3. In the OpenStax College Physics textbook, read the Chapter 5 section 5.1 (Friction). Then for Example 5.1 in that section, change the angle of the ski slope from 25 degrees to 30 degrees; also, change the skier's mass from "62 kg" to “Z kg" where Z is the number computed on the first page of this study guide. Then write the answer to the question in the space below. For your answer, remember to please write down your calculations and explain your reasoning. Draw a sketch if it'd be helpful, and please use your units correctly. If you are using any online resources (such as unit conversion tools) give their URL and webpage title in your answers. Z=49

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3. In the OpenStax College Physics textbook, read the Chapter 5 section 5.1 (Friction). Then for
Example 5.1 in that section, change the angle of the ski slope from 25 degrees to 30 degrees; also,
change the skier's mass from “62 kg" to "Z kg" where Z is the number computed on the first page
of this study guide. Then write the answer to the question in the space below.
For your answer, remember to please write down your calculations and explain your reasoning.
Draw a sketch if it'd be helpful, and please use your units correctly. If you are using any online
resources (such as unit conversion tools) give their URL and webpage title in your answers.
Z=49
Transcribed Image Text:3. In the OpenStax College Physics textbook, read the Chapter 5 section 5.1 (Friction). Then for Example 5.1 in that section, change the angle of the ski slope from 25 degrees to 30 degrees; also, change the skier's mass from “62 kg" to "Z kg" where Z is the number computed on the first page of this study guide. Then write the answer to the question in the space below. For your answer, remember to please write down your calculations and explain your reasoning. Draw a sketch if it'd be helpful, and please use your units correctly. If you are using any online resources (such as unit conversion tools) give their URL and webpage title in your answers. Z=49
Skiing Exercise
A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the
skier if friction is known to be 45.0 N.
Strategy
The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force N
as fk = HkN; thus, the coefficient of kinetic friction can be found if we can find the normal force of the
skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion
perpendicular to the surface, the normal force should equal the component of the skier's weight
perpendicular to the slope. (See the skier and free-body diagram in Figure 5.4.)
Free-body diagram
Figure 5.4 The motion of the skier and friction are parallel to the slope and
so it is most convenient to project all forces onto a coordinate system
where one axis is parallel to the slope and the other is perpendicular (axes
shown to left of skier). N (the normal force) is perpendicular to the slope,
and f (the friction) is parallel to the slope, but w (the skier's weight) has
components along both axes, namely w and W. N is equal in
magnitude to w1, so there is no motion perpendicular to the slope.
However, fis less than W/ in magnitude, so there is acceleration down
the slope (along the x-axis).
That is.
N = w = w cos 25° = mg cos 25°.
Substituting this into our expression for kinetic friction, we get
fk = Hkmg cos 25°,
which can now be solved for the coefficient of kinetic friction lk.
Solution
Solving for uk gives
fik
Uk =
N
fk
w cos 25°
mg cos 25°.
Substituting known values on the right-hand side of the equation,
45.0 N
Hk =
= 0.082.
(62 kg)(9.80 m/s²)(0.906)
Transcribed Image Text:Skiing Exercise A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N. Strategy The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force N as fk = HkN; thus, the coefficient of kinetic friction can be found if we can find the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier's weight perpendicular to the slope. (See the skier and free-body diagram in Figure 5.4.) Free-body diagram Figure 5.4 The motion of the skier and friction are parallel to the slope and so it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N (the normal force) is perpendicular to the slope, and f (the friction) is parallel to the slope, but w (the skier's weight) has components along both axes, namely w and W. N is equal in magnitude to w1, so there is no motion perpendicular to the slope. However, fis less than W/ in magnitude, so there is acceleration down the slope (along the x-axis). That is. N = w = w cos 25° = mg cos 25°. Substituting this into our expression for kinetic friction, we get fk = Hkmg cos 25°, which can now be solved for the coefficient of kinetic friction lk. Solution Solving for uk gives fik Uk = N fk w cos 25° mg cos 25°. Substituting known values on the right-hand side of the equation, 45.0 N Hk = = 0.082. (62 kg)(9.80 m/s²)(0.906)
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