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Chapter 13: Problem 14

An airplane flies in the positive direction along the \(x\) axis, through crosswinds that exert a force \(\mathbf{F}=(a+b x) \hat{\mathbf{x}}+(c+d x) \hat{\mathbf{y}}\). Find the work done by the wind on the plane, and by the plane on the wind, in traveling from the origin to position \(x \)

Short Answer

Expert verified
Work done by wind: \(ax + \frac{b x^2}{2}\); by plane: \(-(ax + \frac{b x^2}{2})\).

Step by step solution

01

Understanding Work Done by Force

Work done by a force is calculated using the integral of the dot product of force \(\mathbf{F}\) and displacement \(d\mathbf{r}\). Here, the displacement is only along the \(x\) axis, i.e., \(d\mathbf{r} = dx \, \hat{\mathbf{x}}\). Thus, the work done \(W\) is given by: \[ W = \int_{x_0}^{x} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{x} [(a + bx) \hat{\mathbf{x}} + (c + dx) \hat{\mathbf{y}}] \cdot \hat{\mathbf{x}} \, dx\] Since \(\hat{\mathbf{y}} \cdot \hat{\mathbf{x}} = 0\), only the component along \(\hat{\mathbf{x}}\) matters for this calculation.
02

Calculate Work Done by the Wind

Substitute the force component along \(\hat{\mathbf{x}}\) into the work integral:\[ W = \int_{0}^{x} (a + bx) \, dx \]Now, we perform the integration:\[ W = \left[ ax + \frac{b x^2}{2} \right]_{0}^{x} \]Simplify the expression by evaluating it at the limits:\[ W = ax + \frac{b x^2}{2} \]
03

Work Done by Plane on the Wind

The work done by the plane on the wind is equal and opposite to the work done by the wind on the plane, according to Newton's third law. Therefore, the work done by the plane on the wind is:\[ W' = - (ax + \frac{b x^2}{2}) \]

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Key Concepts

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

Force and Displacement
When examining work done by forces, it's crucial to understand the relationship between force and displacement. Work is essentially a measure of energy transfer when an object is moved by a force. In our example, a plane travels along the
  • x-axis - meaning its displacement is horizontal.
  • Crosswinds apply force - this force has two components, one along the \(\hat{\mathbf{x}}\) direction and another along \(\hat{\mathbf{y}}\).

The work done, derived from the force and the displacement, is defined by the dot product of these two vectors. In our specific problem, focus is given to one-dimensional horizontal travel along the x-axis.
This means the relevant displacement vector is: \(d\mathbf{r} = dx \, \hat{\mathbf{x}}\).
Since the force's \(\hat{\mathbf{y}}\) component doesn't contribute to the horizontal movement, only the \(\hat{\mathbf{x}}\) component affects the work done. Hence, integration simplifies along just the x-axis.
Integration in Physics
Integration is a powerful tool in physics, used for calculating quantities that are distributed over a continuous range. In our scenario, it helps account for the force acting along varying positions on the x-axis.
Force is represented as a function \((a + bx)\) in the x direction. Hence, to find the work done by this force as displacement varies, we integrate over the span from 0 to x.
The mathematical procedure involves calculating:
  • Formulate Integral: Work is determined by integrating: \[ W = \int_{0}^{x} (a + bx) \, dx \]
  • Evaluate: Performing the integration results in: \[ W = \left[ ax + \frac{b x^2}{2} \right]_{0}^{x} \]
  • Simplify: Evaluate this at the limits to find: \[ W = ax + \frac{b x^2}{2} \]

This process gives a clear measure of work done by the wind on the plane across its journey, showing how both constant and variable components of force contribute to overall work.
Newton's Third Law
Newton's Third Law is key to understanding the interactions of forces. It states that every action has an equal and opposite reaction. In our context, this law applies to the forces exchanged between the plane and the wind.
The wind pushes on the plane, doing work.
  • Wind's Work on Plane: The formulation \( ax + \frac{b x^2}{2} \) describes this work done by the wind as it moves the plane in the x-direction.
  • Plane's Reaction on Wind: Following Newton鈥檚 Third Law, the plane exerts an opposing force on the wind. Consequently, the work done by the plane on the wind is the negative of the wind's work, represented by: \[ W' = - \left(ax + \frac{b x^2}{2}\right) \]

This demonstrates the reciprocal nature of forces. Even as the plane receives energy from the wind, it delivers an equal amount back as a reactionary force.

