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Chapter 7: Problem 40

A can of sardines is made to move along an \(x\) axis from \(x=\) \(0.25 \mathrm{~m}\) to \(x=1.25 \mathrm{~m}\) by a force with a magnitude given by \(F=\) \(\exp \left(-4 x^{2}\right)\), with \(x\) in meters and \(F\) in newtons. (Here exp is the exponential function.) How much work is done on the can by the force?

Short Answer

Expert verified
The work done is approximately 0.243 joules.

Step by step solution

01

Understand the Problem

We need to calculate the work done on the can of sardines as it moves along the x-axis from \(x = 0.25 \, \text{m}\) to \(x = 1.25 \, \text{m}\) by a force \(F = \exp(-4x^2)\). The work done by a variable force along a linear path is given by the integral of the force function with respect to displacement.
02

Set Up the Integral

The work done by the force is given by the integral \[ W = \int_{x_1}^{x_2} F(x) \, dx \] where \(x_1 = 0.25\) m and \(x_2 = 1.25\) m. Substituting the force function, we have \[ W = \int_{0.25}^{1.25} \exp(-4x^2) \, dx. \]
03

Solve the Integral

Solving the integral \[ W = \int_{0.25}^{1.25} \exp(-4x^2) \, dx \] requires substitution or numerical methods, as it doesn't have a simple antiderivative. We can use a numerical integration method, such as Simpson's Rule or Trapezoidal Rule, or a calculator with integration capability to approximate the result. The integral evaluates approximately to 0.243.
04

Interpret the Result

The calculated work done, approximately 0.243 joules, represents the energy transferred to the can as it moves from \(x = 0.25 \, \text{m}\) to \(x = 1.25 \, \text{m}\) under the influence of the given force function.

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Key Concepts

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

Variable Force
Variable forces are forces that change in magnitude or direction as an object moves. Unlike constant forces, variable forces can vary based on the position of the object they act upon. In our exercise, the force acting on the can of sardines is given by the function \( F = \exp(-4x^2) \). This equation shows how the force changes with position \( x \).

Key points about variable forces include:
  • They depend on the position, requiring calculus for their analysis.
  • The force magnitude is not uniform across different points.
  • Understanding the specific relationship of force with position helps us calculate work done more accurately.
For an object experiencing a variable force, like our sardine can, calculating work requires integrating the force function over the distance the object travels.
Integration
Integration is a mathematical process used to find accumulations, such as areas under curves or total quantities, like work. In the context of physics, it integrates variable forces over a specified interval to find the work done by the force. To explore this, consider the integral:\[ W = \int_{0.25}^{1.25} \exp(-4x^2) \, dx \] This integral calculates the total work done from \( x = 0.25 \) to \( x = 1.25 \). Since the force equation \( \exp(-4x^2) \) isn't straightforward to integrate analytically, numerical methods are employed to approximate it.
  • Integration accumulates contributions from the force at each small segment of the path.
  • Setting up the integral involves identifying limits and the correct force function.
  • The solution provides a scalar value, representing total energy transfer.
Numerical Methods
When analytically solving integrals is challenging or impossible, numerical methods come in handy. These methods approximate the value of definite integrals when they do not have a simple antiderivative or are difficult to compute by hand.

There are several common numerical methods:
  • Trapezoidal Rule: Approximates the region under the curve as a series of trapezoids.
  • Simpson's Rule: Uses parabolic segments instead of straight lines for better accuracy.
  • A calculator or software can also be employed to compute complex integrals.
In our exercise, numerical methods help solve the integral \( \int_{0.25}^{1.25} \exp(-4x^2) \, dx \), providing an approximate work value of 0.243 joules.
Physics Concepts
Physics concepts such as work and energy form the basis for understanding the problem at hand. Work is a measure of energy transfer, and when a force acts on an object over a distance, it performs work.
The formula for work done by a constant force \( F \) is \( W = F \cdot d \), where \( d \) is the displacement. For a variable force, the integration approach sums up the small amounts of work done along each tiny part of the path.

Important physics principles here include:
  • Work: The force applied to move an object a certain distance.
  • Energy Transfer: Work results in energy transferred to or from an object.
  • Variable Forces: often found in real-world scenarios, require calculus for accurate calculations.
In this problem, calculating the work done involves integrating the variable force \( \exp(-4x^2) \), underscoring the importance of understanding both the physical and mathematical concepts involved.

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Most popular questions from this chapter

A ice block floating in a river is pushed through a displacement \(\vec{d}=(15 \mathrm{~m}) \hat{\mathrm{i}}-(12 \mathrm{~m}) \hat{\mathrm{j}}\) along a straight embankment by rushing water, which exerts a force \(\vec{F}=(210 \mathrm{~N}) \hat{\mathrm{i}}-(150 \mathrm{~N}) \hat{\mathrm{j}}\) on the block. How much work does the force do on the block during the displacement?

To push a \(25.0 \mathrm{~kg}\) crate up a frictionless incline, angled at \(25.0^{\circ}\) to the horizontal, a worker exerts a force of 209 N parallel to the incline. As the crate slides \(1.50 \mathrm{~m}\), how much work is done on the crate by (a) the worker's applied force, (b) the gravitational force on the crate, and (c) the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?

A force \(\vec{F}=(3.00 \mathrm{~N}) \hat{\mathrm{i}}+(7.00 \mathrm{~N}) \hat{\mathrm{j}}+(7.00 \mathrm{~N}) \hat{\mathrm{k}}\) acts on a \(2.00 \mathrm{~kg}\) mobile object that moves from an initial position of \(\vec{d}_{i}=(3.00 \mathrm{~m}) \hat{\mathrm{i}}-(2.00 \mathrm{~m}) \hat{\mathrm{j}}+(5.00 \mathrm{~m}) \hat{\mathrm{k}}\) to a final position of \(\vec{d}_{f}=-(5.00 \mathrm{~m}) \hat{\mathrm{i}}+(4.00 \mathrm{~m}) \hat{\mathrm{j}}+(7.00 \mathrm{~m}) \hat{\mathrm{k}}\) in \(4.00 \mathrm{~s}\). Find (a) the work done on the object by the force in the \(4.00 \mathrm{~s}\) interval, (b) the average power due to the force during that interval, and (c) the angle between vectors \(\vec{d}_{i}\) and \(\vec{d}_{f}\)

A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by \(1.0 \mathrm{~m} / \mathrm{s}\) and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?

A \(100 \mathrm{~kg}\) block is pulled at a constant speed of \(5.0\) \(\mathrm{m} / \mathrm{s}\) across a horizontal floor by an applied force of \(122 \mathrm{~N}\) directed \(37^{\circ}\) above the horizontal. What is the rate at which the force does work on the block?
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