Expert: James J. Kovalcin - 11/6/2009

Question
The passengers in a roller coaster car feel 40% heavier than their true weight as the car goes through a dip with a 28 m radius of curvature. What is the car's speed at the bottom of the dip?  

Answer
Because this roller coaster is moving in a circular path Newton's 2nd Law applies:
SF=m*a  where SF is the sum of all the external forces acting on a system and where the acceleration is centripetal a=v^2/R.
For the roller coaster at the bottom of its curved path there are two relevant forces acting: the force of gravity Fg down and the normal force Fn of the track up. For an object moving in a circular path the centripetal acceleration is directed towards the center of the circular path. For the roller coaster at the bottom of the loop this acceleration is directed upward.
Therefore, the sum of the external forces acting in teh direction of the acceleration  becomes:
SF=m*a  becomes  Fn-Fg=m*[v^2/R]
According to the problem the normal force [the apparent weight] of the passenger is 40% greater than the the weight and can be written as Fn=1.4*Fg and Newton's 2nd Law becomes:
1.4*Fg-Fg=0.4*Fg=m*v^2/R  solving for  0.4*m*g=m*v^2/R  which becomes  0.4*9.8=v^2/28
Solve for  v=sqrt[0.4*9.8*28)=10.5m/s