Answer:
6859.079 N
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Magnitude of the force of the track on the car when the car is at the top of the loop is 6859.079 N
I believe that B is the answer.
Given values:
Mass of the steel ball, m = 100 g = 0.1 kg
Height of the steel ball, h1 = 1.8 m
Rebound height, h2 = 1.25 m
a. PE= mgh
0.1 x 9.8 x 1.8 =
1.764 Joules
b. KE = PE ->
1.764 Joules
c. KE= 1/2 mv square
so v = square root 2ke/m
square root 2 x 1.764/ 0.1
= 5.93 m/s
d. KE=PE=mgh square
0.1 x 9.8 x 1.21 =
1.186 joules
velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s