Constant friction?

Problem A skier starts from rest at the top of an incline of 20.0 m, as in Figure 5.19. In Part bottom of the slope, the skier encounters a horizontal surface where the coefficient of kinetic friction between the skis and snow is 0.229. I found the speed of the skier to 19.80 by Vb = sqrt (2 * 9.8m / s ^ 2 * 20m) The distance is also in d = 87.35 meters VB 2/2MUkg = ^ 19.8 ^ 2 / (2 *. 229 * 9.8) So my question is "Find the horizontal distance the skier travels before stopping, if the slope also has a coefficient of kinetic friction equal to 0.229 "Picture here: http://i235.photobucket.com/albums/ee48/xsmysticsx/Physics4.jpg

You have found the speed when there is no friction to 19.8 m / s [VB] ^ 2 = 392. If there is no friction to the square of the speed is reduced by a factor (1 – μ / tan θ) = 1 to 0.229 / tan 20 = 0.3708. Vb ^ 2 = 0.3708 x392 = 145.3536 So the horizontal distance = V b ^ 2 / [2 * 0.229 * 9.8] = 32.38 m. ==================================== =========== Note on the factor. V ^ 2 / 2 = GH – μ [g cos 20] dd is the distance traveled along the slope. but d = h / sin 20. Therefore V ^ 2 = 2 * g ((1 – μ / so θ) * g h. This reduces to g ((μ 1 – / tan θ) = =============================

Santa Fe Snow Skiing Pictures 085