Answer:
the speed of the car at the top of the vertical loop [tex]v_{top} = 2.0 \sqrt{gR \ \ }[/tex]
the magnitude of the normal force acting on the car at the top of the vertical loop [tex]F_{N} = 1.47 \ \ N[/tex]
Explanation:
Using the law of conservation of energy ;
[tex]mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R[/tex]
[tex]v_{top} = 2.0 \sqrt{gR \ \ }[/tex]
The magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:
[tex]F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\[/tex]
[tex]F_{N} = 1.47 \ \ N[/tex]
