3. In the OpenStax College Physics textbook, read the Chapter 5 section 5.1 (Friction). Then forExample 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 pageof 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 onlineresources (such as unit conversion tools) give their URL and webpage title in your answers.Z=49 Skiing ExerciseA skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for theskier if friction is known to be 45.0 N.StrategyThe magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force Nas fk = HkN; thus, the coefficient of kinetic friction can be found if we can find the normal force of theskier on a slope. The normal force is always perpendicular to the surface, and since there is no motionperpendicular to the surface, the normal force should equal the component of the skier's weightperpendicular to the slope. (See the skier and free-body diagram in Figure 5.4.)Free-body diagramFigure 5.4 The motion of the skier and friction are parallel to the slope andso it is most convenient to project all forces onto a coordinate systemwhere one axis is parallel to the slope and the other is perpendicular (axesshown 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) hascomponents along both axes, namely w and W. N is equal inmagnitude to w1, so there is no motion perpendicular to the slope.However, fis less than W/ in magnitude, so there is acceleration downthe slope (along the x-axis).That is.N = w = w cos 25° = mg cos 25°.Substituting this into our expression for kinetic friction, we getfk = Hkmg cos 25°,which can now be solved for the coefficient of kinetic friction lk.SolutionSolving for uk givesfikUk =Nfkw cos 25°mg cos 25°.Substituting known values on the right-hand side of the equation,45.0 NHk == 0.082.(62 kg)(9.80 m/s²)(0.906)

Name: 3. In the OpenStax College Physics textbook, read the Chapter 5 secti
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Description: 3. In the OpenStax College Physics textbook, read the Chapter 5 section 5.1 (Friction). Then forExample 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