Question

A student 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?

Answer 1

The apparent weight of the student at the bottom of the loop-the-loop is 730.2 N.

Step-by-step explanation:

To determine the apparent weight of the student at the bottom of the loop-the-loop, we need to consider the forces acting on the student. At the top of the loop, the only force acting on the student is the seat force, which is equal to the normal force. Since the seat force is vertically down, the normal force is also vertically down and has a magnitude of 142 N. As the student reaches the bottom of the loop, the normal force is directed upwards, while the force of gravity is still directed downwards. Therefore, the apparent weight of the student at the bottom of the loop is the difference between the force of gravity and the normal force.

The force of gravity can be calculated by multiplying the mass of the student (89 kg) by the acceleration due to gravity (9.8 m/s^2). So the force of gravity is (89 kg)(9.8 m/s^2) = 872.2 N. The apparent weight of the student at the bottom of the loop is the difference between the force of gravity and the normal force: 872.2 N - 142 N = 730.2 N.

Answer 2

Answer: 17.5 kN

Step-by-step 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