Kinematics

 Conservation of Mechanical Energy

 Solution

Q1: What is the skier's speed at the intermediate point?

Using the conservation of momentum, we have the formulas;

Wnc = Ef - Eo  and  Wnc = (1/2mvf2 + mghf) - (1/2mvo2 + mgho)

Using the given information, it is stated that the skier starts from rest at the highest point. The initial velocity (Vo) is zero and solving for the final velocity (Vf) we get:

Vf = [2(Wnc +mg(ho - hf))/m]1/2

Wnc = -400 kJ + 50 kJ = -350 kJ (-350,000 J)  m = 60.0 kg,  g = 9.8 m/s2, hf = 2,901 m, ho = 3,505 m

Vf = [2({-350,000 J} + {60.0 kg}{9.8m/s2}(3,505 m - 2,901 m))/ {60.0 kg}]1/2

Vf = 13.1 m/s

Q2: What is the skier's final velocity (Vf) at the foot of the Hart Prairie run?

Wnc = -400 kJ - 170 kJ + 50 kJ + 50 kJ = -470 kJ m = 60.0 kg, g = 9.8 m/s2, hf = 2,697 m, ho = 3,505 m

Vf = [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 Wnc and a lower hf, we end up with 13.0 m/s for Vf.

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 = 1/2mv2, GPE = mgh, TE = 1/2mv2+ mgh

 Start Intermediate Final KE 0.000 J 5.022 x 103 J 4.916 x 103 J GPE 2.061 x 106 J 1.706 x 106 J 1.586 x 106 J TE 2.061 x 106 J 1.711x 106 J 1.591 x 106 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 Vf 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 Fnet = ma and Fc = (mv2)/r, we can combine the two to get ma = (mv2)/r. In order to find the necessary velocity to overcome this force we further simplify to; v= (ar)1/2, where a = 9.8m/s2 and r = 5.0 m, v = ({9.8m/s2}{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.

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