Question

The speed of an aeroplane is 1200m/s. 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 3000m/s
relative to the aeroplane. Calculate the thrust of the engine.

Answer 1

Answer on Question #47180 – Physics - Mechanics | Kinematics | Dynamics

The speed of an aeroplane is $1200\mathrm{m/s}$. The engines take in $80\mathrm{kg}$ of air per second and mix it with $40\mathrm{kg}$ of fuel. This mixture is expelled after it ignites and it moves at a velocity of $3000\mathrm{m/s}$ 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 $\square$ 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 $\square$ 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.

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

$v_2 = 3000 \, \text{ms}$ – 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:

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

The thrust of the engine:

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

Answer: 264 kN.

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