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A vehicle travelling at 120 km/hr on a paved surface with an air density of 2.123 Kg/m³ has coefficient of drag of 0.50 and frontal area of 3.0 m². Calculate the air resistance and

Physics

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Answer

1769.17 N\text{1769.17 N}

Step-by-step solution

4 steps

To calculate the air resistance, we use the drag force formula.

  1. Identify the given values and the formula for air resistance.

    The formula for air resistance (drag force) is:

    FD=12ρv2CDAF_D = \frac{1}{2} \rho v^2 C_D A

    Where:

    • FDF_D is the air resistance (drag force)
    • ρ\rho is the air density
    • vv is the velocity of the vehicle
    • CDC_D is the coefficient of drag
    • AA is the frontal area

    Given values:

    • Velocity (vv) = 120 km/hr120\, \text{km/hr}
    • Air density (ρ\rho) = 2.123 kg/m32.123\, \text{kg/m}^3
    • Coefficient of drag (CDC_D) = 0.50
    • Frontal area (AA) = 3.0 m23.0\, \text{m}^2
  2. Convert the velocity to meters per second (m/s).

    v=120 km/hr×1000 m1 km×1 hr3600 sv = 120\, \text{km/hr} \times \frac{1000\, \text{m}}{1\, \text{km}} \times \frac{1\, \text{hr}}{3600\, \text{s}} v=120×10003600 m/sv = \frac{120 \times 1000}{3600}\, \text{m/s} v=1200003600 m/sv = \frac{120000}{3600}\, \text{m/s} v=1003 m/s≈33.333 m/sv = \frac{100}{3}\, \text{m/s} \approx 33.333\, \text{m/s}
  3. Substitute the values into the drag force formula.

    FD=12×(2.123 kg/m3)×(1003 m/s)2×(0.50)×(3.0 m2)F_D = \frac{1}{2} \times (2.123\, \text{kg/m}^3) \times \left(\frac{100}{3}\, \text{m/s}\right)^2 \times (0.50) \times (3.0\, \text{m}^2)
  4. Calculate the air resistance.

    FD=12×2.123×(100009)×0.50×3.0F_D = \frac{1}{2} \times 2.123 \times \left(\frac{10000}{9}\right) \times 0.50 \times 3.0 FD=0.5×2.123×1111.111...×0.5×3.0F_D = 0.5 \times 2.123 \times 1111.111... \times 0.5 \times 3.0 FD=1769.166... NF_D = 1769.166...\, \text{N}

    The air resistance is approximately 1769.17 N\text{1769.17 N}.

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