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14 December, 05:11

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.

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Answers (1)
  1. 14 December, 07:01
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    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.
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