A skier starts from rest at the top of a frictionless incline of height 20.0 m,

Question

A skier starts from rest at the top of a frictionless incline of height 20.0 m, as in Figure 5.19 (above). At the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.218. (a) Find the skier's speed at the bottom. (b) How far does the skier travel on the horizontal surface before coming to rest? Figure 5.19 from book Figure 5.19 The skier slides down the slope and onto a level surface, stopping after traveling a distance d from the bottom of the hill. Strategy Going down the frictionless incline is physically no different than going down a water 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, Equation 5.12, with Wnc = Wfric = -fkd, where fk is the magnitude of the force of friction and d is the distance traveled. Solution (a) Find the skier's speed at the bottom. Write down Equation 5.14, for conservation of energy. 1/2mv_A^2 + mgy_A = 1/2mv_B^2 + mgy_B Insert the values vA = 0 and yB = 0. 0 + mgy_A = 1/2mv_B^2 + 0 Solve for vB and substitute values for g and yA as the skier moves from the top, point circled A, to the bottom, point circled B. vB = sqrt(2gh) = sqrt(2(9.80 text( m/s)^2)(20.0 text( m). m/s (b) Find the distance traveled on the horizontal, rough surface. Apply the work-energy theorem as the skier movers from circled B to circled C. Wnet = -fkd = KE = 1/2mvC2 - 1/2mvB2 Substitute vC = 0 and fk = µkn = µkmg. - µkmgd = -1/2mvB2 Solve for d. d = (v_B^2)/(2mu_kg) = (19.8 text( m/s))^2/(2(0.218)(9.80 text( m/s)^2). Remarks Substituting the symbolic expression vB = sqrt(2gh) into the equation for the distance d shows that d is linearly proportional to h: Doubling the height doubles the distance traveled. Decorative only Decorative only

I HAVE THE ANSWERS TO ALL OTHER PARTS EXCEPT THIS PART HERE: Exercise 5.8 Hints: Getting Started | I'm Stuck Decorative only Find the horizontal distance the skier travels before coming to rest if the incline also has a coefficient of kinetic friction equal to 0.218.

Explanation / Answer

Speed at the bottom V^2=2gh=2×9.8×20=4×98=392

On reaching bottom deceleration ,a=-g=-0.218×9.8=-2.136m/s2

Using. V^2-U^2=2×a×d

0-392=-2×2.136×d

d=91.743 m

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