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Homework Statement
A 50.0kg skier starts at A with an unknown initial velocity. The skier then goes over a hill (point B) which is 4.00m above point A. At point B (top of the hill), the skier experiences a Force normal 5/6 of his/her weight. The hill has a radius of 14.6m. What is the skier's initial velocity?
(Problem continues but only need help for this part)
Homework Equations
K=1/2mv2
G=mgh
Fc=(mv2)/r
The Attempt at a Solution
So since the skier's "apparent weight" is 5/6 of his/her real weight, then force normal is = 5mg/6.
Since only gravity is holding the skier in a circle, then force centripetal = 5/6 force gravity.
([STRIKE]m[/STRIKE]v2)/r = (5/6)([STRIKE]m[/STRIKE]g)
v=sqrt(5/6)(9.81)(14.6)
v=10.92m/s
This is the speed of the skier at the top of the hill
Then we set the conservation of energy equation
Kinitial = Khill + G
K = 1/2mv2 + mgh = (0.5)(50.0)(10.922) + (50.0)(9.81)(4.00)
K = 2983.875J + 1962J = 4945.875J
1/2mv2 = 4945.875J
v= sqrt [(4945.875J)(2)/(50.0kg)]
v=14.07m/s
It seems logical to me but for some reason it is not correct : /
Help appreciated thanks :)