Net force on a particle with potential energy function

[Koscher]

Homework Statement

A particle is moving along the x-axis subject to the potential energy function U(x) = 1/x + x² + x - 1. Determine the x-component of the net force on the particle at the coordinate x= 3.29m.

Homework Equations

U(x) = integral(F(x)dx)

The Attempt at a Solution

To find the net force I took one derivative of the potential energy equation. The resulting (-1/x²) + 2x +1. I then plugged in 3.29 for x and it resulted in 7.487. Which is not the correct answer. Now my though is that if i take the second derivative of the potential energy function and then plug in my value of x=3.29 m, then that would be correct. But I am really not sure.

 

[cepheid]

Homework Statement

A particle is moving along the x-axis subject to the potential energy function U(x) = 1/x + x² + x - 1. Determine the x-component of the net force on the particle at the coordinate x= 3.29m.

Homework Equations

U(x) = integral(F(x)dx)

Actually, the definition of the potential energy function for this 1D case is$$U(x) = -\int F(x)\,dx$$The negative sign is important. As a result, $$F(x) = - \frac{dU}{dx}$$ A motivation for this definition is so that the force points in the direction of maximum decrease in potential energy, which, if you think about it, is consistent with examples of conservative forces that you may have encountered already, such as gravity.

The Attempt at a Solution

To find the net force I took one derivative of the potential energy equation. The resulting (-1/x²) + 2x +1. I then plugged in 3.29 for x and it resulted in 7.487. Which is not the correct answer. Now my though is that if i take the second derivative of the potential energy function and then plug in my value of x=3.29 m, then that would be correct. But I am really not sure.

No. If potential energy is the negative integral of force, then force is the negative derivative of potential energy, since integration and differentiation are inverse operations. So, basically, taking the second derivative makes no sense at all.

 

[Koscher]

Thank you. I completely forgot the negative sign on the equation. When i took it into account I got the correct answer. Thank you.

 

[cepheid]

Thank you. I completely forgot the negative sign on the equation. When i took it into account I got the correct answer. Thank you.

No problem, you're welcome.

 

[asteriatic]

I would like to clarify that the net force on a particle is the sum of all the forces acting on it. In this case, the potential energy function U(x) is not a force itself, but rather a representation of the work done by conservative forces on the particle as it moves along the x-axis. In order to find the net force on the particle at a specific point, we would need to consider other factors such as the particle's mass and any other forces acting on it.

Assuming that there are no other forces acting on the particle, the x-component of the net force can be calculated by taking the derivative of the potential energy function with respect to x, which would be the negative of the force function F(x). So, the correct approach would be to take the first derivative of U(x) and then plug in the value of x=3.29m to find the x-component of the net force.

Taking the second derivative of U(x) would give us the force function, but it would not be the correct approach in this scenario as it would result in a different value for the net force. It is important to note that the potential energy function and the force function are related, but they are not the same.

In conclusion, to find the x-component of the net force on the particle at x=3.29m, we would need to take the first derivative of U(x) and then plug in the value of x=3.29m to get the correct answer. It is also important to consider other factors and forces acting on the particle in order to fully understand its motion.