Question

The weight of passengers on a roller coaster increases by 60 % as the car goes through a dip with a 37 m radius of curvature. what is the car's speed at the bottom of the dip?

Answer 1

The force acting on the passenger is equal to:
$F=N-mg$
Where N is the normal force and mg is a force of gravity. If a person feels like their weight is increased by 60% at the bottom that means normal force is 60% stronger than the force of gravity:
$N-mg=0.6mg$
We know that net force must be equal to the centripetal force:
$0.6mg=\frac{mv^2}{R}\\ v^2=0.6gR\\ v=\sqrt{0.6gR}=14.75\frac{m}{s}$