A 60.0-kg skier starts from rest at the top of a ski slope 65.0 m high. (a) If friction forces do - 10.5 kJ of work on her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of soft snow where µk = 0.20. If the patch is 82.0 m wide and the average force of air resistance on the skier is 160 N, how fast is she going after crossing the patch? (c) The skier hits a snowdrift and penetrates 2.5 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Solution 62P a)The skier’s kinetic energy at the bottom can be found from the potential energy at the top minus the work done by friction, K 1 mgh W f = (60 kg × 9.8 m/s × 65 m) 10500 J or K = 38200 J 10500 J = 27720 J 1 2K1 Then v = 2 m = 2(27720 J 60 kg = 30.4 m/s b)K 2 K 1(W + f air = 27720 J ( mgd + f d k air = 27720 J (0.2 × 588 N × 82 m) + (160N × 82 m) 0r K2= 27720 J 22763 = 4957 J Then v = 2 2K2 m 4957 J = 60 kg = 12.9 m/s c) According to work energy theorem work done = change in kinetic energy W = K K F = d 4957 = 2.5 m = 200 N