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

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The potential energy equation U(x) = .333(x-4)^3 - (x^2/4) + 5 describes the motion of a 5 kg mass subjected to forces derived from this potential. The force at x=2 is calculated to be -3 N, indicating a leftward direction. When released from x=4, the mass is expected to move in the positive x direction due to a force of 2 N at that point. The maximum speed of the mass is determined to be approximately 1.576 m/s, and the location where it comes to rest again is incorrectly calculated as 7.162 m, which does not satisfy the condition U(xfinal)=U(4). Understanding the relationship between potential energy and force is crucial for analyzing motion in this context.
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:

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

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 F(x) = -dU/dx
I don't get intuitively what you mean.
 
Last edited:
darksyesider said:
I came up with F(x) = -(x-4)^2+\dfrac{x}{2}
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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