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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 \( F_x = 3x^2 - 4x - 3 \). The force equals zero at the roots of the quadratic equation \( 3x^2 - 4x - 3 = 0 \). Stable and unstable equilibrium points can be observed in the plot of \( U(x) \) versus \( x \) at the local minima and maxima, respectively, where the force crosses zero.

Step by step solution

01

Determine the force as a function of position

The force as a function of position can be found by taking the negative derivative of the potential energy function. The force is given by the equation: \( F_x = -\frac{dU}{dx} \). So, calculate \( F_x \) by differentiating \( U(x) = -x^{3} + 2x^{2} + 3x \).
02

Calculate the derivative of the potential energy

Differentiate \( U(x) \) with respect to \( x \) to get \( F_x \). The derivative of \( -x^3 \) is \( -3x^2 \), the derivative of \( 2x^2 \) is \( 4x \), and the derivative of \( 3x \) is \( 3 \). Summing these up, we get \( F_x = -(-3x^2 + 4x + 3) = 3x^2 - 4x - 3 \).
03

Find the values of x where the force is zero

To find the values of \( x \) for which the force is zero, solve the equation \( F_x = 3x^2 - 4x - 3 = 0 \). This is a quadratic equation which can be solved using factorization, completing the square, or the quadratic formula.
04

Solve the quadratic equation for x

Applying the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 3 \), \( b = -4 \), and \( c = -3 \), we get the two solutions for \( x \).
05

Plot U(x) versus x and Fx versus x

Plotting the graphs of \( U(x) \) and \( F_x \) versus \( x \) will require graphing software or paper and pencil. Points of stable equilibrium correspond to minima on the \( U(x) \) graph, where \( F_x = 0 \), and unstable equilibrium points correspond to maxima.

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Key Concepts

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

Force as a Function of Position
Understanding how force changes as an object moves is crucial in physics. In the context of potential energy, the force is directly related to the object's position within a field, such as a gravitational or electric field. The potential energy function, denoted as \(U(x)\), describes the energy stored due to the object's position.

The force exerted by the potential energy field is equal to the negative gradient (slope) of the potential energy function with respect to position. Mathematically, force \(F_{x}\) can be calculated by taking the negative derivative of the potential energy function \(U(x)\) with respect to \(x\). This relationship is given by the formula: \( F_x = -\frac{dU}{dx} \).

Illustrative Example

For our potential energy function \(U(x) = -x^{3} + 2x^{2} + 3x\), differentiating with respect to \(x\) yields the force function \(F_x = 3x^2 - 4x - 3\). Here, the force depends on the position \(x\) following a polynomial relationship. As \(x\) changes, the force experienced by the object also changes according to this function.
Differentiation in Physics
Differentiation is a powerful tool in physics that helps us analyze how a function changes at any given point. For potential energy functions, differentiation allows us to find the rate at which the potential energy changes with respect to position, which is essential in determining the force experienced by an object.

When we differentiate the potential energy function \(U(x)\), we're essentially asking 'how much does \(U\) change as \(x\) changes by an infinitesimally small amount?'. This rate of change is what gives us the force function. Physics often involves this concept to analyze motion, force, energy, and other dynamic systems.

Applying Differentiation

In the given exercise, we differentiate the cubic function \(U(x)\) to derive the force function \(F_x\). Each term of the equation \(U(x)\) is treated separately, and standard differentiation rules are applied to find the resulting expression.
Equilibrium in Physics
Equilibrium is a state where the net force acting on an object is zero, meaning there's no acceleration and, if initially at rest, the object will remain at rest. In potential energy terms, points of equilibrium occur where the force function \(F_{x}\) is equal to zero. These points are significant because they signify stability or instability depending on whether they're minima or maxima of the potential energy function.

The exercise highlights the importance of finding the values of \(x\) for which the force is zero, which in turn indicates the equilibrium positions. Stable equilibrium points are found at the minima of \(U(x)\) because here a small displacement leads to a force that pushes the object back to its initial position. Conversely, maxima correspond to unstable equilibrium because any small displacement further destabilizes the object.

Visualizing Equilibrium

By plotting the function \(U(x)\) and \(F_{x}\) against \(x\), you can visualize the stability of the equilibrium points. Minima on the \(U(x)\) graph are indicated where \(F_{x}\) crosses the x-axis from positive to negative, marking stable equilibrium, while maxima correspond to the opposite, indicating unstable equilibrium.

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Most popular questions from this chapter

An 80.0 -kg skydiver jumps out of a balloon at an altitude of 1000 \(\mathrm{m}\) and opens the parachute at an altitude of 200 \(\mathrm{m}\) . (a) Assuming that the total retarding force on the diver is constant at 50.0 \(\mathrm{N}\) with the parachute closed and constant at 3600 \(\mathrm{N}\) with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will be injured? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 \(\mathrm{m} / \mathrm{s} ?\) (d) How realistic is the assumption that the total retarding force is constant? Explain.

At time \(t_{i}\) , the kinetic energy of a particle is 30.0 \(\mathrm{J}\) and the potential energy of the system to which it belongs is 10.0 \(\mathrm{J}\) . At some later time \(t\) , the kinetic energy of the particle is 18.0 \(\mathrm{J}\) . (a) If only conservative forces act on the particle, what are the potential energy and the total energy at time \(t_{f} ?\) (b) If the potential energy of the system at time \(t_{f}\) is 5.00 \(\mathrm{J}\) , are there any nonconservative forces acting on the particle? Explain.

A single conservative force acting on a particle varies as \(\mathbf{F}=\left(A x+B x^{2}\right) \hat{\mathbf{i}} \mathrm{N},\) where \(A\) and \(B\) are constants and \(x\) is in meters. (a) Calculate the potential-energy function \(U(x)\) associated with this force, taking \(U=0\) at \(x=0 .\) (b) Find the change in potential energy and the change in kinetic energy as the particle moves from \(x=2.00 \mathrm{m}\) to \(x=3.00 \mathrm{m} .\)

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 \(5.00-\mathrm{kg}\) block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is held fixed. The spring is compressed 0.100 \(\mathrm{m}\) from equilibrium and released. The speed of the block is 1.20 \(\mathrm{m} / \mathrm{s}\) when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which the coefficient of kinetic friction is 0.300 . Determine the speed of the block at the equilibrium position of the spring.
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