Graph of potential energy

In summary, the conversation discusses finding the force, motion, and maximum speed of a mass following a potential energy equation. It also includes calculating where the mass will come to rest again. The equation for the force is -dU/dx and the mass will start moving in the positive x direction when released at x=4. The location where the force reverses direction is at approximately x=5.68 m. The final location where the mass will come to rest is when the potential energy at that point is equal to the initial potential energy at x=4.
  • #1
darksyesider
63
0
[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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  • #2
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)?
 
  • #3
I came up with [tex]F(x) = -dU/dx[/tex]
I don't get intuitively what you mean.
 
Last edited:
  • #4
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.
 
  • #5


I cannot provide a specific response to this content without knowing the specific units and variables used in the graph of potential energy. However, I can provide some general information about potential energy and its relationship to force and motion.

Potential energy is a measure of the energy that an object possesses due to its position or configuration. It is often represented by the symbol U and is measured in joules (J). In the graph of potential energy, the x-axis typically represents the position of the object, while the y-axis represents the potential energy at that position.

To answer the first question, you correctly used the equation for potential energy, U(x), to find the force on the mass at a specific position, x=2. The negative sign indicates that the force is acting in the opposite direction of the displacement.

For the second question, it is important to note that the potential energy is at its minimum value at x=4 m. This means that the force acting on the mass at that position is 0. Therefore, the mass will remain at rest if placed at x=4 m.

For the third question, you correctly used the equation for kinetic energy, 1/2mv^2, to find the velocity of the mass at x=4 m. However, it is important to note that the maximum speed of the mass will occur at the minimum point of the potential energy curve, which is at x=4 m. So, the maximum speed of the mass would be 1.576 m/s.

For the fourth question, you correctly used the conservation of energy principle to determine the position, x_0, where the mass will come to rest again. The potential energy at x_0 will be equal to the potential energy at x=4 m. However, it is important to note that the mass will not necessarily come to rest at x_0, as there could be external forces acting on the mass that could affect its motion.

Overall, your approach and answers seem correct, but it would be helpful to have more information about the specific units and variables used in the graph of potential energy. This would ensure that the calculations are accurate and the conclusions are valid.
 

1. What is a graph of potential energy?

A graph of potential energy is a visual representation of the change in potential energy of a system as a function of its position or configuration. It can also show the relationship between potential energy and other variables, such as time or temperature.

2. How is potential energy represented on a graph?

Potential energy is typically represented on the y-axis of a graph, while the variable affecting potential energy (such as position or temperature) is represented on the x-axis. The shape of the curve on the graph can provide information about the nature of the potential energy, such as its stability or the presence of energy barriers.

3. What are the different types of potential energy shown on a graph?

The most common types of potential energy shown on a graph are gravitational potential energy, elastic potential energy, and chemical potential energy. However, potential energy can also be represented for other types of forces, such as electric or magnetic forces.

4. How is a graph of potential energy useful in understanding physical systems?

A graph of potential energy can provide valuable insights into the behavior and properties of a physical system. It can help identify stable and unstable configurations, determine the amount of energy required to overcome energy barriers, and predict the motion and behavior of objects within the system.

5. How is potential energy related to kinetic energy on a graph?

On a graph of potential energy, kinetic energy is represented as the slope or derivative of the potential energy curve. This means that the steeper the slope of the curve, the greater the kinetic energy of the system. Additionally, at points where the curve is flat (slope = 0), the potential energy is at a maximum and kinetic energy is at a minimum.

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