7:00 PM 40% E spock.physast.uga.edu C Course Contents » Problem Set 9» (8) oo T-

Question

7:00 PM 40% E spock.physast.uga.edu C Course Contents » Problem Set 9» (8) oo T-Mobile : Timer Notes S Evaluate Feedback PrintInfo A 69.9 kg skier encounters a dip in the snow's surface that has a circular cross section with radius of curvature of = 12.4 m. If the skier's speed at point A in the figure below is 8.04 m/s, what is the normal force exerted by the snow on the skier at point B? As shown,= 1.87 m. Ignore frictional forces. 1060.kN Submit AnswerIncorrect. Tries 5/8 Previous Tries Post Discussion Send Feedback

Explanation / Answer

Given,

m = 69.9 kg ; r = 12.4 m ; v1 = 8.04 m/s ; h = 1.87 m

The skier has kinetic energy at A

KE1 = 1/2 m v1^2

Let v2 be the speed at B,

KE2 = 1/2 m v2^2

From the conservation of energy,

KE2 - KE1 - PE = 0

1/2 m v2^2 = 1/2 m v1^2 + m g h

1/2 v2^2 = 0.5 x 8.04^2 + 9.8 x 1.87

1/2 v2^2 = 50.65 =>v2 = 10.06 m/s

The centripital force at B is:

F = mv^2/r also F = ma

a = v^2/r = 10.06^2/12.4 = 8.16 m/s^2

The net force will be:

ma = F - mg

F = m (a + g)

F = 69.9 (8.16 + 9.81) = 1256.103 N

Hence, F = 1256.103 N

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