Answer:
The normal force exerted on the ant is 0.75 N.
Explanation:
Given;
diameter of the ball, D = 40 cm = 0.4m
radius of the ball, r = 0.2m
mass of the beach ball, m₁ = 300 g = 0.3 kg
mass of the ant, m₂ = 4 x 10⁻⁶ kg
speed of the ball, v = 4 m/s
The area of the ball, assuming spherical ball is given by;
A = 4πr²
A = 4π(0.2)² = 0.5027 m²
The drag force (resistance) experienced by the spherical ball is given as;
[tex]F_D = \frac{1}{2}C\rho Av^2[/tex]
where;
C is the drag coefficient of the spherical ball = 0.45
ρ is density of air = 1.21 kg/m³
[tex]F_D = \frac{1}{2}C\rho Av^2\\\\F_D = \frac{1}{2}(0.45)(1.21) (0.5027)(4)^2\\\\F_D = 2.19 \ N[/tex]
The downward force of the ball due to its weight and that of the ant is given by;
[tex]F_g = mg\\\\F_g =g(m_{ant} + m_{ball})\\\\F_g = g(4*10^{-6} \ kg\ + \ 0.3\ kg)\\\\F_g = g(0.300004 \ kg) \ \ \ (mass \ of \ the \ ant \ is \ insignificant)\\\\F_g = 9.8(0.3)\\\\F_g = 2.94 \ N[/tex]
The net downward force experienced by the ball is given by;
[tex]F_{net} = F_g - F_D\\\\F_{net} = 2.94 \ N - 2.19 \ N\\\\F_{net} = 0.75 \ N[/tex]
This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.
Thus, the normal force exerted on the ant is 0.75 N.
