A 30. N stone is dropped from a height of 10. m, and strikes the ground with a velocity of 13. m/s. What average force of air friction acts on it as it falls?
A) 4.1 N
B) 2.9 N
C) 0.13 KN
D) 7.2 N
E) 1.2 N

Question

Quizplus · Accepted Answer

The average force of air friction can be calculated using the work-energy principle. The work done by the air friction is equal to the difference between the gravitational potential energy at the start and the kinetic energy just before the stone hits the ground. First, calculate the initial potential energy (PE_initial = mgh) and the final kinetic energy (KE_final = 0.5 * m * v^2), where m is the mass in kg (30 N corresponds to approximately 3 kg, since g ≈ 10 m/s^2), g is the acceleration due to gravity (10 m/s^2 for simplicity), h is the height (10 m), and v is the velocity (13 m/s).PE_initial = 3 kg * 10 m/s^2 * 10 m = 300 JKE_final = 0.5 * 3 kg * (13 m/s)^2 = 253.5 JThe work done by air friction is the difference: 300 J - 253.5 J = 46.5 J. This work is also equal to the force of air friction times the distance over which it acts (W = Fd), where d is the distance of 10 m. Solving for F gives F = W/d = 46.5 J / 10 m = 4.65 N. The closest answer to this calculation is 4.1 N, acknowledging some rounding in the process.

A 30 N Stone Is Dropped from a Height of 10