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Expert-verified Found in: Page 148 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A ski that is placed on snow will stick to the snow. However, when the ski is moved along the snow, the rubbing warms and partially melts the snow, reducing the coefficient of kinetic friction and promoting sliding. Waxing the ski makes it water repellent and reduces friction with the resulting layer of water. A magazine reports that a new type of plastic ski is especially water repellent and that, on a gentle 200 m slope in the Alps, a skier reduced his top-to-bottom time from 61 s with standard skis to 42 s with the new skis. Determine the magnitude of his average acceleration with (a) the standard skis and (b) the new skis. Assuming a ${\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{°}}$ slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.

$\left(a\right){a}_{1}=0.11m/{s}^{2}\phantom{\rule{0ex}{0ex}}\left(b\right){a}_{2}=0.23m/{s}^{2}\phantom{\rule{0ex}{0ex}}\left(c\right){\mu }_{k1}=0.041\phantom{\rule{0ex}{0ex}}\left(d\right){\mu }_{k2}=0.029$

See the step by step solution

## Step 1: Given

New ski is 200 m slope in the Alps, a skier reduced his top-to-bottom time from 61 s with standard skis to 42 s .

## Step 2: Understanding the concept

Frictional force is given by the product of coefficient of friction and the normal reaction. Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass. Using these concepts, the problem can be solved.

Formula:

${{\mathbf{F}}}_{{\mathbf{net}}}{\mathbf{=}}{\mathbf{mg}}{\mathbf{}}{\mathbf{sin\theta }}{\mathbf{-}}{{\mathbf{f}}}_{{\mathbf{k}}}$

## Step 3:  Calculation for average acceleration with the standard skis

(a)

If the distance is L during time t with initial speed zero and a constant acceleration a, then

$L=a{t}^{2}/2,$

Which gives the acceleration for the first (old) pair of skis:

${a}_{1}=\frac{2L}{{t}_{1}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\left(200m\right)}{{\left(61s\right)}^{2}}\phantom{\rule{0ex}{0ex}}=0.11m/{s}^{2}$

## Step 4:  Calculation for average acceleration with the new skis

(b)

The acceleration for the second (new) pair is

${a}_{2}=\frac{2L}{{t}_{1}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\left(200m\right)}{{\left(42s\right)}^{2}}\phantom{\rule{0ex}{0ex}}=0.23m/{s}^{2}$

## Step 5: Assuming a 3.0° slope, compute the coefficient of kinetic friction for the standard skis

(c)

The net force along the slope acting on the skier of mass m is

${F}_{net}=mg\mathrm{sin}\theta -{f}_{k}\phantom{\rule{0ex}{0ex}}=mg\left(\mathrm{sin}\theta -{\mu }_{k}\mathrm{cos}\theta \right)\phantom{\rule{0ex}{0ex}}=ma$

Which we solve for for the first pair of skis:

${\mu }_{k1}=\mathrm{tan}\theta -\frac{{a}_{1}}{g\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left(3°\right)-\frac{0.11m/{s}^{2}}{\left(9.8m/{s}^{2}\right)\mathrm{cos}\left(3°\right)}\phantom{\rule{0ex}{0ex}}=0.041$

## Step 6: Assuming a 3.0° slope, compute the coefficient of kinetic friction for the new skis

(d)

For second pair, we have

${\mu }_{k2}=\mathrm{tan}\theta -\frac{{a}_{2}}{g\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left(3°\right)-\frac{0.23m/{s}^{2}}{\left(9.8m/{s}^{2}\right)\mathrm{cos}\left(3°\right)}\phantom{\rule{0ex}{0ex}}=0.029$ ### Want to see more solutions like these? 