A skier of mass 77.7 kg, starting from rest, slides down a slope at an angle $ heta$ of 35.7° with the horizontal. The coefficient of kinetic friction is 0.08. What is the net work, i.e. net gain in kinetic energy, (in J) done on the skier in the first 10.9 s of descent?

Question

A skier of mass 77.7 kg, starting from rest, slides down a slope at an angle $	heta$ of 35.7° with the horizontal. The coefficient of kinetic friction is 0.08. What is the net work, i.e. net gain in kinetic energy, (in J) done on the skier in the first 10.9 s of descent?

Henry · Answer

Ws = m*g = 77.7kg * 9.8N/kg = 761.5 N = Wt. of skier.

Fp = 761.5*sin35.7 = 444.3 N. = Force parallel to slope.

Fn = 761.5*cos35.7 = 618.4 = Normal = Force perpendicular to slope.

Fk = u*Fn = 0.08 * 618.4 = 49.47 N. =
Force of kinetic friction.

Fp-Fk = m*a
a=(Fp-Fk)/m=(444.3-49.47)/77.7=5.08m/s^2
V = Vo + a*t
V = 0 + 5.08*10.9 = 55.39 m/s.

Work=m*V^2/2 = 77.7*55.39^2/2=119,194 J.

A skier of mass 77.7 kg, starting from rest, slides down a slope at an angle $\theta$ of 35.7° with the horizontal. The coefficient of kinetic friction is 0.08. What is the net work, i.e. net gain in kinetic energy, (in J) done on the skier in the first 10.9 s of descent?