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Chapter 7: Problem 86

A jet plane is flying at 130 m/s relative to the ground. There is no wind. The engines take in \(81 \mathrm{kg}\) of air per second. Hot gas (burned fuel and air) is expelled from the engines at high speed. The engines provide a forward force on the plane of magnitude \(6.0 \times 10^{4} \mathrm{N} .\) At what speed relative to the ground is the gas being expelled? [Hint: Look at the momentum change of the air taken in by the engines during a time interval \(\Delta t .\) ] This calculation is approximate since we are ignoring the \(3.0 \mathrm{kg}\) of fuel consumed and expelled with the air each second.

Short Answer

Expert verified
Answer: The speed of the hot gas being expelled from the engines relative to the ground is approximately 742 m/s.

Step by step solution

01

Write the momentum conservation equation.

We start by writing the conservation of momentum equation: \(\Delta momentum = m_{air}(v_{final} - v_{initial})\) where \(\Delta momentum\) is the change in momentum, \(m_{air}\) is the mass of the air taken in by the engines per second, \(v_{final}\) is the speed of the expelled gas relative to the ground, and \(v_{initial}\) is the initial speed of the jet plane.
02

Calculate the change in momentum.

Since we know that the engines provide a forward force of \(6.0 \times 10^{4} \mathrm{N}\), we can use the equation: \(\Delta momentum = F_{engine}\Delta t\) Where \(F_{engine}\) is the forward force of the engines, and \(\Delta t\) is the time interval. Therefore, \(\Delta momentum = m_{air}(v_{final} - v_{initial}) = 60,000 \mathrm{N}\ \Delta t\)
03

Solve for the speed of expelled gas relative to the ground.

Next, we'll solve for \(v_{final}\) in the conservation of momentum equation: \(v_{final} = \frac{6.0 \times 10^{4} \mathrm{N}\Delta t}{81\ \mathrm{kg}} +130\ \mathrm{m/s}\) We know that \(81\ \mathrm{kg}\) of air is taken in by the engines per second, so the mass flow rate of the gas remains constant. Thus, we can divide both sides of the equation by \(\Delta t\) and solve for \(v_{final}\): \(v_{final} = \frac {6.0 \times 10^{4}\ \mathrm{N}}{81\ \mathrm{kg/s}} + 130\ \mathrm{m/s}\)
04

Calculate the speed of expelled gas relative to the ground.

Now, we just plug in the numbers and calculate \(v_{final}\): \(v_{final} = \frac {6.0 \times 10^{4}\ \mathrm{N}}{81\ \mathrm{kg/s}} + 130\ \mathrm{m/s}= 741.98\ \mathrm{m/s}\) Therefore, the speed of the hot gas being expelled from the engines relative to the ground is approximately \(742\ \mathrm{m/s}\).

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