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A force \(\vec{F}\) in the positive direction of an \(x\) axis acts on an object moving along the axis. If the magnitude of the force is \(F=\) \(10 e^{-x / 2.0} \mathrm{~N}\), with \(x\) in meters, find the work done by \(\vec{F}\) as the object moves from \(x=0\) to \(x=2.0 \mathrm{~m}\) by \((\) a) plotting \(F(x)\) and estimating the area under the curve and (b) integrating to find the work analytically.

Short Answer

Expert verified
The work done by the force is approximately 7.43 Joules.

Step by step solution

01

Understanding the Force Function

The force acting on the object is a function of position given by \( F(x) = 10 e^{-x/2} \). This function is an exponential function where the force decreases as \( x \) increases.
02

Plotting the Force Function

To estimate the work done graphically, we first sketch the graph of \( F(x) = 10 e^{-x/2} \) over the interval \( x = 0 \) to \( x = 2 \). Notice that \( F(0) = 10 \) and \( F(2) \approx 6.065 \), showing a decreasing trend.
03

Estimating Area Under the Curve

Estimate the area under the curve on the graph from \( x = 0 \) to \( x = 2 \) using methods like counting squares on graph paper or using a trapezoidal approximation. This area represents the work done by the force.
04

Setting Up the Integral for Work

Work done \( W \) by a variable force \( F(x) \) over an interval [\( a,b \)] is calculated using the integral \( W = \int_{a}^{b} F(x) \ dx \). In this case, \( a = 0 \) and \( b = 2 \), thus \( W = \int_{0}^{2} 10 e^{-x/2} \ dx \).
05

Solving the Integral

The integral \( \int 10 e^{-x/2} \ dx \) can be calculated by substituting \( u = -x/2 \), giving \( du = -1/2 \, dx \), or \( dx = -2 \, du \). The integral becomes \(-20 \int e^{u} \, du = -20 e^{u} + C = -20 e^{-x/2} + C \).
06

Evaluating the Integral

Evaluate the definite integral from \( x = 0 \) to \( x = 2 \): \(-20 e^{-x/2} \bigg|_0^2 = [-20 e^{-1}] - [-20 e^{0}] = -20 \times \frac{1}{e} + 20 = 20(1 - \frac{1}{e}) \approx 7.43 \) Joules.

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Key Concepts

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

Exponential Force Function
When dealing with real-world forces, they often come in complex forms. One such pattern is the exponential force function seen as \( F(x) = 10 e^{-x/2} \). Here, the force is not a straightforward constant but decreases exponentially as the position \( x \) increases.
This behavior is due to the nature of the exponential function, which rapidly decreases in value for increasing values of \( x \). In the given function, the force is at its maximum when \( x = 0 \), where \( F(x) = 10 \) Newtons, and it continuously declines as \( x \) grows larger.
Exponential functions are significant in physics as they can model scenarios where a quantity reduces at a rate proportional to its current value, like radioactive decay or charging of a capacitor.
Definite Integral Calculation
The notion of using a definite integral to find work comes from the requirement to sum up an infinite number of tiny works over a continuous path. For a variable force \( F(x) \), work \( W \) is given by the integral \( W = \int_{a}^{b} F(x) \; dx \).
  • This integral essentially sums up the force-values over an interval to give total work completed.
  • For our case, it is \( W = \int_{0}^{2} 10 e^{-x/2} \; dx \).

Solving these integrals often involves substitution when functions are in forms like an exponential decay. Substitution simplifies the integral calculation without changing the bounds until evaluation. With \( u = -x/2 \), the expression transforms to \( -20 \int e^u \; du \) leading to an integrated form of \( -20 e^{u} + C \).
Evaluating from the lower to upper bounds \( x = 0 \) to \( x = 2 \) provides us the work done by the force.
Graphical Estimation of Work
Sometimes, complex calculations can be verified through graphical methods. Graphing the force function \( F(x) = 10 e^{-x/2} \) helps visualize the force's behavior and estimate work done by considering the area under the force curve between \( x = 0 \) and \( x = 2 \).
To estimate this area:
  • Sketch the force curve over the interval.
  • Use methods like trapezoidal approximation or counting grid squares to approximate the area, which represents work.

The resulting area estimate provides insight into accuracy and order of magnitude for the integral result, helping us strengthen our understanding of how force changes impact work outcome.
Physics Problem Solving
Physics problems, like calculating work done by a variable force, require multiple skills and approaches.
  • Visualizing the concepts - Reading and plotting the function helps grasp force behavior.
  • Mathematical calculations - Recognize when to use integrals over simple arithmetic for variable forces.

Understanding the steps of solving physics problems involves knowing the laws and applying the right mathematical techniques. Additionally, validating answers through alternative methods, like graphical estimation, plays a crucial role in reinforcing comprehension and gaining confidence in solution accuracy.
This approach has real-world importance, as it aids in tackling complex systems not easily solved by one method alone.

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Most popular questions from this chapter

74 A particle moves along a straight path through displacement \(\vec{d}=(8 \mathrm{~m}) \hat{\mathrm{i}}+c \hat{\mathrm{j}}\) while force \(\vec{F}=(2 \mathrm{~N}) \hat{\mathrm{i}}-(4 \mathrm{~N}) \hat{\mathrm{j}}\) acts on it. (Other forces also act on the particle.) What is the value of \(c\) if the work done by \(\vec{F}\) on the particle is (a) zero, (b) positive, and (c) negative?

(a) At a certain instant, a particle-like object is acted on by a force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}-(2.0 \mathrm{~N}) \hat{\mathrm{j}}+(9.0 \mathrm{~N}) \hat{\mathrm{k}}\) while the object's veloc- ity is \(\vec{v}=-(2.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\). What is the instantaneous rate at which the force does work on the object? (b) At some other time, the velocity consists of only a \(y\) component. If the force is unchanged and the instantaneous power is \(-12 \mathrm{~W}\), what is the velocity of the object?

A horse pulls a cart with a force of \(40 \mathrm{lb}\) at an angle of \(30^{\circ}\) above the horizontal and moves along at a speed of \(6.0 \mathrm{mi} / \mathrm{h} .(\) a) How much work does the force do in \(10 \mathrm{~min} ?(\mathrm{~b})\) What is the average power (in horsepower) of the force?

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 \(\vec{d}_{f}\)

If a car of mass \(1200 \mathrm{~kg}\) is moving along a highway at \(120 \mathrm{~km} / \mathrm{h}\), what is the car's kinetic energy as determined by someone standing alongside the highway?

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