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lsu777
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A 69.2-kg skier encounters a dip in the snow's surface that has a circular cross section with radius of curvature of r = 13.2 m. If the skier's speed at point A in the figure below is 8.13 m/s, what is the normal force exerted by the snow on the skier at point B?
http://educog.com/res/prenhall/walker/Physics_3E/Chap08/graphics/walk0850.gif"
As shown, h = 1.63 m. Ignore frictional forces.
ok so we are given:
mass=69.2kg
radius=13.2m
Velocity1=8.13 m/s
Change in Y=1.63m
gravity=9.81m/s^2
we know KE1 + PE1=KE2 + PE2
with potential energy2= O
ok so first thing I did was try and find velocity2 with the formula
so 1/2*m*v1^2 + 0= 1/2*m*V2^2 + m*g*Y
so V2=sqrt(V1^2 - 2*G*Y)
I got V2=5.84092
Then I did sum of Forces Y=(normal force - m*g)= mass * Arad
Arad= V2^2/R
so normal force=(mass*(V2^2/R))+m*g
so normal force = 709.473 N
can somebody smarter then me please check this. I think I got it wrong the first time because of math error. ****ing lon-capa sucks and doesn't give you any info. I have one more shot to get it right.
http://educog.com/res/prenhall/walker/Physics_3E/Chap08/graphics/walk0850.gif"
As shown, h = 1.63 m. Ignore frictional forces.
ok so we are given:
mass=69.2kg
radius=13.2m
Velocity1=8.13 m/s
Change in Y=1.63m
gravity=9.81m/s^2
we know KE1 + PE1=KE2 + PE2
with potential energy2= O
ok so first thing I did was try and find velocity2 with the formula
so 1/2*m*v1^2 + 0= 1/2*m*V2^2 + m*g*Y
so V2=sqrt(V1^2 - 2*G*Y)
I got V2=5.84092
Then I did sum of Forces Y=(normal force - m*g)= mass * Arad
Arad= V2^2/R
so normal force=(mass*(V2^2/R))+m*g
so normal force = 709.473 N
can somebody smarter then me please check this. I think I got it wrong the first time because of math error. ****ing lon-capa sucks and doesn't give you any info. I have one more shot to get it right.
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