Problem 7.62 Part A A skier of mass 60.0 kg starts from rest at the top of a ski

Question

Problem 7.62 Part A A skier of mass 60.0 kg starts from rest at the top of a ski slope of height 69.0 nm If frictional forces do -1.02x10 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 = 9.80 m/s2 . m/s Submit My Answers Give Up Part B Now moving horizontally, the skier crosses a patch of soft snow where the coefficient of friction is 14-0.250. If the patch is of width 65.0 m and the average force of air resistance on the skier is 180 N, how fast is she going after crossing the patch? ! ? m/s Submit My Answers Give Up Part C After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.00 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her? Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

A)

using conservation of energy

Potential energy at the top = work done by friction + kinetic energy

mgh = 1.02 x 104 + (0.5) m v2

(60) (9.8) (69) = 1.02 x 104 + (0.5) (60) v2

v = 31.8 m/s

b)

Fk = frictional force during the patch = uk mg = 0.25 x 60 x 9.8 = 147 N

using conservation of energy

initial KE = final kinetic energy + work done by friction + work done by resistance

(0.5) m vi2 = (0.5) m vf2 + Fk d + F d

(0.5) (60) (31.8)2 = (0.5) (60) vf2 + (147) (65) + (180) (65)

vf = 17.4 m/s

C)

Vi = 17.4 m/s

Vf = 0 m/s

m = mass = 60 kg

d = penetration distance = 2 m

F = force

using the equation

F d = (0.5) m (Vf2 - Vi2)

- F (2) = (0.5) (60) ((0)2 - (17.4)2)

F = 4541.4 N

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