Q1: What is the skier's
speed at the intermediate point?
Using the conservation of momentum, we have the formulas;
W_{nc} = E_{f}  E_{o}_{
and }W_{nc} = (1/2mv_{f}^{2}
+ mgh_{f})  (1/2mv_{o}^{2} + mgh_{o})
Using the given information, it is stated that the skier starts
from rest at the highest point. The initial velocity (V_{o})
is zero and solving for the final velocity (V_{f}) we get:
V_{f} = [2(W_{nc} +mg(h_{o}
 h_{f}))/m]^{1/2}
Wnc = 400 kJ + 50 kJ = 350 kJ (350,000 J) m
= 60.0 kg, g = 9.8 m/s2, h_{f} = 2,901 m, h_{o}
= 3,505 m
V_{f} = [2({350,000 J} + {60.0 kg}{9.8m/s^{2}}(3,505
m  2,901 m))/ {60.0 kg}]^{1/2}
V_{f} = 13.1 m/s
Q2: What
is the skier's final velocity (V_{f}) at the foot of the
Hart Prairie run?
W_{nc} = 400 kJ  170 kJ + 50 kJ + 50 kJ = 470 kJ
m = 60.0 kg, g = 9.8 m/s^{2}, h_{f} = 2,697 m, h_{o}
= 3,505 m
V_{f} = [2({470 kJ} + {60.0 kg}{9.8m/s2}(3,505 m 
2,697 m))/{60.0 kg}]^{1/2}
Using the same formulas from the previous equation with a larger
W_{nc} and a lower h_{f}, we end up with 13.0 m/s
for V_{f}.
Q3: Please
draw a crude sketch calculating Kinetic energy (KE), Gravitational
potential energy (GPE), and total energy at the three points
labeled Agassiz (start), Intermediate, and Hart Prairie (final).
Use SI (m, k, and s) units throughout. Remember: [1.0 m = 3.281
ft]
KE = 1/2mv^{2}, GPE = mgh, TE = 1/2mv^{2}+
mgh

Start

Intermediate

Final

KE

0.000 J

5.022 x 10^{3} J

4.916 x 10^{3} J

GPE

2.061 x 10^{6}^{} J

1.706 x 10^{6}^{} J

1.586 x 10^{6}^{} J

TE

2.061 x 10^{6}^{} J

1.711x 10^{6}^{} J

1.591 x 10^{6}
J

Advanced Question: Near
the very foot of Hart Prairie run there is a small circular 'speed
bump' hill of radius 5.0 m. At
what minimum V_{f} will the skier become airborne? Does
this particular skier become airborne?
(Draw a FBD)
Riding this hill involves centripetal force due to the
circular shape of it. Knowing that
F_{net} = ma and F_{c}
= (mv^{2})/r, we can combine the two to get
ma = (mv^{2})/r.
In order to find the necessary velocity to overcome this
force we further simplify to;
v= (ar)^{1/2}, where a = 9.8m/s^{2}
and r = 5.0 m,
v = ({9.8m/s^{2}}{5.0 m})^{1/2} = 7 m/s
So if the minimum speed to launch is 7 m/s, then
the skier traveling at 13.0 m/s will become airborne.


