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Ch 10: Interactions and Potential Energy
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 10, Problem 61a

A particle that can move along the x-axis is part of a system with potential energy U(x) = A/x2 − B/x where A and B are positive constants. Where are the particle's equilibrium positions?

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Step 1: Recall that equilibrium positions occur where the force acting on the particle is zero. The force is related to the potential energy by the equation F(x) = -dU(x)/dx. Begin by calculating the derivative of the given potential energy function U(x) = A/x^2 - B/x with respect to x.
Step 2: Differentiate U(x) with respect to x. Using the power rule for derivatives, the derivative of A/x^2 is -2A/x^3, and the derivative of -B/x is B/x^2. Combine these results to find dU(x)/dx = -2A/x^3 + B/x^2.
Step 3: Set the force equal to zero to find the equilibrium positions. This means solving the equation -dU(x)/dx = 0, or equivalently, -2A/x^3 + B/x^2 = 0.
Step 4: Factorize the equation to simplify. Rewrite the equation as (B/x^2) - (2A/x^3) = 0. Factor out 1/x^2, resulting in (1/x^2)(B - 2A/x) = 0. Since 1/x^2 cannot be zero, solve the remaining equation B - 2A/x = 0.
Step 5: Solve for x to find the equilibrium positions. Rearrange the equation B - 2A/x = 0 to isolate x, giving x = 2A/B. This is the position where the particle is in equilibrium.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Potential Energy

Potential energy is the energy stored in a system due to its position or configuration. In this case, the potential energy U(x) is defined as a function of the particle's position along the x-axis. Understanding how potential energy varies with position is crucial for analyzing the forces acting on the particle and determining its equilibrium positions.
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Equilibrium Position

An equilibrium position occurs when the net force acting on a particle is zero, meaning the particle is in a state of balance. For a particle moving along the x-axis, this can be found by setting the derivative of the potential energy function U(x) to zero. At these points, the particle experiences no net force and can remain at rest or move with constant velocity.
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Force from Potential Energy

The force acting on a particle can be derived from the potential energy function using the relationship F(x) = -dU/dx. This means that the force is equal to the negative gradient of the potential energy with respect to position. By calculating this derivative, one can identify the points where the force is zero, which correspond to the equilibrium positions of the particle.
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Related Practice
Textbook Question

CALC The potential energy for a particle that can move along the x-axis is U = Ax2 + B sin(πx/L), where A, B, and L are constants. What is the force on the particle at (a) x = 0, (b) x = L/2, and (c) x = L?

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Textbook Question

A system has potential energy U(x)=(10J)[1sin((3.14rad/m)x)]U(x) = (10 \, \(\text{J}\)) [1 - \(\sin\)((3.14 \, \(\text{rad/m}\)) x)] as a particle moves over the range 0 m x3 m0\(\text{ m }\]\le\) x\(\le\)3\(\text{ m}\). For each, is it a point of stable or unstable equilibrium?

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Textbook Question

CALC An object moving in the xy-plane is subjected to the force F=(2xyi^+x2j^)N\(\vec{F}\) = (2xy\,\(\hat{i}\) + x^2\,\(\hat{j}\))\,\(\text{N}\), where x and y are in m. The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis. How much work does the force do?

Textbook Question

A 100 g particle experiences the one-dimensional, conservative force Fx shown in FIGURE P10.60. Suppose the particle is shot to the right from x = 1.0 m with a speed of 25 m/s. Where is its turning point?

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Textbook Question

A clever engineer designs a 'sprong' that obeys the force law Fx=−q(x−xeq)³ , where xeq is the equilibrium position of the end of the sprong and q is the sprong constant. For simplicity, we'll let xeq = 0 m .Then Fx = −qx³. Find an expression for the potential energy of a stretched or compressed sprong.

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Textbook Question

A 100 g particle experiences the one-dimensional, conservative force Fx shown in FIGURE P10.60. Let the zero of potential energy be at x = 0 m . What is the potential energy at x = 1.0, 2.0, 3.0, and 4.0 m? Hint: Use the definition of potential energy and the geometric interpretation of work.

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