After a long race, a cross country skier (m = 79 kg) reaches the final section o

Question

After a long race, a cross country skier (m = 79 kg) reaches the final section of the course before the finish line that is a long slope which is 15 degrees below horizontal. Going down the slope, just 13 meters from the finish line, the skier is travelling 3.1 m/s. The coefficient of kinetic friction between the skier and the hill is 0.15. Work out all answers for the last 13 meters of the course. Show all necessary equations and substitutions as well as any diagrams in each part.
(a) Find the magnitude of the kinetic friction force exerted on the skier.
112.17 N

(b) Compute the work done by the friction force.
-1458.12 J

(c) Compute the change in the skiers gravitational potential energy.

J

(d) Compute the final speed of the skier as he crosses the finish line.
m/s

Explanation / Answer

a) f = N = mgcos = 0.15 X 71 X 9.81x cos15 = 100.92 N

b) W = f.d = 100.92 X 11 = - 1110.08 J    (force and motion are in oppsoite direction)

c) P.E. = mgh = mg( - 11sin15) = - 1982.97 J

d) mv2 / 2 + 1982.97 - 1110.08 = 71 X 2.52 /2

mv2 = 1302.03

v = 4.28 m/s

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