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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 7 / Problem 62P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: A 60.0-kg skier starts from rest at the top of a

Chapter 7, Problem 62P

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QUESTION:

Problem 62P

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?

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QUESTION:

Problem 62P

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?

ANSWER:

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,

${K}_{1}=mgh-{W}_{f}$

$=(60\ kg\times{}9.8\ m/{s}^{2}\times{}65\ m)-10500\ J\ \ \ or$

${K}_{1}=38200\ J-10500\ J=27720\ J$

$Then\ \ \ \ {v}_{2}=\sqrt{\frac{{2K}_{1}}{m}}$

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