38. A rock of mass 0.400 kg is released from the surface and sinks in the ocean. As the rock descends it is acted upon by three forces: gravity, buoyancy, and drag. The buoyancy is an upward force equal to half its weight. Drag from the water can be modeled by F = kv, where k = 0.650 kg/s. (a) Determine the terminal speed of the sinking rock. (b) Determine its depth, speed, and acceleration 1.50 seconds after it is released. (c) At what depth will it be at 99.0% of its terminal speed?
To solve this problem, we need to consider the three forces acting on the sinking rock: gravity, buoyancy, and drag.
(a) To determine the terminal speed of the sinking rock, we need to find the point at which the drag force equals the gravitational force acting on the rock. At terminal speed, the net force on the rock is zero.
The gravitational force acting on the rock is given by the formula: F_gravity = m * g, where m is the mass of the rock and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = 0.400 kg * 9.8 m/s^2
F_gravity = 3.92 N
The buoyant force is equal to half the weight of the rock: F_buoyancy = 0.5 * F_gravity
F_buoyancy = 0.5 * 3.92 N
F_buoyancy = 1.96 N
The drag force is given by the formula: F_drag = k * v, where k is the drag constant and v is the velocity of the rock.
At terminal speed, F_drag = F_gravity + F_buoyancy
k * v_terminal = F_gravity + F_buoyancy
0.650 kg/s * v_terminal = 3.92 N + 1.96 N
v_terminal = (3.92 N + 1.96 N) / 0.650 kg/s
v_terminal = 9.54 m/s
Therefore, the terminal speed of the sinking rock is 9.54 m/s.
(b) To determine the depth, speed, and acceleration of the rock 1.50 seconds after it is released, we can use the equations of motion.
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time:
Initial velocity, u = 0 m/s (rock is released from the surface)
Acceleration, a = (F_gravity - F_buoyancy - F_drag) / m
a = (3.92 N - 1.96 N - (0.650 kg/s * 0 m/s)) / 0.400 kg
a = 3.92 N / 0.400 kg
a = 9.8 m/s^2
Using a = 9.8 m/s^2 and t = 1.50 s in the equation of motion:
v = 0 m/s + (9.8 m/s^2 * 1.50 s)
v = 14.7 m/s
The speed of the rock 1.50 seconds after release is 14.7 m/s.
To determine the depth, we need to use the equation of motion, h = ut + (1/2)at^2, where h is the depth.
h = 0 m/s * 1.50 s + (1/2)(-9.8 m/s^2)(1.50 s)^2
h = -11.025 m
The depth of the rock 1.50 seconds after release is -11.025 m (negative because it is sinking).
(c) To determine the depth at which the rock will be at 99.0% of its terminal speed, we need to find the depth at which the speed of the rock is 99.0% of the terminal speed.
Let's assume the depth at which the rock reaches 99.0% of its terminal speed is h_terminal.
Using the equation of motion, v^2 = u^2 + 2gh, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and h is the depth:
v_terminal^2 = 0 m/s^2 + 2 * 9.8 m/s^2 * h_terminal
(9.54 m/s)^2 = 19.6 m/s^2 * h_terminal
h_terminal = (9.54 m/s)^2 / 19.6 m/s^2
h_terminal = 4.6128 m
Therefore, the rock will be at 99.0% of its terminal speed at a depth of approximately 4.6128 m.