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Thread: mountain angle and kinetic energy theorem

  1. #1
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    mountain angle and kinetic energy theorem

    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.
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  2. #2
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    Quote Originally Posted by rcmango View Post
    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 $\displaystyle w$ be the weight of the skier
    $\displaystyle F_N$ be the normal force
    $\displaystyle F_a$ the accelrating force and
    $\displaystyle F_f$ the friction

    Then you have:

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

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

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

    With a uniformly accelrated movement you have:

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

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

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

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

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

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

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

    Therefore the final speed is
    $\displaystyle v = v_v + 9.81 \cdot 0.7725~\implies~v=7.3168 \cdot \sin(25^\circ) + 9.81 \cdot 0.7725 \approx 10.67\ \frac ms$. Approximately 38.4 km/h
    Attached Thumbnails Attached Thumbnails mountain angle and kinetic energy theorem-ski_schussfahrt.gif  
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