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.