Astudent of mass m = 89 kg takes a ride in a frictionless loop-the-loop at an amusement park. the radius of the loop-the-loop is r = 19 m. the force due to the seat on the student at the top of the loop-the-loop is fn = 142 n and is vertically down. what is the apparent weight of the student at the bottom of the loop-the-loop?

17.5 kN

Explanation:

force downward on student: 
mg+142-mv^2/r=0


mg=mv^2/r-142


mv^2/r=mg+142 

at bottom, apparent weight
weight=mg+mv^2/r=mg+mg+142


apparent weight= 89(2g)+142



for force at bottom, reverse v^2/r


force=m(g+v^2/r) = 89 ( 9.8 + v^2 / 19 )

F = mg + mv^2 / r

F = mg + mg + 142 = 2mg + 142 = 2 x 89 x 9.8 + 142 = 17.5 kN