A rocket sled of \(2.5 \mathrm{Mg}\) (tare) burns \(90 \mathrm{~kg}\) of fuel a second and the uniform exit velocity of the exhaust gases relative to the rocket is \(2.6 \mathrm{~km} \cdot \mathrm{s}^{-1}\). The total resistance to motion at the track on which the sled rides and in the air equals \(K V\), where \(K=1450 \mathrm{~N} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}\) and \(V\) represents the velocity of the sled. Assuming that the exhaust gases leave the rocket at atmospheric pressure, calculate the quantity of fuel required if the sled is to reach a maximum velocity of \(150 \mathrm{~m} \cdot \mathrm{s}^{-1}\).

Understanding the thrust force is essential for rocket sled propulsion. Thrust force propels the sled forward.
It's calculated using the rate of change of mass (fuel burn rate) and the exhaust velocity. Here's the key formula you'll need:
\( F_{thrust} = \frac{dm}{dt} \times v_{exhaust} \)
This formula tells us that the thrust force depends directly on how fast the fuel burns and how fast the exhaust gases exit.
For our rocket sled:
\( \frac{dm}{dt} = 90 \: kg/s \)
\( v_{exhaust} = 2600 \: m/s \)
Plug these values in to get:
\( F_{thrust} = 90 \: kg/s \times 2600 \: m/s = 234,000 \: N \)
This means that our sled experiences a thrust force of 234,000 Newtons.
Applying this force will help the sled overcome resistance and accelerate.
This is a powerful concept and a foundational principle for rocket sled propulsion.

In real-world scenarios, resistance force always acts against the motion of objects.
For our rocket sled, the resistance force comes from the track and air friction.
The key formula used here is:
\( F_{resistance} = K \times V \)
where \( K \) is the resistance constant and \( V \) is the velocity.
For our example:
\( K = 1450 \: N \cdot \frac{m^{-1}}{s} \) and \( V = 150 \: m/s \)
Plugging in these values:
\( F_{resistance} = 1450 \: N/m \cdot \frac{1}{s} \times 150 \: m/s = 217,500 \: N \)
This means that at the target velocity of 150 m/s, the sled faces an opposing force of 217,500 Newtons.
It's crucial for students to grasp that resistance grows with speed.
This concept helps in understanding why more power (thrust) is needed as speed increases.

Newton's Second Law of Motion is pivotal in understanding rocket sled dynamics.
It connects the forces acting on an object to its resulting acceleration.
The fundamental equation is:
\( F = ma \)
Here, \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.
For our sled problem, reaching terminal velocity means \( a = 0 \) and hence \( F_{net} = 0 \).
Substituting the forces we get:
\( F_{thrust} = F_{resistance} \)
Using the values:
\( 234,000 \: N = 217,500 \: N + ma \)
Solving for \( ma \) indicates:
\( ma = 16,500 \: N \)
A solid grasp of Newton's Second Law helps students understand how forces balance out when an object reaches a steady speed.
It highlights the tension between thrust and resistance in motion.

Terminal velocity is the speed where the net force acting on an object is zero.
In our rocket sled example, it means the thrust force equals the resistance force.
The sled stops accelerating and travels at constant speed.
For our calculations, we found:
\( F_{thrust} = F_{resistance} \)
Substituting values:
\( 234,000 \: N = 217,500 \: N \)
The sled reaches a terminal velocity of 150 m/s.
Understanding terminal velocity helps in seeing how forces interact.
It also illustrates why no matter how much fuel burns, the sled won't go faster once terminal velocity is hit.
This is a balanced state, crucial for designing efficient propulsion systems and knowing their limits.

Fuel burn rate directly influences the thrust force of a rocket sled.
It's how quickly fuel is consumed per second.
In our sled's case:
\( \frac{dm}{dt} = 90 \: kg/s \)
This means every second, 90 kg of fuel burns to produce thrust.
The importance of fuel burn rate lies in its impact on overall thrust.
If the rate changes, so does the thrust force:
Higher burn rate = Higher thrust
Lower burn rate = Lower thrust
Monitoring fuel burn rate helps in understanding the dynamics.
It ensures the sled maintains necessary thrust to overcome resistance.
Accurate calculation is key for efficient and effective propulsion.