A skier of mass 70.4 kg, starting from rest, slides down a slope at an angle $ heta$ of 36.8° with the horizontal. The coefficient of kinetic friction is 0.14. What is the net work, i.e. net gain in kinetic energy, (in J) done on the skier in the first 8.4 s of descent?

Question

A skier of mass 70.4 kg, starting from rest, slides down a slope at an angle $	heta$ of 36.8° with the horizontal. The coefficient of kinetic friction is 0.14. What is the net work, i.e. net gain in kinetic energy, (in J) done on the skier in the first 8.4 s of descent?

Henry · Accepted Answer

M*g = 70.4kg * 9.8N/kg = 689.9 N. = Wt.
of skier.

Fp = 689.9*sin36.8 = 413.3 N. = Force
parallel to the incline.
Fn = 689.9*Cos36.8 = 552.4 N. = Normal
force = Force perpendicular to the incline.

Fk = u*Fn = 0.14 * 552.4 = 77.34 N. =
Force of kinetic friction.

a = (Fp-Fk)/M = (413.3-77.34)/70.4 = 4.77 m/s^2.

V = Vo + a*t
V = 0 + 4.77*8.4 = 40.07 m/s

W = KE = M*V^2/2 =
70.4*40.07^2/2 = 56,517 J.

A skier of mass 70.4 kg, starting from rest, slides down a slope at an angle $\theta$ of 36.8° with the horizontal. The coefficient of kinetic friction is 0.14. What is the net work, i.e. net gain in kinetic energy, (in J) done on the skier in the first 8.4 s of descent?