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Most popular questions from this chapter

A car accelerates from rest. At low speeds, its acceleration is limited by static friction, so that if we press too hard on the gas, we will "burn rubber" (or, for many newer cars, a computerized traction-control system will override the gas pedal). At higher speeds, the limit on acceleration comes from the power of the engine, which puts a limit on how fast kinetic energy can be developed. (a) Show that if a force \(F\) is applied to an object moving at speed \(v\), the power required is given by \(P=v F\). (b) Find the speed \(v\) at which we cross over from the first regime described above to the second. At speeds higher than this, the engine does not have enough power to burn rubber. Express your result in terms of the car's power \(P\), its mass \(m\), the coefficient of static friction \(\mu_{s}\), and \(g\). (c) Show that your answer to part b has units that make sense. (d) Show that the dependence of your answer on each of the four variables makes sense physically. (e) The 2010 Maserati Gran Turismo Convertible has a maximum power of \(3.23 \times 10^{5} \mathrm{~W}\) (433 horsepower) and a mass (including a 50\(\mathrm{kg}\) driver ) of \(2.03 \times 10^{3} \mathrm{~kg}\). (This power is the maximum the engine can supply at its optimum frequency of 7600 r.p.m. Presumably the automatic transmission is designed so a gear is available in which the engine will be running at very nearly this frequency when the car is moving at \(v .\) ) Rubber on asphalt has \(\mu_{s} \approx 0.9\). Find \(v\) for this car. Answer: \(18 \mathrm{~m} / \mathrm{s}\), or about 40 miles per hour. (f) Our analysis has neglected air friction, which can probably be approximated as a force proportional to \(v^{2} .\) The existence of this force is the reason that the car has a maximum speed, which is 176 miles per hour. To get a feeling for how good an approximation it is to ignore air friction, find what fraction of the engine's maximum power is being used to overcome air resistance when the car is moving at the speed \(v\) found in part e. Answer: \(1 \%\)

In the power stroke of a car's gasoline engine, the fuel-air mixture is ignited by the spark plug, explodes, and pushes the piston out. The exploding mixture's force on the piston head is greatest at the beginning of the explosion, and decreases as the mixture expands. It can be approximated by \(F=a / x\), where \(x\) is the distance from the cylinder to the piston head, and \(a\) is a constant with units of \(\mathrm{N} \cdot \mathrm{m}\). (Actually \(a / x^{1.4}\) would be more accurate, but the problem works out more nicely with \(a / x !\) ) The piston begins its stroke at \(x=x_{1}\), and ends at \(x=x_{2}\). The 1965 Rambler had six cylinders, each with \(a=220 \mathrm{~N} \cdot \mathrm{m}, x_{1}=1.2 \mathrm{~cm}\), and \(x_{2}=10.2 \mathrm{~cm}\) (a) Draw a neat, accurate graph of \(F\) vs \(x\), on graph paper. (b) From the area under the curve, derive the amount of work done in one stroke by one cylinder. (c) Assume the engine is running at 4800 r.p.m., so that during one minute, each of the six cylinders performs 2400 power strokes. (Power strokes only happen every other revolution.) Find the engine's power, in units of horsepower \((1 \mathrm{hp}=746 \mathrm{~W})\) (d) The compression ratio of an engine is defined as \(x_{2} / x_{1}\). Explain in words why the car's power would be exactly the same if \(x_{1}\) and \(x_{2}\) were, say, halved or tripled, maintaining the same compression ratio of \(8.5 .\) Explain why this would not quite be true with the more realistic force equation \(F=a / x^{1.4}\).

A certain binary star system consists of two stars with masses \(m_{1}\) and \(m_{2}\), separated by a distance \(b\). A comet, originally nearly at rest in deep space, drops into the system and at a certain point in time arrives at the midpoint between the two stars. For that moment in time, find its velocity, \(v\), symbolically in terms of \(b, m_{1}\), \(m_{2}\), and fundamental constants.

In 1935 , Yukawa proposed an early theory of the force that held the neutrons and protons together in the nucleus. His equation for the potential energy of two such particles, at a center-tocenter distance \(r\), was \(P E(r)=g r^{-1} e^{-r / a}\), where \(g\) parametrizes the strength of the interaction, \(e\) is the base of natural logarithms, and \(a\) is about \(10^{-15} \mathrm{~m}\). Find the force between two nucleons that would be consistent with this equation for the potential energy.

Does it make sense to say that work is conserved?
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