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 220 m slope in the Alps, a skier reduced his top-to-bottom time from 61.0 s with standard skis to 43.0 s with the new skis. Determine the magnitude of his average acceleration with (a) the standard skis and (b) the new skis. Assuming a 3.40° slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.

Question

Jasmin Long

Physics

28 June, 15:12

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 220 m slope in the Alps, a skier reduced his top-to-bottom time from 61.0 s with standard skis to 43.0 s with the new skis. Determine the magnitude of his average acceleration with (a) the standard skis and (b) the new skis. Assuming a 3.40° slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.

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Answers (1)

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Lyla Marquez · Answer 1

a) a1 = 0.118 m/s^2

b) a2 = 0.238 m/s^2

c) μk1 = 0.047

d) μk2 = 0.035

Explanation:

The acceleration for the first pair of (old) skis is equal to:

a1 = (2*L) / t1^2, where

a1 = acceleration of old skis

L = distance = 220 m

t = time = 61 s

Replacing values:

a)

a1 = (2*220) / (61^2) = 0.118 m/s^2

b)

For the new skis we have:

a2 = (2*220) / (43^2) = 0.238 m/s^2

c)

the equation for the net force along the slope equals:

F = m*g*sinθ - fk = m*g (sinθ - μk*cosθ) = m*a

Clearing μk and solve for the old skis:

μk1 = tanθ - (a1 / (g*cosθ)) = tan (3.4°) - (0.118 / (9.8*cos (3.4))) = 0.047

d) for the new skis:

μk2 = tan (3.4°) - (0.238 / (9.8*cos (3.4))) = 0.035