Potential Energy of a single particle

[yb1013]

Homework Statement

A single conservative force acting on a particle varies as Fvec = (-Ax + Bx5)i-hat N, where A and B are constants and x is in meters. Accurately round coefficients to three significant figures.

(a) Calculate the potential energy function U(x) associated with this force, taking U = 0 at x = 0. (Use A, B, and x as appropriate.)

U = ____________

(b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 1.20 m to x = 3.80 m. (Use A, B, and x as appropriate.)

ΔU = ____________

ΔK = ____________

Homework Equations

Not sure if there are relevant equations but PE = mgh

The Attempt at a Solution

I really have never been this confused about a physics problem... I think the answer needs to be in an equation form? I am really not sure on this one guys, I tried to understand it, its just not coming to me. I did all the rest of the homework completely fine other than this one.

 

[Vuldoraq]

To get you started: Potential energy is just the integral of the force.

 

[yb1013]

So, Like -Ax^2+Bx^6?

 

[Vuldoraq]

So, Like -Ax^2+Bx^6?

Just so. Don't forget to divide by the power though. Since the question gives you some initial conditions you can also find the constant of integration.

 

[yb1013]

ok so for the first question the formula would be -Ax^2+B/5x^6 i think. But then they give you U and x are = 0 ... So i guess you can't just plug x in because that would be 0. How do you solve for potential energy?

 

[Vuldoraq]

ok so for the first question the formula would be -Ax^2+B/5x^6 i think. But then they give you U and x are = 0 ... So i guess you can't just plug x in because that would be 0. How do you solve for potential energy?

Try to use the LaTeX for equations, it's much clearer. For the potential you use,

\int F dx= U

For any indefinite integral you will have a constant of integration, for example,

\int x^{n} dx= \frac{x^{n+1}}{n+1} +C

When it gives you the value of U when x is zero it is basically saying plug these vaules into your P.E. equation to find your integration constant.

 

[yb1013]

ok well i found part a.. but when i go to plug in 3.8 and 1.2 to get the answer, I come out with 7.94A-502.3B , and it comes up incorrect..

 

[Vuldoraq]

ok well i found part a.. but when i go to plug in 3.8 and 1.2 to get the answer, I come out with 7.94A-502.3B , and it comes up incorrect..

Maybe you have a minus out of place? Is this what you are evaluating?

[\frac{-Ax^{2}}{2}+\frac{Bx^{6}}{6}]_{1.2}^{3.2}

 

[yb1013]

Im evaluating A/2x^2-B/6x^6 , I am sorry, i tried using latex but I've never used it before and i just couldn't figure it out.. But yea that's what I am evaluating, you had to multiply the whole thing by -1 because of the formula. So with A/2x^2-B/6x^6 i am taking (3.8) - (1.2), and i get what i told you before

 

[Vuldoraq]

Check your calculation, I think you added the A's rather than subtracted the second from the first.

 

[yb1013]

aahhhh, thank you sir, very wise!
haha alright well thanks again man and have a great night.

 

[Vuldoraq]

Thanks and your welcome.