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Chapter 8: Problem 58

The potential energy function for a system is given by \(U(x)=-x^{3}+2 x^{2}+3 x\) (a) Determine the force \(F_{x}\) as a function of \(x\) (b) For what values of \(x\) is the force equal to zero? (c) Plot \(U(x)\) versus \(x\) and \(F_{x}\) versus \(x,\) and indicate points of stable and unstable equilibrium.

Short Answer

Expert verified
The force function is \(F_x=3x^{2}-4x-3\). It is zero at \(x=-0.4244\) and \(x=2.4244\). The first point represents unstable equilibrium, and the second point represents stable equilibrium.

Step by step solution

01

Compute the Force Function

The force experienced by a system is given by the negative derivative of the potential energy function. It can be computed using formula: \(F_x=-\frac{d}{dx} U(x)\). Therefore, differentiating the given potential energy function \(U(x)\) with respect to \(x\), will give the force function: \(F_x=-\frac{d}{dx}(-x^{3}+2x^{2}+3x)=3x^{2}-4x-3\).
02

Determine Where the Force Is Zero

The force function \(F_x\) equals zero at points of stable or unstable equilibrium. These can be found by setting \(F_x=0\) and solving the resulting equation. Therefore, solving the equation \(3x^{2}-4x-3=0\) gives two solutions: \(x_{1,2}=\frac{4\pm\sqrt{4^2-4*3*(-3)}}{2*3}\), approximately \(x_1=-0.4244\) and \(x_2=2.4244 \).
03

Plotting \(U(x)\) and \(F_x\)

Plot the functions \(U(x)=-x^{3}+2x^{2}+3x\) and \(F_x=3x^{2}-4x-3\) over a convenient range of \(x\) values, like \(-2鈮鈮�5\). Points where \(F_x=0\) (found in the previous step) signify equilibrium. If the force changes sign from negative to positive at this point (indicating a change from repulsion to attraction), then the point is one of stable equilibrium. If the force changes from positive to negative (indicating a change from attraction to repulsion), then the point is one of unstable equilibrium.
04

Discuss Equilibrium Points

The plots reveal that the two solutions found \(x=-0.4244\) and \(x=2.4244\) represent a point of unstable and stable equilibrium respectively, as the force function changes its sign from positive to negative at \(x=-0.4244\) and from negative to positive at \(x=2.4244\).

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Key Concepts

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

Equilibrium Points
In physics, equilibrium points are key to understanding the stability and potential motion of a system. An equilibrium point is where the total force acting on an object is zero, indicating no net change in motion. In terms of potential energy, equilibrium points occur where the graph of the potential energy function, U(x), has a horizontal tangent鈥攎eaning its derivative with respect to position (dU/dx) is zero. This condition unveils the presence of stable or unstable equilibrium.

When you're dealing with a potential energy function like U(x) = -x^3 + 2x^2 + 3x, the equilibrium points are found by setting its derivative to zero and solving for x. This corresponds to the points where the force (F_x) is zero since force is the negative gradient of potential energy. If the slope of the potential energy graph goes from negative to positive at these points, it indicates a valley鈥攖his is a stable equilibrium because if the system is perturbed, it will return to this point. Conversely, if the slope goes from positive to negative, it indicates a hilltop鈥攁 unstable equilibrium, as any displacement will lead the system to move away from this point. These concepts are essential as they define where an object at rest may stay at rest (stable) or not (unstable).
Force Function
The force function, often denoted as F(x), plays a central role in dynamics and is intimately linked with potential energy. The force acting on an object is defined as the negative derivative of its potential energy with respect to position. Algebraically expressed as F(x) = -dU/dx.

To understand why the negative sign is important, imagine you're on a hill. The direction where you would naturally roll downhill is the direction of increasing potential energy. Conversely, the force you experience鈥攇ravity鈥攑ulls you in the opposite direction, hence the negative sign. Therefore, when the potential energy function for a system is given, such as U(x) = -x^3 + 2x^2 + 3x, calculating its derivative with respect to x reveals the force function: F(x) = -(-3x^2 + 4x + 3) or F(x) = 3x^2 - 4x - 3. Identifying where this force function is zero will point us to the equilibrium points which are so crucial for the system's behavior.
Plotting Potential Energy
To visualize the behavior of a system, plotting potential energy as a function of position can be highly informative. A graph of U(x) will depict how potential energy varies across different positions. This isn't just a snapshot of energy values; it's a roadmap of the system's possible motions and stable configurations.

By plotting the potential energy function given by, say, U(x) = -x^3 + 2x^2 + 3x, we can observe the shape of the function's graph. Peaks and valleys will represent unstable and stable equilibrium points, respectively. Similarly, plotting the force function alongside (F(x) = 3x^2 - 4x - 3) helps us see where the force is zero (equilibrium points) and whether the equilibrium is stable or unstable based on the slope change at those points.

Graphical visualization is an indispensable tool in physics because it translates complicated algebraic expressions into a picture that can often be understood intuitively. Recognizing the connection between the graph's features and physical properties like equilibrium provides a deeper comprehension of the system's dynamics.

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Most popular questions from this chapter

A potential-energy function for a two-dimensional force is of the form \(U=3 x^{3} y-7 x\). Find the force that acts at the point \((x, y)\).

The world's biggest locomotive is the MK5000C, a behemoth of mass 160 metric tons driven by the most powerful engine ever used for rail transportation, a Caterpillar diesel capable of 5000 hp. Such a huge machine can provide a gain in efficiency, but its large mass presents challenges as well. The engineer finds that the locomotive handles differently from conventional units, notably in braking and climbing hills. Consider the locomotive pulling no train, but traveling at \(27.0 \mathrm{m} / \mathrm{s}\) on a level track while operating with output power \(1000 \mathrm{hp} .\) It comes to a \(5.00 \%\) grade (a slope that rises \(5.00 \mathrm{m}\) for every \(100 \mathrm{m}\) along the track). If the throttle is not advanced, so that the power level is held steady, to what value will the speed drop? Assume that friction forces do not depend on the speed.

In her hand a softball pitcher swings a ball of mass \(0.250 \mathrm{kg}\) around a vertical circular path of radius \(60.0 \mathrm{cm}\) before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude \(30.0 \mathrm{N}\) in the direction of motion around the complete path. The speed of the ball at the top of the circle is \(15.0 \mathrm{m} / \mathrm{s} .\) If she releases the ball at the bottom of the circle, what is its speed upon release?

A \(1.50-\mathrm{kg}\) object is held \(1.20 \mathrm{m}\) above a relaxed massless vertical spring with a force constant of \(320 \mathrm{N} / \mathrm{m} .\) The object is dropped onto the spring. (a) How far does it compress the spring? (b) What If? How far does it compress the spring if the same experiment is performed on the Moon, where \(g=1.63 \mathrm{m} / \mathrm{s}^{2} ?\) (c) What If? Repeat part (a), but this time assume a constant air-resistance force of \(0.700 \mathrm{N}\) acts on the object during its motion.

A block of mass \(0.250 \mathrm{kg}\) is placed on top of a light vertical spring of force constant \(5000 \mathrm{N} / \mathrm{m}\) and pushed downward so that the spring is compressed by \(0.100 \mathrm{m}\). After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?
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