A skier of mass 79.1 kg, starting from rest, slides down a slope at an angle q of 38 with the horizontal. The coefficient of kinetic friction, m, is 0.09. What is the net work (the net gain in kinetic energy) done on the skier in the first 8.9 s of descent?

Answer 1

m•a= μ•m•g•cosα
a= μ•g•cosα
v=at = μ•g•cosα•t
W= ΔKE=KE2-KE1 =mv²/2 -0= mv²/2