Question

The speed of an aeroplane is 1200ms−1. The engines take in 80 kg of air per second and
mix it with 40 kg of fuel. This mixture is expelled after it ignites and it moves at a
velocity of 3000ms−1 relative to the aeroplane. Calculate the thrust of the engine

Answer 1

Answer on Question #41984 – Physics - Mechanics | Kinematics | Dynamics

The speed of an aeroplane is 1200ms⁻¹. The engines take in 80 kg of air per second and mix it with 40 kg of fuel. This mixture is expelled after it ignites and it moves at a velocity of 3000ms⁻¹ relative to the aeroplane. Calculate the thrust of the engine

Solution:

Thrust is a reaction force described quantitatively by Newton's second and third laws. When a system expels or accelerates mass in one direction, the accelerated mass will cause a force of equal magnitude but opposite direction on that system.

From Newton's second law of motion a force F on an object is equal to the rate of change of its momentum:

$$F = \frac{dp}{dt} = \frac{d(mv)}{dt}$$

In our case a force F can be expressed as

$$F = \frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t}$$

$m_1$ – mass of air per second, $m_2$ – mass of fuel per second.

$v_1 = 1200 \frac{\text{m}}{\text{s}}$ – air relative to the airplane, initial velocity of a fuel relative to the airplane.

$v_2 = 3000 \frac{\text{m}}{\text{s}}$ – final velocity of a fuel and an air relative to the airplane, initial velocity of a fuel relative to the airplane is 0.

Change of momentum in 1 second:

$$\Delta p = m_1 (v_2 - v_1) + m_2 (v_2 - 0) = v_2 (m_1 + m_2) + m_1 v_1$$

The thrust of the engine:

$$F = \frac{\Delta p}{\Delta t} = \frac{v_2 (m_1 + m_2) + m_1 v_1}{\Delta t} = \frac{3000 \frac{\text{m}}{\text{s}} \cdot (80 \text{kg} + 40 \text{kg}) + 80 \text{kg} \cdot 1200 \frac{\text{m}}{\text{s}}}{1 \text{s}} = 264 \text{ kN}$$

Answer: thrust of the engine is equal to $264 \text{ kN}$.

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