Question

2. 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. Of r = 21.5 m, how fast is the roller coaster traveling at the bottom of the dip?

Answer 1

Answer:$v=\sqrt{4gr}$

Explanation:

Given

Passenger feels the seat of the car Pushing upward with a force equal to five times her weight

Therefore

$N-mg=\frac{mv^2}{r}$

it is given $N=5mg$

where N=normal reaction

$5mg-mg=\frac{mv^2}{r}$

thus $v=\sqrt{4gr}$