A skier starts from rest at the top of a incline of height 20.0 m, as shown in t

Question

A skier starts from rest at the top of a incline of height 20.0 m, as shown in the figure. The coefficient of kinetic friction between skis and snow is 0.188. Neglect air resistance.

Find the horizontal distance the skier travels between points B and C before coming to rest. Assume that ? = 20.0?.

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.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.