What is the speed of a skier at the bottom of a circular hill?

[RedPhoenix]

A skier weighing 0.80 kN comes down a frictionless ski run that is circular (R = 30 m) at the bottom, as shown. If her speed is 12 m/s at point A, what is her speed at the bottom of the hill (point B)?

http://www.physics.gatech.edu/people/faculty/larry/images/16212.gif

I keep getting mixed results.

I did 30(cos(40))*-1 to get the distance from A to B... but now what?

 

[Doc Al]

I did 30(cos(40))*-1 to get the distance from A to B... but now what?

I assume you mean 30*(1 - cos40). What's conserved?

 

[RedPhoenix]

I assume you mean 30*(1 - cos40). What's conserved?

Energy is conserved. But I honestly do not have a clue where to proceed. I am obviously missing the portion of this in the book. The answer is in front of me, but I am being blind.

 

[Doc Al]

Yes, mechanical energy (KE + gravitational PE) is conserved. So how would you express that mathematically?

 

[RedPhoenix]

Yes, mechanical energy (KE + gravitational PE) is conserved. So how would you express that mathematically?

PE = mgy
KE = .5mv^2

so... 800N / 9.8 = 81.63

(.5 x 81.63 x 12^2) + (800 x 30(1-cos40)) = 11492.29/1000.. still wrong. where did I mess up?

 

[Doc Al]

(.5 x 81.63 x 12^2) + (800 x 30(1-cos40)) = 11492.29/1000

The left side is perfectly OK, but I'm not sure what you're doing on the right. Set the left side equal to .5mv^2 and solve for v.

 

[RedPhoenix]

The left side is perfectly OK, but I'm not sure what you're doing on the right. Set the left side equal to .5mv^2 and solve for v.

I was converting to kN... it did not matter, I get it.

Bingo!


(.5 x 81.63 x 12^2) + (800 x 30(1-cos40)) = 11492.29

11492.29/.5/81.63 = 281.57 ; 281.57^1/2 = 16.78m/s

Excellent, thank you for your help!