A rock with mass m = 4.00 kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of F = 17.4 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 = 2.16 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 0.1 a0.
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.

a) 4.35 m/s²

b) 2.73 m/s²

c) 7.25 m/s

d) 8.06 m/s

e) At t = 2 s

x = 16.5 m

v = 7.88 m/s

a = 0.099 m/s²

f) t = 0.743 s

Explanation:

Force balance on the rock

ma = 17.4 - F

4a = 17.4 - kv

4a = 17.4 - 2.16v

a) At the initial instant, F = kv = 0

4a = 17.4

a = 4.35 m/s²

b) When v = 3 m/s

4a = 17.4 - (2.16)(3) = 10.92

a = 2.73 m/s²

c) a₀ = 4.35 m/s²

0.1 a₀ = 0.435 m/s²

4a = 17.4 - 2.16v

4(0.435) = 17.4 - 2.16v

1.74 = 17.4 - 2.16v

2.16v = 15.66

v = 7.25 m/s

d) Terminal speed is when the body stops accelerating in the fluid

When a = 0

0 = 17.4 - 2.16v

2.16 v = 17.4

v = 8.06 m/s

e) 4a = 17.4 - 2.16v

a = 4.35 - 0.54 v

But a = dv/dt

(dv/dt) = 4.35 - 0.54v

∫ dv/(4.35 - 0.54v) = ∫ dt

Integrating the left hand side from 0 to v and the right hand side from 0 to t

- 1.852 In (4.35 - 0.54v) = t

In (4.35 - 0.54v) = - 0.54 t

4.35 - 0.54v = e⁻⁰•⁵⁴ᵗ

0.54v = 4.35 - e⁻⁰•⁵⁴ᵗ

v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

Then, v = dx/dt

(dx/dt) = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

dx = (8.06 - 0.54 e⁻⁰•⁵⁴ᵗ) dt

∫ dx = ∫ (8.06 - 0.54 e⁻⁰•⁵⁴ᵗ) dt

Integrating the left hand side from 0 to x and the right hand side from 0 to t

x = 8.06t + e⁻⁰•⁵⁴ᵗ

Acceleration too can be obtained as a function of time

since v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ and a = dv/dt

a = 0.54² e⁻⁰•⁵⁴ᵗ = 0.2916 e⁻⁰•⁵⁴ᵗ

At t = 2 s

Coordinate

x = 8.06t + e⁻⁰•⁵⁴ᵗ

x = (8.06)(2) + e^(-1.08) = 16.5 m down into the fluid.

Velocity

v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

v = 8.06 - 0.54 e^(-1.08) = 7.88 m/s

Acceleration

a = 0.2916 e⁻⁰•⁵⁴ᵗ

a = 0.2916 e^(-1.08) = 0.099 m/s²

f) t = ? When v = 0.9 × 8.06 = 7.254 m/s

v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

7.254 = 8.06 - 0.54e⁻⁰•⁵⁴ᵗ

- 0.806 = - 0.54 e⁻⁰•⁵⁴ᵗ

e⁻⁰•⁵⁴ᵗ = 1.493

0.54t = In 1.493 = 0.401

t = 0.743 s.

f is prop to m1xm2 double m1 double force