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. As the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is0.181.

Explanation / Answer

(a) At the bottom of the incline KE = PE. In other words: (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. (b) 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) - (0.185)(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 = -(0.185)(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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