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Need the exercise portion Hit the Ski Slopes EXAMPLE 5.8 GOAL Combine conservation of mechanical energy...

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Hit the Ski Slopes EXAMPLE 5.8 GOAL Combine conservation of mechanical energy with the work-energy theorem involving friction(B) Find the distance traveled on the horizontal, rough surface. Apply the work-energy theorem as the B Wnet fd = AKE =mv- ^mEXERCISE HINTS: GETTING STARTED IM STUCK! Use the values from PRACTICE IT to help you work this exercise. Find the horizontaNeed the exercise portion

Hit the Ski Slopes EXAMPLE 5.8 GOAL Combine conservation of mechanical energy with the work-energy theorem involving friction on a horizontal surface. h = 20.0 m у A skier starts PROBLEM from rest at the top of a frictionless incline of height (В С) 20.0 m, as in the figure. At the bottom of the incline, the skier The skier slides down the slope and onto a level surface, stopping after encounters a horizontal traveling a distance d from the bottom of the hill. surface where the coefficient of kinetic friction between skis and snow is 0.210. (a) Find the skier's speed at the bottom. (b) How far does the skier travel on the horizontal surface before coming to rest? Neglect air resistance. STRATEGY Going down the frictionless incline is physically no different than going down a frictionless slide and is handled the same way, using conservation of mechanical energy to find the speed vB at the bottom. On the flat, rough surface, use the work-energy theorem with Wnc = Wfric = -fkd, where f is the magnitude of the force of friction and d is the distance traveled on the horizontal surface before coming to rest SOLUTION (A) Find the skier's speed at the bottom. Write down the equation for = 19.8 m/s Ув 3D V 2gh V2(9.80 m/s2)(20.0 m) conservation of energy, insert the values vA 0 and yB = 0, solve for vB and substitute values for g and yB as the skier moves from the top, point , to the bottom, point
(B) Find the distance traveled on the horizontal, rough surface. Apply the work-energy theorem as the B Wnet fd = AKE =mv- ^mvB2 skier moves from to Substitute vc = 0 and -mgd mB2 fk Hkn = Hkmg. Solve for d (19.8 m/s)2 2(0.210)(9.8 m/s2) VB2 = 95.2 m d = 2Шкg LEARN MORE REMARKS Substituting the symbolic expression vB = V2gh into the equation for the distance d shows that d is linearly proportional to h: Doubling the height doubles the distance traveled. QUESTION Give two reasons why skiers typically assume a crouching position when going down a slope. (Select all that apply.) Crouching decreases the skier's inertia. In the crouching position there is less air resistance. The acceleration of gravity g is increased by crouching Crouching decreases the mass of the skier. Crouching lowers the skier's center of mass, making it easier to balance PRACTICE IT Use the worked example above to help you solve this problem. A skier starts from rest at the top of a frictionless incline of height 20.0 m, as shown in the figure. At the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.222. Neglect air resistance. (a) Find the skier's speed at the bottom. m/s 19.8 (b) How far does the skier travel on the horizontal surface before coming to rest? 91.0
EXERCISE HINTS: GETTING STARTED I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. Find the horizontal distance the skier travels before coming to rest if the incline also has a coefficient of kinetic friction equal to 0.222. Assume that 0 20.0° 36.0 What is the initial energy of the skier? How much work is done as the skier slides down the incline? How much work is done as the skier slides to a stop on the horizontal surface? In other words, write expressions for these three amounts of energy, then relate them to each other. m Need Help? Talk to a Tutor
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W PEKE = PE tKE ini Elost Otmgh 2 m Vs t Fd A mahveg cose d 2 Vo 2 g h-ugdcose) h dsine 2(48x20-0222x98 cot20 xho) SinB 2 x G

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