How Does the Force on a Particle Change with Position in a Potential Well?

[bakin]

Homework Statement

A particle is trapped in a potential well described by U(x)= 1.8 x²-b where U is in joules, x is in meters, and b = 4.5 J.

Find the force on the particle when it's at x = 2.8m
Find the force on the particle when it's at x = 0 m
Find the force on the particle when it's at x = -1.4 m.

Homework Equations

F(x) = -dU(x) / dx

The Attempt at a Solution

so I just differentiated U(x), slapped a negative sign, and ended up with F(x) = 3.6x

Plugged in 2.8m, ended up with -10.08, and it ain't right. Any ideas why? I'm thinking it might have to do with the phrase "potential well".

 

[bakin]

Never mind

F(x) = U(x) / x

edit: (never mind)² by using this, part B came out correct, (the answer would be zero), but it doesn't make sense. By plugging zero in there, you're dividing by zero, which is ILLEGAL. so i guess I am back here for help, lol.

edit: (never mind)³ /sigh the way i had it originally was correct, but online I put in the wrong number of significant figures. oh well.

 

[nlightner]

I would like to provide a more detailed response to the content. The potential well described by U(x) = 1.8x^2 - b represents a system where the particle is confined within a specific region due to the presence of a potential energy barrier. The value of b represents the depth of the potential well, which is a measure of the maximum potential energy that the particle can have within the system.

To find the force on the particle at a specific position x, we can use the formula F(x) = -dU(x)/dx, where dU(x)/dx represents the derivative of the potential energy function with respect to x. This formula is based on the concept of the conservative force, where the force acting on a particle is equal to the negative gradient of the potential energy function.

When we differentiate U(x) = 1.8x^2 - b, we get F(x) = 3.6x, which is the correct formula for the force on the particle at any position x within the potential well. However, it is important to note that this formula only applies within the region where the particle is confined within the potential well. Outside of this region, the force on the particle will be different, as the particle is no longer subject to the potential energy barrier.

Now, let's look at the specific values of x given in the homework statement. At x = 2.8m, the force on the particle is F(2.8) = 3.6(2.8) = 10.08 N. This means that the particle is experiencing a force of 10.08 N in the negative direction (towards x = 0) due to the potential energy barrier. Similarly, at x = 0m, the particle is experiencing a force of F(0) = 3.6(0) = 0 N, as it is at the bottom of the potential well and there is no potential energy barrier. At x = -1.4m, the force on the particle is F(-1.4) = 3.6(-1.4) = -5.04 N, indicating that the particle is experiencing a force of 5.04 N in the positive direction (towards x = 0) due to the potential energy barrier.

In summary, the force on the particle at a specific position within the potential well can be calculated using the