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Chapter 8: Problem 26

A conservative force \(\vec{F}=(6.0 x-12) \hat{i} \mathrm{~N}\), where \(x\) is in meters, acts on a particle moving along an \(x\) axis. The potential energy \(U\) associated with this force is assigned a value of \(27 \mathrm{~J}\) at \(x=0\). (a) Write an expression for \(U\) as a function of \(x\), with \(U\) in joules and \(x\) in meters. (b) What is the maximum positive potential energy? At what (c) negative value and (d) positive value of \(x\) is the potential energy equal to zero?

Short Answer

Expert verified
a) \(-3x^2 + 12x + 27\); b) 39 J at \(x=2\); c) \(x \approx -1.606\); d) \(x \approx 5.605\).

Step by step solution

01

Understand the Relationship Between Force and Potential Energy

When a force is conservative, the potential energy, \( U \), associated with it is related through the negative integral of the force. In mathematical terms, \( U(x) = -\int F(x) \, dx + C \), where \( C \) is the integration constant.
02

Integrate the Force to Find Potential Energy

For the given force \( \vec{F} = (6.0x - 12) \hat{i} \, \mathrm{~N} \), integrate to find the potential energy:\[U(x) = -\int (6x - 12) \, dx = -(3x^2 - 12x) = -3x^2 + 12x + C.\]
03

Apply the Initial Condition to Find the Integration Constant

We know \( U(0) = 27 \, \mathrm{J} \). Substituting \( x = 0 \) into the equation:\[U(0) = -3(0)^2 + 12(0) + C = 27.\]Therefore, \( C = 27 \).
04

Write the Full Expression for Potential Energy

Substituting \( C = 27 \) back into the equation, we get:\[U(x) = -3x^2 + 12x + 27.\]
05

Find the Maximum Positive Potential Energy

To find the maximum potential energy, take the derivative of \( U(x) \) and set it to zero to find critical points:\[\frac{dU}{dx} = -6x + 12 = 0.\]Solve for \( x \):\[x = 2.\]Substituting \( x = 2 \) back into the potential energy expression:\[U(2) = -3(2)^2 + 12(2) + 27 = 39 \, \mathrm{J}.\]
06

Find Negative and Positive Values of x for Zero Potential Energy

Set the potential energy \( U(x) = 0 \) and solve for \( x \):\[-3x^2 + 12x + 27 = 0.\]Solve the quadratic equation using the quadratic formula where \( a = -3, \, b = 12, \, c = 27 \):\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-12 \pm \sqrt{12^2 - 4(-3)(27)}}{-6}.\]Calculate the discriminant:\[12^2 - 4(-3)(27) = 144 + 324 = 468.\]Substitute back into formula:\[x = \frac{-12 \pm \sqrt{468}}{-6}.\]Solve for the two values:\[\sqrt{468} \approx 21.63.\]Thus:\[x = \frac{-12 + 21.63}{-6} \approx -1.606, \quad x = \frac{-12 - 21.63}{-6} \approx 5.605.\]
07

Summarize the Results

a) The expression for potential energy is \( U(x) = -3x^2 + 12x + 27 \).b) The maximum positive potential energy occurs at \( x = 2 \) with \( U = 39 \, \mathrm{J} \).c) The negative value of \( x \) where \( U = 0 \) is \( x \approx -1.606 \).d) The positive value of \( x \) where \( U = 0 \) is \( x \approx 5.605 \).

