Answer:
Part a)
![F_n = 306 N](https://img.qammunity.org/2020/formulas/physics/college/1awy4i0poegy5krgq1048u0eetkmc2ixkv.png)
Part B)
![v = 12.1 m/s](https://img.qammunity.org/2020/formulas/physics/college/uevcoa6bexn218camnptifd7mnhuvzl7wk.png)
Step-by-step explanation:
Part A)
At the top of the hump the force on the rider is
1) Normal force
2) weight
so here we know that
![mg - F_n = (mv^2)/(R)](https://img.qammunity.org/2020/formulas/physics/college/jx9fh9sy558ix5fptgs994059ai3lk6y1h.png)
![F_n = mg - (mv^2)/(R)](https://img.qammunity.org/2020/formulas/physics/college/ho7c55alnstzbc4izc4n6fvct993aozq82.png)
![F_n = (100)(9.81) - (100(9^2))/(12)](https://img.qammunity.org/2020/formulas/physics/college/d3smb7f8fzh7z6bnk412111d0heb6sjqr1.png)
![F_n = 306 N](https://img.qammunity.org/2020/formulas/physics/college/1awy4i0poegy5krgq1048u0eetkmc2ixkv.png)
Part B)
At the top of the loop we will have
![F_n + mg = (mv^2)/(R)](https://img.qammunity.org/2020/formulas/physics/college/oe1ldpprwzgvdsmb5sq6ahy247iyf5unab.png)
in order to remain in contact the normal force must be just greater than zero
so we will have
![mg = (mv^2)/(R)](https://img.qammunity.org/2020/formulas/physics/middle-school/a6wbf6p74ilwa42gc4qj2hflcq2cxyoeag.png)
![v = √(Rg)](https://img.qammunity.org/2020/formulas/physics/middle-school/fzxawb6uc5jljhdjraojzq7pegkuz5iat9.png)
![v = √(15* 9.81)](https://img.qammunity.org/2020/formulas/physics/college/a636bx7f5mut18aiwgwy49476pvmztu1xf.png)
![v = 12.1 m/s](https://img.qammunity.org/2020/formulas/physics/college/uevcoa6bexn218camnptifd7mnhuvzl7wk.png)