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A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip. If r=20.0 m, how fast is the roller coaster traveling at the bottom of the dip?

I'm having some real troubles answering this problem. Thus far, here's

what I've come up with.

Fc = ( mv^2 ) / r

FN - 2mg = ( mv^2 ) / r

2mg - mg = ( mv^2 ) / r

(2mg - mg )(r) = ( mv^2 )

(2mg - mg )(r) = ( v^2 )

sqrt( (2mg - mg )(r) ) = v

sqrt( ( 19.6 - 9.8 m/s^2 )( 20.0m ) ) = v

14 m/s = v

I've been informed that 14 m/s is not the correct answer and that to one digit the correct answer is 10 m/s.

It seems to me that I have all the right pieces, but I may be missing something. Will someone help, please? Big thanks!