A skier of mass 60.0kg starts from rest at the top of a ski slope of height 61.0

Question

A skier of mass 60.0kg starts from rest at the top of a ski slope of height 61.0m.

a) If frictional forces do -1.09 x 10^4 J of work on her as she descends, how fast is she going at the bottom of the slope?
Take free fall acceleration to be g=9.80m/s^2.

b) Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is uk=0.150. If the patch is of width 67.0m and the average force of air resistance on the skier is 160N, how fast is she going after crossing the patch?

c) After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.60minto it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Explanation / Answer

b) so work is defined to be the amount of force (f) times the distance (d), so work = Fd. In this problem, it tells you, there is a force of air resistance of 160 N. We now need to find the frictional force. We know that frictional force (Ff) is = to coefficient of friction (uk) times normal force (force perpendicular to surface, Fn), so Ff=uk*Fn. We need to find the normal force. The normal force is equal to the weight (mg) of the skier because mg force is acting down and Fn is acting up (perpendicular to surface) and since the skier is not randomly flying off up the surface or sinking into the ground, you know Fn = mg. Thus, you get mg to = (60kg*9.8m/s^2)=. Now you can use this to find your frictional force since you're given the coefficient => .15 (588 kg*m/s^2) = 88.2 N. Now to find the work, you want total force, so air resistance + frictional (88.2N + 160N = 248.2N). She travels a distance of 67 m so work = 248.2N * 67m = 16629.4 J. Now you know the skier loses this much energy when going across that patch of snow. You found before that she is going 28.8 m/s at the bottom. Now using the energy equation (1/2mv^2

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