A 1300 kg car carrying four 84 kg people travels over a “washboard” dirt road with corrugations 3.2 m apart. The car bounces with maximum am

Question

A 1300 kg car carrying four 84 kg people travels over a “washboard” dirt road with corrugations 3.2 m apart. The car bounces with maximum amplitude when its speed is 13 km/h. When the car stops, and the people get out, by how much does the car body rise on its suspension?

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Eirian 4 years 2021-07-19T05:50:56+00:00 2 Answers 18 views 0

Answers ( )

  1. Delwyn
    0
    2021-07-19T05:52:12+00:00
    Reply

    Answer: 0.0392 m

    Explanation:

    Given

    Mass of the car, M = 1300 kg

    Mass of each person, m = 84 kg

    Distance of corrugation, d = 3.2 m

    Speed of car, v = 13 km/h = 3.61 m/s

    To solve this, we would be doing some derivations.

    If, T = d/v, the angular frequency of Simple Harmonic Motion is,

    w = √(k / M + 4m), but we also know that,

    w = 2π/T. Now we substitute for w

    2π/T = √(k / M + 4m), here again, we substitute for T

    2πv/d = √(k / M + 4m), making subject of formula, we have

    k = (M + 4m) [2πv/d]²

    the vertical displacement of the car with respect to the ground is given by, F = kx. We also know that the mass is M + 4m, so that

    (M + 4m) g = kx(i)

    x(i) = (M + 4m) g / k, we can also write

    Mg = kx(f)

    x(f) = Mg / k

    x(i) – x(f) = 4mg / k

    x(i) – x(f) = 4mg / (M + 4m) [2πv/d]²

    x(i) – x(f) = 4mg/M + 4m * (d/2πv) ², now we substitute all the values into the equation to have

    x(i) – x(f) = (4 * 84 * 9.8)/(1300 + 4 * 84) * (3.2/2 * 3.142 * 3.61)

    x(i) – x(f) = (3292.8/1636) * (0.14)²

    x(i) – x(f) = 2 * 0.0196

    x(i) – x(f) = 0.0392 m

  2. Vodka
    0
    2021-07-19T05:52:40+00:00
    Reply

    Answer:

    0.04m

    Explanation:

    Mass of car (M) = 1300kg

    Mass of people (m) = 84kg

    Distance (d) = 3.2m

    Speed (v) = 13km/h

    V = 13km/h = x m/s

    (13 * 1000) / 3600 = 3.6m/s

    The distance travelled between the two maximums is the distance travelled during the period.

    T = v / d ……. equation (i)

    Angular frequency of simple harmonic motion

    ω = √(k / M + 4m)

    but ω = 2π /T

    2π / T = √(k / M + 4m)

    put T = v / d

    2πv / d = √(k / M + 4m

    solving for k,

    K = (M + 4m) * (2πv / d)²

    The vertical displacement of the car with respect to ground and force constant K =

    F = Kx

    F = force of gravity

    F(i) = (M + 4m)g

    (M + 4m)g = KX(i)

    X(i) = (M + 4m)g / I

    Force without people F(f) = m*g

    mg = KX(f)

    X(f) = Mg / k

    X(i) – X(f) = [(M + 4m)g / k – Mg / k]

    X(i) – X(f) = 4mg / k

    But k = (M + 4m) * (2πv / d)²

    X(i) – X(f) = [4mg / (M + 4m) ] * (d /2π v)²

    X(i) – X(f) = [4 * 84 * 9.8 / (1300 + 4*84)] * [(3.2 / 2Π*3.6 ]²

    X(i) – X(f) = (3292.8 / 1636) * 0.020

    X(i) – X(f) = 2.01 * 0.020

    x(i) – x(f) = 0.04m

    The car body rise on its suspension by 0.04m

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