\frac12 \cdot 9.81 \cdot t^2$ solve for t. I've got $t \approx 0.7725 \ s$

Therefore the final speed is
$v = v_v + 9.81 \cdot 0.7725~\implies~v=7.3168 \cdot \sin(25^\circ) + 9.81 \cdot 0.7725 \approx 10.67\ \frac ms$ . Approximately 38.4 km/h

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