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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 63

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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Step 1: Recall the relationship between force and potential energy. The force on a particle is given by the negative gradient of the potential energy function: F(x) = -dU/dx.
Step 2: Differentiate the given potential energy function U(x) = Ax² + B sin(πx/L) with respect to x. Use the chain rule for the sine term. The derivative is: dU/dx = 2Ax + B(π/L)cos(πx/L).
Step 3: For part (a), substitute x = 0 into the expression for the force: F(0) = -[2A(0) + B(π/L)cos(π(0)/L)]. Simplify the expression to find the force at x = 0.
Step 4: For part (b), substitute x = L/2 into the expression for the force: F(L/2) = -[2A(L/2) + B(π/L)cos(π(L/2)/L)]. Simplify the expression to find the force at x = L/2.
Step 5: For part (c), substitute x = L into the expression for the force: F(L) = -[2A(L) + B(π/L)cos(π(L)/L)]. Simplify the expression to find the force at x = L.

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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 an object due to its position in a force field, such as gravitational or elastic fields. In this case, the potential energy U is a function of position x, defined by the equation U=Ax²+B sin(πx/L). Understanding how potential energy varies with position is crucial for analyzing the forces acting on the particle.
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Force and Potential Energy Relationship

The force acting on a particle can be derived from the potential energy function using the relation F = -dU/dx. This means that the force is equal to the negative gradient of the potential energy with respect to position. This concept is essential for determining the force at specific positions along the x-axis, as it directly links the potential energy to the force experienced by the particle.
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Differentiation

Differentiation is a fundamental mathematical process used to find the rate at which a quantity changes. In the context of this problem, differentiation is applied to the potential energy function U with respect to x to calculate the force. Mastery of differentiation techniques is necessary to accurately compute the force at the specified positions x=0, x=L/2, and x=L.
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Related Practice
Textbook Question

In a physics lab experiment, a compressed spring launches a 20 g metal ball at a 30° angle. Compressing the spring 20 cm causes the ball to hit the floor 1.5 m below the point at which it leaves the spring after traveling 5.0 m horizontally. What is the spring constant?

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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. Is this a conservative force?

Textbook Question

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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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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