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28 September, 05:17

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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Answers (2)
  1. 28 September, 07:28
    0
    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
  2. 28 September, 08:23
    0
    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
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