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Chapter 7: Problem 12

The force acting on a particle is \(F_{x}=(8 x-16) \mathrm{N},\) where \(x\) is in meters. (a) Make a plot of this force versus \(x\) from \(x=0\) to \(x=3.00 \mathrm{m} .\) (b) From your graph, find the net work done by this force on the particle as it moves from \(x=0\) to \(x=3.00 \mathrm{m}.\)

Short Answer

Expert verified
The net work done by the force on the particle as it moves from \(x=0\) to \(x=3.00 \mathrm{m}\) is \(-12 \mathrm{J}\)

Step by step solution

01

Sketch a graph of the force vs displacement

Firstly, the force-function \(F_{x}=(8 x-16) \mathrm{N}\) must be plotted against \(x\) in terms of meters from \(x=0\) up to \(x=3.00\) meters. This can be performed using any graph-plotting software or manually. The graph is a straight line starting at \(-16 \mathrm{N}\) at \(x=0\) and reaching \(8 \cdot 3.00 \mathrm{m} -16 = 8 \mathrm{N}\) at \(x=3.00 \mathrm{m}\).
02

Calculate the work done

The work done from \(x=0\) to \(x=3.00 \mathrm{m}\) is the area under the curve in the force vs displacement graph. Because the force is a linear function of displacement, the graph is a trapezoid, and its area can be calculated using the formula for the area of a trapezoid: \[ A = 0.5 * (base1 + base2) * height \] In this case, the bases are the forces at \(x=0\) and \(x=3.00 \mathrm{m}\), and the height is the displacement of \(3.00 \mathrm{m}\). Substituting the given values yields \( W = 0.5 * (-16 N + 8 N ) * 3.00 m \) .
03

Solve for work

Solve the above equation for work yields \(W = -12 \mathrm{J}\). This indicates that the net work done by the force on the particle is negative, meaning the force is doing work against the motion of the particle.

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Key Concepts

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

Force-Displacement Graph
A force-displacement graph is a visual tool used to illustrate how force varies with displacement. In the provided exercise, the force acting on a particle is given by the equation \( F_{x} = 8x - 16 \text{ N} \). Here, \( x \) represents the displacement in meters.

To understand how this force behaves, you can plot it on a graph with force values on the y-axis and displacement on the x-axis. For the given linear force equation:
  • When \( x = 0 \), \( F_{x} = -16 \text{ N} \).
  • When \( x = 3 \text{ m} \), \( F_{x} = 8 \cdot 3 - 16 = 8 \text{ N} \).
The graph is a straight line starting at \( -16 \text{ N} \) and ending at \( 8 \text{ N} \) over the interval from \( x = 0 \) to \( x = 3 \text{ m} \). This line shows how the force changes with displacement, being negative initially and increasing until it's positive.
Net Work Done
The net work done by a force on an object is the total energy transferred to that object by the force. It can be found by calculating the area under the force-displacement graph.

In this context, the graph forms a shape known as a trapezoid. The work done (or energy transferred) is the area of this trapezoid, which can be calculated using the formula:\[ A = 0.5 \times (\text{base1} + \text{base2}) \times \text{height} \]Here:
  • \( \text{base1} = -16 \text{ N} \) (force at \( x = 0 \))
  • \( \text{base2} = 8 \text{ N} \) (force at \( x = 3 \text{ m} \))
  • \( \text{height} = 3 \text{ m} \) (displacement)
Substituting these values results in:\[ W = 0.5 \times (-16 + 8) \times 3 = -12 \text{ J} \]This calculation shows that the net work is \(-12 \text{ J}\), indicating that work is done against the motion of the particle.
Linear Force Function
A linear force function describes how force varies with respect to displacement in a straight-line relationship. The equation \( F_{x} = 8x - 16 \text{ N} \) is an example of such a function, where:
  • \( 8 \) is the rate of change of force per unit of displacement.
  • \(-16 \) is the initial force value at zero displacement.

This linear relationship is characterized by a constant slope, indicating consistent changes in force with displacement.

When you plot this on a graph, it results in a straight line, meaning the force increases uniformly as the particle moves from \( x = 0 \) to \( x = 3 \text{ m} \). Understanding linear force functions helps in determining how force behaves linearly over a range of displacements, making it easier to predict and calculate work done, as seen in this exercise.

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Most popular questions from this chapter

A 40.0 -kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of \(130 \mathrm{N} .\) If the coefficient of friction between box and floor is \(0.300,\) find \((a)\) the work done by the applied force, (b) the increase in internal energy in the box- floor system due to friction, (c) the work done by the normal force, (d) the work done by the gravitational force, (e) the change in kinetic energy of the box, and (f) the final speed of the box.

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as \(1 \mathrm{kcal}=4186 \mathrm{J} .\) Metabolizing one gram of fat can release 9.00 kcal. A student decides to try to lose weight by exercising. She plans to run up and down the stairs in a football stadium as fast as she can and as many times as necessary. Is this in itself a practical way to lose weight? To evaluate the program, suppose she runs up a flight of 80 steps, each \(0.150 \mathrm{m}\) high, in \(65.0 \mathrm{s}\). For simplicity, ignore the energy she uses in coming down (which is small). Assume that a typical efficiency for human muscles is \(20.0 \%\) This means that when your body converts \(100 \mathrm{J}\) from metabolizing fat, \(20 \mathrm{J}\) goes into doing mechanical work (here, climbing stairs). The remainder goes into extra internal energy. Assume the student's mass is \(50.0 \mathrm{kg} .\) (a) How many times must she run the flight of stairs to lose one pound of fat? (b) What is her average power output, in watts and in horsepower, as she is running up the stairs?

A 650 -kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of \(1.75 \mathrm{m} / \mathrm{s} .\) (a) What is the average power of the elevator motor during this period? (b) How does this power compare with the motor power when the elevator moves at its cruising speed?

A \(2.00-\mathrm{kg}\) block is attached to a spring of force constant \(500 \mathrm{N} / \mathrm{m}\) as in Figure \(7.10 .\) The block is pulled \(5.00 \mathrm{cm}\) to the right of equilibrium and released from rest. Find the speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is 0.350

You can think of the work-kinetic energy theorem as a second theory of motion, parallel to Newton's laws in describing how outside influences affect the motion of an object. In this problem, solve parts (a) and (b) separately from parts \((\mathrm{c})\) and \((\mathrm{d})\) to compare the predictions of the two theories. In a rifle barrel, a \(15.0-\mathrm{g}\) bullet is accelerated from rest to a speed of \(780 \mathrm{m} / \mathrm{s}\). (a) Find the work that is done on the bullet. (b) If the rifle barrel is \(72.0 \mathrm{cm}\) long, find the magnitude of the average total force that acted on it, as \(F=W /(\Delta r \cos \theta) .\) (c) Find the constant acceleration of a bullet that starts from rest and gains a speed of \(780 \mathrm{m} / \mathrm{s}\) over a distance of \(72.0 \mathrm{cm} .\) (d) If the bullet has mass \(15.0 \mathrm{g},\) find the total force that acted on it as \(\Sigma F=m a.\)
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