A 147 g baseball is dropped from a tree 12.0 m above the ground. If it actually hits the ground with a speed of 8.00 m/s, what is the magnitude of the average force of air resistance exerted on it?
The air resistance is not constant, but increases in proportion to V^2.
The best you can do is compute an average air resistance during the fall. You would do this by computing the work done against air friction and dividing it by the distance fallen.
Friction work = M g H - (1/2) M V^2
where V is the final velocity of 8.00 m/s
W = 0.147)[9.81*12 - 8^2/2] = 12.6 Joules
F(average) = 12.6J/12 m = 1.05 N
To find the magnitude of the average force of air resistance exerted on the baseball, we need to consider the work-energy principle. The energy of the falling baseball is converted from potential energy (due to its initial height) to kinetic energy (due to its final velocity).
First, let's calculate the potential energy of the baseball when it is 12.0 m above the ground. The potential energy (PE) of an object with mass (m) at a height (h) is given by the formula: PE = mgh, where g is the acceleration due to gravity (approximately 9.8 m/s²).
PE = (147 g) * (9.8 m/s²) * (12.0 m)
PE = 17299.2 J
Next, let's calculate the kinetic energy (KE) of the baseball when it hits the ground. The kinetic energy is given by the formula: KE = 0.5mv², where v is the velocity of the ball.
KE = 0.5 * (147 g) * (8.00 m/s)²
KE = 4704 J
According to the work-energy principle, the change in potential energy is equal to the work done by all forces, including the force of air resistance. Therefore, we can find the magnitude of the average force of air resistance (F) by subtracting the kinetic energy from the initial potential energy:
F * distance = PE - KE
F * 12.0 m = 17299.2 J - 4704 J
F * 12.0 m = 12595.2 J
F = 12595.2 J / 12.0 m
F ≈ 1049.60 N
So, the magnitude of the average force of air resistance exerted on the baseball is approximately 1049.60 N.