Textbook Question
A system consists of interacting objects A and B, each exerting a constant 3.0 N pull on the other. What is (delta)U for the system if A moves 1.0 m toward B while B moves 2.0 m toward A?
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Ch 10: Interactions and Potential Energy
Chapter 10, Problem 10
A particle moving along the y-axis is in a system with potential energy U = 4y^3 J, where y is in m. What is the -component of the force on the particle at y = 0 m, 1 m, and 2 m?
Verified step by step guidance1
Identify the expression for potential energy, U, which is given as U = 4y^3 J.
Recall that the force, F, on a particle in a potential field is given by the negative gradient of the potential energy, F = -\nabla U. For motion along the y-axis, this simplifies to F_y = -\frac{dU}{dy}.
Differentiate the potential energy function U = 4y^3 with respect to y to find \frac{dU}{dy}. The derivative of 4y^3 with respect to y is 12y^2.
Substitute the derivative into the force equation to get F_y = -12y^2. This equation gives the y-component of the force as a function of y.
Evaluate F_y at y = 0 m, 1 m, and 2 m using the expression F_y = -12y^2 to find the force at each of these points.
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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 = 4y^3 J indicates that the energy depends on the position y of the particle along the y-axis. Understanding how potential energy varies with position is crucial for analyzing the forces acting on the particle.
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Potential Energy Graphs
Force and Potential Energy Relationship
The force acting on a particle can be derived from the potential energy function using the relationship F = -dU/dy. 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 by calculating the derivative of the given potential energy function.
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Calculating Derivatives
Calculating derivatives is a fundamental mathematical skill used to find rates of change. In the context of this problem, taking the derivative of the potential energy function U = 4y^3 with respect to y allows us to find the force at different positions. Mastery of differentiation is necessary to solve for the force values at y = 0 m, 1 m, and 2 m.
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Related Practice
Textbook Question
A system has potential energy U(x)=(10 J)[1−sin((3.14 rad/m) x)] as a particle moves over the range 0 m≤x≤3 m
b. For each, is it a point of stable or unstable equilibrium?
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Textbook Question
CALC 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.
a. Where are the particle's equilibrium positions?
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Textbook Question
CALC 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.
b. Find an expression for the potential energy of a stretched or compressed sprong.
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Textbook Question
FIGURE EX10.24 is the potential-energy diagram for a 500 g particle that is released from rest at A. What are the particle's speeds at B, C, and D?
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Textbook Question
FIGURE EX10.25 is the potential-energy diagram for a 20 g particle that is released from rest at x = 1.0m. (b) What is the particle's maximum speed? At what position does it have this speed?
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