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A \(25,000-\mathrm{kg}\) rocket blasts off vertically from the earth's sur- face with a constant acceleration. During the motion considered in the problem, assume that \(g\) remains constant (see Chapter 12 ) Inside the rocket, a \(15.0-\mathrm{N}\) instrument hangs from a wire that can support a maximum tension of 35.0 \(\mathrm{N}\) . (a) Find the minimum time for this rocket to reach the sound barrier \((330 \mathrm{m} / \mathrm{s})\) without breaking the inside wire and the maximum vertical thrust of the rocket engines under these conditions. (b) How far is the rocket above the earth's surface when it breaks the sound barrier?

Short Answer

Expert verified
Minimum time: 25.24 s; Maximum thrust: 572,000 N; Distance: 4164.56 m above the surface.

Step by step solution

01

Determine Maximum Acceleration

The maximum tension that the wire can sustain is 35.0 N, and the weight of the instrument is 15.0 N. Use the formula for net force, where the tension in the wire equals the weight of the instrument plus the force due to acceleration.\[ T = W + ma \]where:- \( T = 35.0 \) N (maximum tension)- \( W = 15.0 \) N (weight of the instrument)- \( m \) is the mass of the instrument- \( a \) is the acceleration of the rocketWe can rearrange this equation to solve for acceleration:\[ a = \frac{T - W}{m} \]Given that \( W = mg \), where \( g \approx 9.81 \) m/s², we solve for \( m = \frac{15.0}{9.81} \approx 1.53 \) kg:\[ a = \frac{35.0 - 15.0}{1.53} = \frac{20.0}{1.53} \approx 13.07 \text{ m/s}^2 \]
02

Calculate Minimum Time to Reach Sound Barrier

Use the kinematics equation to solve for time:\[ v = u + at \]where:- \( v = 330 \) m/s (speed of sound barrier)- \( u = 0 \) m/s (initial velocity)- \( a \approx 13.07 \) m/s² (as found earlier)Rearrange to solve for time \( t \):\[ t = \frac{v - u}{a} = \frac{330}{13.07} \approx 25.24 \text{ seconds} \]
03

Calculate Maximum Vertical Thrust

The vertical thrust must counteract both the gravitational force on the rocket and provide the net acceleration. The formula for thrust \( F_t \) is:\[ F_t = mg + ma \]where:- \( m = 25000 \) kg (mass of the rocket)- \( g = 9.81 \) m/s² (acceleration due to gravity)- \( a \approx 13.07 \) m/s²Substitute the values:\[ F_t = 25000 (9.81 + 13.07) = 25000 \times 22.88 = 572000 \text{ N} \]
04

Calculate Distance Traveled when Sound Barrier is Breached

Use the kinematics equation for displacement:\[ s = ut + \frac{1}{2}at^2 \]where:- \( u = 0 \) m/s- \( a \approx 13.07 \) m/s²- \( t \approx 25.24 \) sSubstitute the values:\[ s = 0 + \frac{1}{2} \times 13.07 \times (25.24)^2 \approx \frac{1}{2} \times 13.07 \times 637.06 \approx 4164.56 \text{ m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing that motion. It's all about describing how things move. In this exercise, kinematics helps us determine how long it takes for a rocket to reach the speed of sound and how far it travels during this time.

We start with the formula for motion at constant acceleration:
  • Velocity: \( v = u + at \)
  • Displacement: \( s = ut + \frac{1}{2}at^2 \)
In these equations:
  • \( v \) is the final velocity, \( u \) is the initial velocity.
  • \( a \) is the acceleration and \( t \) is the time taken.
  • \( s \) is the distance traveled.
In this situation, the rocket starts from rest, so \( u = 0 \). By applying these equations, we calculate the time needed to reach a velocity of 330 m/s (the speed of sound) and find out how far it has traveled by then.
Acceleration
Acceleration is the rate of change of velocity over time. It's a key factor in analyzing rocket motion and is often expressed in meters per second squared (m/s²). The acceleration of a rocket is crucial because it must overcome both the force of gravity and any resistance forces such as air resistance.

For the rocket, it’s important to note that the acceleration needs to be high enough to reach the desired velocity without exceeding the tension limits of components like wires inside. The exercise calculates the maximum permissible acceleration by considering the tensions acting on an instrument linked to the rocket.
  • Net Force: \( T = W + ma \)
  • Rearrange to find: \( a = \frac{T - W}{m} \)
Here, \( T \) is the maximum tension the wire can handle, \( W \) is the weight of the instrument, and \( m \) its mass.
Thrust
Thrust is the force that moves a rocket through the air and into space. It's the push exerted by the engines. For a rocket to accelerate upwards, its thrust must be greater than the gravitational pull pulling it downwards.

The thrust must not only provide the force needed to accelerate the rocket upwards, but it must also counteract the force of gravity acting on the rocket's mass. The formula linking thrust to force is:
  • Thrust Force: \( F_t = mg + ma \)
Where:
  • \( F_t \) is the thrust force produced by the engines.
  • \( m \) is the mass of the rocket.
  • \( g \) is the acceleration due to gravity.
  • \( a \) is the additional acceleration needed to exceed gravitational pull.
The calculation shows that sufficient thrust ensures the rocket's ascent at the given acceleration.
Force Analysis
Force analysis involves examining all the forces acting on the rocket to ensure each component works safely within its limits. Proper analysis is crucial to design rockets that can withstand forces during launch and flight.

In the context of this exercise, understanding forces is vital in calculating the maximum vertical thrust the rocket engines need to exert. Each force's contribution must be carefully analyzed as follows:
  • Gravitational Force: Acts downward, calculated by \( mg \).
  • Tension in the Wire: Must not exceed 35 N to avoid breaking.
  • Net Force for Acceleration: Derives from subtracting gravitational force from the total thrust.
For successful launch, ensure the net force can support necessary acceleration without exceeding structural limits. This force balance is central to understanding how the rocket's propulsion and structure interact.

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Most popular questions from this chapter

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