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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
All textbooksKnight Calc 5th EditionCh 10: Interactions and Potential EnergyProblem 60b
Chapter 10, Problem 60b

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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Step 1: Understand the concept of a turning point. A turning point occurs when the particle's kinetic energy is completely converted into potential energy, meaning its velocity becomes zero at that position.
Step 2: Use the principle of conservation of mechanical energy. The total mechanical energy (kinetic energy + potential energy) of the particle remains constant because the force is conservative. Write the equation: \( E_{total} = K + U \), where \( K \) is the kinetic energy and \( U \) is the potential energy.
Step 3: Calculate the initial total mechanical energy. The initial kinetic energy is given by \( K = \frac{1}{2} m v^2 \), where \( m \) is the mass of the particle (convert 100 g to kg: \( m = 0.1 \, \text{kg} \)) and \( v \) is its initial velocity (25 m/s). The initial potential energy \( U \) at \( x = 1.0 \, \text{m} \) can be determined from the force diagram or the potential energy function provided in the problem.
Step 4: Determine the potential energy \( U \) at different positions \( x \) using the relationship between force and potential energy: \( F_x = -\frac{dU}{dx} \). Integrate the force function \( F_x \) to find \( U(x) \), ensuring you account for the constant of integration based on the reference point.
Step 5: Solve for the turning point. At the turning point, the kinetic energy is zero, so \( E_{total} = U(x) \). Set the total mechanical energy equal to the potential energy function \( U(x) \) and solve for \( x \). This will give the position where the particle stops and reverses direction.

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Key Concepts

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

Conservative Forces

Conservative forces are forces that do not dissipate energy and depend only on the position of an object. The work done by a conservative force on an object moving between two points is independent of the path taken. Examples include gravitational and elastic forces. In this context, the conservative force Fx will determine the potential energy associated with the particle's position.
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Kinetic and Potential Energy

Kinetic energy is the energy of an object due to its motion, calculated as KE = 1/2 mv², where m is mass and v is velocity. Potential energy, on the other hand, is the stored energy based on an object's position in a force field, such as gravitational or elastic potential energy. The conservation of mechanical energy principle states that the total mechanical energy (kinetic + potential) remains constant in a conservative system, allowing us to find the turning point of the particle.
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Turning Point

The turning point of a particle in motion is the position where its velocity becomes zero, indicating a change in direction. At this point, all kinetic energy has been converted into potential energy. To find the turning point, one must analyze the energy conservation between kinetic and potential energy, using the initial speed and the potential energy function derived from the conservative force acting on the particle.
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Related Practice
Textbook Question

A 1.0 kg mass that can move along the x-axis experiences the potential energy U = (x²−x) J, where x is in m. The mass has velocity vx = 3.0 m/s at position x = 1.0 m. At what position has it slowed to 1.0 m/s?

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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)[1−sin((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 ≤x≤3 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

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

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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