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 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 0.186. Neglect air resistance.

Find the horizontal distance the skier travels before coming to rest if the incline also has a coefficient of kinetic friction equal to 0.186. Assume that ? = 20.0

Explanation / Answer

(1/2)mv^2 = mgh ==> v = v(2gh). Thus, the speed of the skier on the bottom is: v = v(2gh) = v[2(9.8 m/s^2)(20.0 m)] = 19.8 m/s. Taking friction into account, the speed of the skier at the bottom of the incline is: KE = PE - W(friction) = PE - F(friction)d = PE - u(k)mgd*cos?. Since the angle makes an angle of 20.0° with the ground, we see that the skier traveled a distance of: sin(20.0°) = 20/d ==> d = 20/sin(20.0°) = 58.5 m down the incline. With KE = (1/2)mv^2 and PE = mgh, we see that: (1/2)mv^2 = mgh - u(k)mgd*cos? ==> (1/2)v^2 = gh - u(k)gd*cos?, as the masses cancel out ==> v = v{2[gh - u(k)gd*cos?]}. Thus, the true speed of the skier at the bottom is: v = v{2[gh - u(k)gd*cos?]} = v{2[(9.8 m/s^2)(20.0 m) - (0.186)(9.8 m/s^2)(58.5 m)cos(20.0°)]} = 13.9 m/s. Now, we need to find the deceleration of the skier due to friction. We see that: F(friction) = -u(k)*F(n) = -u(k)mg. Since F(friction) = ma, we have: ma = u(k)mg ==> a = -u(k)g. So, the skier decelerates at a rate of: a = -u(k)g = -(0.186)(9.8 m/s^2) = -1.81 m/s^2. Using (v_f)^2 = (v_i)^2 + 2ad, we see that: d = [(v_f)^2 - (v_i)^2]/(2a). Hence, the total distance that the skier travels is: d = [(v_f)^2 - (v_i)^2]/(2a) = [(0.00 m/s)^2 - (13.9 m/s)^2]/[2(-1.81 m/s^2)] = 53.4 m.

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