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Chapter 7: Problem 13

A 17,000 -kg jet takes off from an aircraft carrier via a catapult (Fig. \(7-22 \mathrm{a}\) ). The gases thrust out from the jet's engines exert a constant force of \(130 \mathrm{kN}\) on the jet; the force exerted on the jet by the catapult is plotted in Fig. \(7-22 \mathrm{~b}\). Determine: (a) the work done on the jet by the gases expelled by its engines during launch of the jet; and \((b)\) the work done on the jet by the catapult during launch of the jet.

Short Answer

Expert verified
(a) \( 130,000 \times d \) Work by engines; (b) \( A \) Work by catapult.

Step by step solution

01

Convert the thrust force from kN to N

The force exerted by the gases is given as 130 kN. To convert this into Newtons (N), multiply by 1000 because 1 kN = 1000 N: \[ 130 \text{ kN} = 130,000 \text{ N} \]
02

Identify the distance traveled by the jet

We need to know the distance over which the force is applied to calculate work. Here, it is not provided, so we're assuming we know it or we'll refer to given or drawn diagrams in a full problem setup.
03

Find the work done by the engine's thrust

Work is calculated using the formula: \[ W = F \cdot d \cdot \cos(\theta) \]where \( F \) is the force applied, \( d \) is the distance over which the force acts, and \( \theta \) is the angle between the force and the direction of the motion. For engine thrusting in the direction of motion: \[ \theta = 0 \text{ degrees} \Rightarrow \cos(\theta) = 1 \] Thus, the work done by the gases is: \[ W_{\text{engine}} = 130,000 \times d \]
04

Analyze the force versus distance plot for the catapult

Assume we have a force vs. distance (work-energy) graph for the catapult. The work done by the catapult can be calculated as the area under the force-distance graph.
05

Calculate the work done by the catapult

Find the area under the force-distance graph using appropriate methods (e.g., area of rectangles, triangles, or trapezoids, using integration if required). Let's denote this area (or integration result) as \( A \). \[ W_{\text{catapult}} = A \]
06

Compile expressions or results

Summarize with: - The work done by the gases is \( 130,000 \times d \). - The work done by the catapult is \( A \). If distances or areas were missing but later obtained, plug these values in for final numerical answers.

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Key Concepts

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

Force and Motion
Force and motion are fundamental concepts in physics that explain how objects move and interact with forces.
Force is a push or pull upon an object resulting from its interaction with another object. Motion, on the other hand, refers to the change in position of an object over time. When a force is applied to an object, it causes the object to move, change direction, or deform. The relationship between force and motion is described by Newton's Laws of Motion. These laws help us understand:
  • How forces affect an object's motion and velocity.
  • The concept of inertia or an object's resistance to changes in its state of motion.
  • The principle of action and reaction, wherein forces always occur in pairs.
Understanding these concepts enables us to analyze various physics problems, like determining the work done on a jet by different forces during its launch.
The formula for work, which is a key component here, is given by:\[W = F \cdot d \cdot \cos(\theta)\]Where:
  • \(W\) is the work done,
  • \(F\) is the force applied,
  • \(d\) is the displacement over which the force is applied,
  • \(\theta\) is the angle between the force and the direction of motion.
Jet Propulsion
Jet propulsion involves the movement of a jet due to the forces produced by its engines. A jet engine works by expelling gases at high speed, generating a thrust in the opposite direction according to Newton's Third Law. This principle is pivotal in understanding how jets achieve motion and significant speed.An essential aspect of jet propulsion is the thrust force. This is the force exerted by the expelled gases on the jet, propelling it forward. In the context of a physics problem like our example, calculations often require us to determine the work done by this force. The thrust force and the work it performs can be represented and calculated with:
  • The conversion of force units for accurate calculations, like converting kilonewtons (kN) to newtons (N).
  • The application of formulas to find work, such as \( W = F \cdot d \), assuming the thrust is aligned with the motion.
Understanding jet propulsion helps in solving problems that involve calculating forces and understanding the mechanisms of motion involved in high-speed jet launches.
Physics Problems
Physics problems often require a solid understanding of concepts like force, motion, and the specific mechanisms involved. These problems typically involve applying physics principles to calculate unknowns. In the example of a jet's launch: The main tasks in solving physics problems of this nature include:
  • Identifying and converting the given forces into consistent units.
  • Understanding the motion involved and the forces causing this motion.
  • Using diagrams or force vs. distance plots to extract necessary data for calculations.
  • Applying formulae and mathematical concepts to find the work done by various forces.
In such problems, solving for the work done might involve:
  • Calculating the work from engine thrust, ensuring the force is along the direction of movement.
  • Using graphical methods, like finding the area under a force-distance graph, for calculating the work done by variable forces such as a catapult.
Physics problems enhance analytical skills, enabling students to dissect a problem, understand each component, and solve step by step. These problems can be challenging, but they provide a practical application of physics concepts to real-world scenarios.

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Most popular questions from this chapter

A cyclist starts from rest and coasts down a \(4.0^{\circ}\) hill. The mass of the cyclist plus bicycle is \(85 \mathrm{~kg} .\) After the cyclist has traveled \(250 \mathrm{~m},\) (a) what was the net work done by gravity on the cyclist? (b) How fast is the cyclist going? Ignore air resistance.

A constant force \(\overrightarrow{\mathbf{F}}=(2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}) \mathbf{N}\) acts on an object as it moves along a straight- line path. If the object's displacement is \(\overrightarrow{\mathbf{d}}=(1.0 \hat{\mathbf{i}}+5.0 \mathbf{j}) \mathrm{m},\) calculate the work done by \(\overrightarrow{\mathbf{F}}\) using these alternate ways of writing the dot product: (a) \(W=F d \cos \theta\) \(;\) (b) \(W=F_{x} d_{x}+F_{y} d_{y}\)

(I) One car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by \(7.0 \mathrm { m } / \mathrm { s } ,\) they then have the same kinetic energy. What were the original speeds of the two cars?

A train is moving along a track with constant speed \(v_{1}\) relative to the ground. A person on the train holds a ball of mass \(m\) and throws it toward the front of the train with a speed \(v_{2}\) relative to the train. Calculate the change in kinetic energy of the ball \((a)\) in the Earth frame of reference, and (b) in the train frame of reference. (c) Relative to each frame of reference, how much work was done on the ball? (d) Explain why the results in part \((b)\) are not the same for the two frames-after all, it's the same ball.

(III) We usually neglect the mass of a spring if it is small compared to the mass attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length \(\ell\) and mass \(M _ { \text { s uniformly } }\) distributed along the length of the spring. A mass \(m\) is attached to the end of the spring. One end of the spring is fixed and the mass \(m\) is allowed to vibrate horizontally without friction (Fig. \(30 ) .\) Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed \(v _ { 0 } ,\) the midpoint of the spring moves with speed \(v _ { 0 } / 2 .\) Show that the kinetic energy of the mass plus spring when the mass is moving with velocity \(v\) is \(K = \frac { 1 } { 2 } M v ^ { 2 }\) where \(M = m + \frac { 1 } { 3 } M _ { \mathrm { S } }\) is the "effective mass" of the system. [Hint: Let \(D\) be the total length of the stretched spring. Then the velocity of a mass \(d m\) of a spring of length \(d x\) located at \(x\) is \(v ( x ) = v _ { 0 } ( x / D ) .\) Note also that \(d m = d x \left( M _ { \mathrm { S } } / D \right) . ]\)
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