Suggested languages for you:

Americas

Europe

Q100P

Expert-verifiedFound in: Page 148

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 ${\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\xb0}}$**** slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.**

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

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 .

**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:**

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

(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:

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

(b)

The acceleration for the second (new) pair is

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

(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:

${\mu}_{k1}=\mathrm{tan}\theta -\frac{{a}_{1}}{g\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left(3\xb0\right)-\frac{0.11m/{s}^{2}}{\left(9.8m/{s}^{2}\right)\mathrm{cos}\left(3\xb0\right)}\phantom{\rule{0ex}{0ex}}=0.041$

(d)

For second pair, we have

${\mu}_{k2}=\mathrm{tan}\theta -\frac{{a}_{2}}{g\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left(3\xb0\right)-\frac{0.23m/{s}^{2}}{\left(9.8m/{s}^{2}\right)\mathrm{cos}\left(3\xb0\right)}\phantom{\rule{0ex}{0ex}}=0.029$

94% of StudySmarter users get better grades.

Sign up for free