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 five times her weight as she goes through the dip. If r = 17.0 m, how fast is the roller coaster traveling at the bottom of the dip

16.3 m/s

Explanation:

Let m = mass of the passenger,

v = speed of the roller coaster

Forces on the passenger are

1. Force by seat = 2.18 mg in upward direction.

2. Weight mg in downward direction

Therefore, net upward force = 2.18 mg - mg = 1.18 mg

Also, net upward force = centripetal force = mv^2/r

Therefore,

1.18 mg = mv^2/r

Dividing by m,

1.18 g = v^2/r

Or v^2 = 1.18 gr

Or v = sqrt (1.18 gr)

= sqrt (1.18 * 9.81 * 22.9)

= sqrt (265)

= 16.3 m/s