An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.40 kN, and the radius of the circle is 9.60 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v

Complete Question

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.40 kN, and the radius of the circle is 9.60 m. At the top of the circle,

(a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v = 5.10 m/s?

(b) What is FB if v = 15.0 m/s? Use g=9.80 m/s2.

The force FB at v= 5.10 m/s is

The force FB at  v=15.0 m/s is      

Explanation:

The diagram of this setup is shown on the first uploaded image

From the question we are told that

     The weight of the car and the rider is

       The radius of the circle is  

At the top the force acting on the car can be mathematically represented as

Where is normal upward force acting on the car , v is the linear velocity acting on the car , m is the mass of the (car+riders) , g is the acceleration due to gravity

making  the subject  we have  

This equation can be modified to

Which is equivalent to the previous equation

    So

  Now at  

   Now  at  

ill do it for 100, also this isnt 80 its 40