Question

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0o with
the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts
down a distance of 10.4 m before coming to the edge of a cliff. Without slowing down, she skis off the
cliff and lands downhill at a point whose vertical distance is 3.50 m below the edge. How fast is she
going just before she lands?

Answer 1

Fp = Mg*sin25 = 0.423Mg.

Fn = Mg*Cos25 = 0.906Mg.

Fk = u*Fn = 0.2 * 0.906Mg = 0.181Mg.

Fp-Fk = M*a.
0.423Mg-0.181Mg = M*a.
Divide both sides by M:
0.423g-0.181g = a,
a = 0.242g = 0.242 * 9.8 = 2.37 m/s^2.

V^2 = Vo^2 + 2a*d.
V^2 = 0 + 4.74*10.4 = 49.3,
V = 7.02 m/s.

V^2 = Vo^2 + 2g*h.
V^2 = 0 + 19.6*3.5 = 68.6,
V = 8.28 m/s.

Answer 2

Note: Fp = Force parallel with the incline.

Fn = Normal force. = Force perpendicular to the incline.

Answer 3

Diagram please

Answer 4

Hi Mark!!!!!!

Answer 5

Mark do your homework or I'm going to come over and drink your lemonade.

Answer 6

Please don't Mr. Thom I want to keep my lemonade!!!

Answer 7

Waffles I know you are watching, please do your HW.