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## Homework Statement

The radius of curvature of a loop-the-loop for a roller coaster is 14.0 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.35mg. Find the speed of the roller coaster at the top of the loop. (Enter your answers from smallest to largest.)

*(This question has two answers, one of which, I found = the speed at the top of the loop. I'm guessing the 2nd answer is the speed at the bottom of the loop?)*

## Homework Equations

∑F

_{r}= ma

_{r}

F

_{sp}+ mg = m(v

^{2}/R)

a = v

^{2}/R

## The Attempt at a Solution

F

_{sp}= m[(v

^{2}/R) - g]

.35mg = m[(v

^{2}/R) - g]

.35g = (v

^{2}/R) - g

1.25g = v

^{2}/R

v

_{_top}= √(gR1.35) ≈ 13.62 m/s

Now to find the 2nd answer, I think I need to calculate a (since I believe this is constant):

a = v

^{2}/R ≈ 13.25 m/s

^{2}

But this can't be correct I'll just get the same answer for velocity.

Does anyone know how I am supposed to find the 2nd answer?