Question

A rock with mass m = 3.00 kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of 18.0 N (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force f = kv, where v is the speed in m/s and k = 2.20 N s/m.

(a) Find the initial acceleration a0.
(b) Find the acceleration when the speed is 3.00 m/s.
(c) Find the speed when the acceleration equals 0.1a0.
(d) Find the terminal speed vt
(e) Find the coordinate, speed, and acceleration 2.00 s after the start of the motion.
(f) Find the time required to reach a speed of 0.9 vt.

Answer 1

The initial acceleration is 6.00 m/s². The acceleration when the speed is 3.00 m/s is 2.20 m/s². The speed when the acceleration equals 0.1 times the initial acceleration is 15.0 m/s.

Step-by-step explanation:

(a) To find the initial acceleration (0), we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is 18.0 N and the mass is 3.00 kg, so we can write:

Fnet = m0

18.0 N = 3.00 kg 0

Simplifying the equation, we find that the initial acceleration is 6.00 m/s².

(b) To find the acceleration when the speed is 3.00 m/s, we can use the formula for fluid resistance force: f = kv. Rearranging the formula, we have:

a = f/m = kv/m = (2.20 N s/m)(3.00 m/s)/3.00 kg = 2.20 m/s².

(c) To find the speed when the acceleration equals 0.1a0, we substitute the given values into the equation:

a = f/m = kv/m = (2.20 N s/m)v/3.00 kg = 0.1(6.00 m/s²)

Simplifying the equation, we find that the speed is 15.0 m/s.

(d) The terminal speed (vt) of an object falling in a viscous medium is the speed at which the fluid resistance force is equal in magnitude to the net force on the object. In this case, the net force is 18.0 N and the fluid resistance force is kvt. So we can write:

kvt = 18.0 N

Simplifying the equation, we find that the terminal speed is 8.18 m/s.

(e) To find the coordinate, speed, and acceleration 2.00 s after the start of the motion, we can use the equations of motion. The coordinate (y) can be found using the formula:

y = y0 + v0t + (1/2)at²

where y0 is the initial coordinate, v0 is the initial speed, t is the time, and a is the acceleration. Substituting the given values into the equation:

y = 0 + 0(2.00 s) + (1/2)(6.00 m/s²)(2.00 s)² = 12.0 m

The speed (v) can be found using the formula:

v = v0 + at

Substituting the given values into the equation:

v = 0 + (6.00 m/s²)(2.00 s) = 12.0 m/s

The acceleration can be found using the given value of −9.80 m/s².

(f) To find the time required to reach a speed of 0.9vt, we can use the formula:

v = v0 + at

Substituting the given values into the equation:

0.9vt = 0 + (6.00 m/s²)t

Simplifying the equation, we find that the time is 0.9vt/6.00 s.