91影视

A 50-cm-diameter, 400 g beach ball is dropped with a 4.0 mg ant riding on the top. The ball experiences air resistance.

the ball鈥檚 speed is 2.0 m/s

diameter of the ball, D = 50 cm = 0.5m

radius of the ball, r = 0.25m

mass of the beach ball, m鈧� = 400 g = 0.4 kg

mass of the ant, m鈧� = 4 x 10鈦烩伓 kg

speed of the ball, v = 2 m/s

The area of the ball, assuming spherical ball is given by;

A = 4蟺r虏

A = 4蟺(0.25)虏 = 0.7854 m虏

The drag force (resistance) experienced by the spherical ball is given as;

$F_D=\frac{1}{2}C蚁A{v}^2$

where;

C is the drag coefficient of the spherical ball = 0.45

蚁 is density of air = 1.21 kg/m鲁

$\begin{array}{r}{F}_D=\frac{1}{2}C蚁A{v}^2\\ F_D=\frac{1}{2}\left(0.45\right)\left(1.21\right)\left(0.7854\right)(2{)}^2\\ F_D=0.855\mathrm{N}\end{array}$

The downward force of the ball due to its weight and that of the ant is given by;

role="math" localid="1650545692020" $\begin{array}{r}{F}_g=mg\\ F_g=g\left(m_{ant}+m_{ball}\right)\\ F_g=g\left(4鈭�{10}^{鈭�6}\mathrm{kg}+0.4\mathrm{kg}\right)\\ \left.F_g=g\left(0.400004\mathrm{kg}\right)\text{(mass of the ant is insignificant}\right)\\ F_g=9.8\left(0.4\right)\\ F_g=3.92\mathrm{N}\end{array}$

The net downward force experienced by the ball is given by;

$\begin{array}{r}{F}_{net}=F_g鈭�F_D\\ F_{net}=3.92\mathrm{N}鈭�0.855\mathrm{N}\\ F_{net}=3.07\mathrm{N}\end{array}$

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 3.07N.