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Key Concepts

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

Potential Energy Equation
The potential energy equation links force with potential energy when dealing with a conservative force. If you have a conservative force, like in our exercise where \( \vec{F} = (6.0x - 12) \hat{i} \mathrm{~N} \), you can find the potential energy \( U \) using the formula: \( U(x) = -\int F(x) \, dx + C \). This equation means that potential energy is the negative integral of the force function plus an integration constant (\( C \)).
Potential energy represents stored energy, which can be converted to kinetic energy. The goal is to find an expression for potential energy \( U(x) \). We carry out the integration of the force to discover the specific form of \( U(x) \), keeping in mind that \( C \) will be determined by the initial conditions.
Integration Constant
The integration constant is vital in forming the full potential energy equation. After integrating the force function, we always have an additional term: the constant \( C \). This constant represents the starting potential energy when the position \( x \) is known. Applying the initial condition to the resulting potential energy expression allows us to calculate \( C \).
In this context, we were given that \( U(0) = 27 \) J when \( x = 0 \). By substituting \( x = 0 \) into the integrated potential energy equation \( U(x) = -3x^2 + 12x + C \), and setting the expression equal to 27 J, we found that \( C = 27 \).
This step ensures the potential energy formula is uniquely tailored to the conditions of our problem.
Maximum Potential Energy
Maximum potential energy is the highest energy stored due to the conservative force and occurs at a specific position \( x \). To find this, we derive the potential energy equation and set the derivative to zero. This process finds critical points where potential energy can be at its maximum or minimum.
For \( U(x) = -3x^2 + 12x + 27 \), the derivative is \( \frac{dU}{dx} = -6x + 12 \). Setting this result to zero gives \( x = 2 \).
Substituting \( x = 2 \) back into the potential energy expression results in the maximum potential energy value of \( U(2) = 39 \) J. Understanding how to locate this peak value is crucial, as it signifies where the greatest potential energy transformation can occur.
Zero Potential Energy Points
Zero potential energy points are specific values of \( x \) where the potential energy becomes zero. To find these points, solve the potential energy equation \( U(x) = 0 \). This involves using the quadratic formula for expressions like \( -3x^2 + 12x + 27 = 0 \).
The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) helps determine the values of \( x \) where potential energy equals zero. For this equation, \( a = -3 \), \( b = 12 \), and \( c = 27 \).
Calculating gives us two points: \( x \approx -1.606 \) and \( x \approx 5.605 \). These points signify where the stored energy prediction changes, crucial for analyzing ranges where different forces are at play.

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Most popular questions from this chapter

A \(5.0 \mathrm{~g}\) marble is fired vertically upward using a spring gun. The spring must be compressed \(8.0 \mathrm{~cm}\) if the marble is to just reach a target \(20 \mathrm{~m}\) above the marble's position on the compressed spring. (a) What is the change \(\Delta U_{g}\) in the gravitational potential energy of the marble-Earth system during the \(20 \mathrm{~m}\) ascent? (b) What is the change \(\Delta U_{s}\) in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

A stone with a weight of \(5.29 \mathrm{~N}\) is launched vertically from ground level with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\), and the air drag on it is \(0.265 \mathrm{~N}\) throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

A block of mass \(6.0 \mathrm{~kg}\) is pushed up an incline to its top by a man and then allowed to slide down to the bottom. The length of incline is \(10 \mathrm{~m}\) and its height is \(5.0 \mathrm{~m}\). The coefficient of friction between block and incline is \(0.40\). Calculate (a) the work done by the gravitational force over the complete round trip of the block, (b) the work done by the man during the upward journey, (c) the mechanical energy loss due to friction over the round trip, and (d) the speed of the block when it reaches the bottom.

At a lunar base, a uniform chain hangs over the edge of a horizontal platform. A machine does \(1.0 \mathrm{~J}\) of work in pulling the rest of the chain onto the platform. The chain has a mass of \(2.0\) \(\mathrm{kg}\) and a length of \(3.0 \mathrm{~m}\). What length was initially hanging over the edge? On the Moon, the gravitational acceleration is \(1 / 6\) of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\).

Tarzan, who weighs \(688 \mathrm{~N}\), swings from a cliff at the end of a vine \(18 \mathrm{~m}\) long (Fig. 8-28). From the top of the cliff to the bottom of the swing, he descends by \(3.2 \mathrm{~m}\). The vine will break if the force on it exceeds \(950 \mathrm{~N}\) (a) Does the vine break? (b) If it does not, what is the greatest force on it during the swing? If it does, at what angle with the vertical does it break?
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