Finding normal force from kinetic and potential energies

AI Thread Summary
The discussion focuses on calculating the normal force exerted by snow on a skier at point B after descending a dip. Given the skier's mass, initial speed, and height change, the kinetic and potential energy equations are applied to find the speed at point B. The calculated speed at point B was found to be lower than the initial speed, indicating a potential error in the sign convention used for height. Participants emphasize the importance of correctly identifying the height reference points and signs in the calculations. The final normal force was computed to be approximately 709.473 N, pending verification of the calculations.
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.
 
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I can't see the diagram without a username and password. Is B lower than A, or higher? Is v1 the speed at A, or B? Make sure everything is consistent.
 
ideasrule said:
I can't see the diagram without a username and password. Is B lower than A, or higher? Is v1 the speed at A, or B? Make sure everything is consistent.

A is original heigth, B is the bottom of the circle, below A.

V1= Velocity at A


picture was attached.
 

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lsu777 said:
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
Is v1 the speed at A, or B? Is Y the y coordinate at A, or B? Your method is correct, but you have a sign issue. (One way to see that you made a mistake is to note that V2 is smaller than the initial speed, which doesn't make sense.)
 
ideasrule said:
Is v1 the speed at A, or B? Is Y the y coordinate at A, or B? Your method is correct, but you have a sign issue. (One way to see that you made a mistake is to note that V2 is smaller than the initial speed, which doesn't make sense.)

v1 is speed at A, Y is the height at B to the level of A. can you tell me where I made a mistake on my signs?
 
If your height at A is 0, your height at B should be negative, not positive.
 
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