What Does the Potential Energy Equation U(x) Tell Us About Motion and Forces?

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Homework Help Overview

The discussion revolves around a mass of 5 kg and its potential energy described by the equation U(x) = .333(x-4)^3 - (x^2/4) + 5. Participants explore the implications of this potential energy on the force acting on the mass, its motion when released from a specific position, and the calculation of maximum speed and subsequent resting positions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of force from the potential energy equation and question the direction of the force when the mass is placed at x=4. There are attempts to derive the force function and explore the implications of the potential energy on motion.

Discussion Status

Some participants have provided calculations for force and velocity, while others seek clarification on the intuitive understanding of these results. There is ongoing exploration of the relationship between potential energy and force, with no explicit consensus reached on the correctness of the calculations presented.

Contextual Notes

Participants are working under the constraints of the problem statement and the equations provided, with some questioning the assumptions made regarding the potential energy's implications on motion and force direction.

darksyesider
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[solved] graph of potential energy

Homework Statement



A mass of 5 kg follows the potential energy shown by the equation:

[tex]U(x) = .333(x-4)^3 - \dfrac{x^2}{4} + 5[/tex]

1. Find the force on the mass at x=2
2. Describe the motion qualitatively if the mass is placed at x=4 m and released
3. With part 1, determine the max speed of the mass
4. With #1, determine where the mass will again come to rest if at all.

Homework Equations



mgh
1/2mv^2

The Attempt at a Solution




1. I got this pretty easily, as -3N
2. I really had no idea for this. I thought that it would eventually settle at x=5.6 m.
3. U(4)-U(5.68) = .5mv^2 so v=1.576
4. u(4)-U(5.868) = U(x_0) => x_0 = 7.162 m

Is this correct?
 
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I assume you equation for the potential is in Joules, correct? From the equation for the potential, what is the equation for the force? To get a better understanding of what happens in part 2-4, plot a graph of the force as a function of distance. Or at least tell us the direction of the force when the mass is placed at x = 4. Will it continue moving in this direction? How far does it have to go before the force reverses direction (if at all)?
 
I came up with [tex]F(x) = -dU/dx[/tex]
I don't get intuitively what you mean.
 
Last edited:
darksyesider said:
I came up with [tex]F(x) = -(x-4)^2+\dfrac{x}{2}[/tex]
I don't get intuitively what you mean.
At x = 4, F = 2 N, so that, once the mass is released, it is going to start moving in the + x direction. You didn't show the details, so I didn't know whether you were doing it right, but, in part 3, it looks like you correctly calculated the location at which the force goes to zero (and reverses sign), and also calculated the velocity correctly. In part 4, you should have set U(xfinal)=U(4) to find location when the mass comes to rest again. It doesn't look like 7.162 satisfies this.
 

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