A 63 kg skier encounters a dip in the snow's surface that has a circular cross section with a radius of curvature of 12 m. If the skier's speed at point A in Figure 825 is 7.9 m/s, what is the normal force exerted by the snow on the skier at point B? Ignore frictional forces.
radius=12
h of dip=1.75m
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1 answer

We don't have access to your Figure 825. The normal force extered by the snow on the skier must equal the centripetal force MV^2/R , plus the component of the weight (M g) in the normal direction. Use conservation of energy to get V at point B.
R = 12 m 👍
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