Potential energy functions of a particle

AI Thread Summary
The discussion revolves around calculating the potential energy function U(x) for a particle influenced by a conservative force F = (-Ax + Bx^4)i N. The user initially attempts to derive U(x) by integrating the force, suggesting U(x) = -1/2Ax^2 + 1/4Bx^5, but questions the accuracy of their approach. There is confusion regarding the correct integration process, leading to a proposed alternative of U(x) = -1/2Ax^2 + 1/5Bx^5. The need for clarity on the final form of the potential energy function is emphasized. The thread highlights the importance of correctly applying integration to derive potential energy from a variable force.
ramenmeal
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Homework Statement



A single conservative force acting on a particle varies as Fvec = (-Ax + Bx4)i 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.)

(b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 1.80 m to x = 3.40 m. (Use A, B, and x as appropriate.)

Homework Equations



w = f x d

The Attempt at a Solution



ive only attempted part a since i think i will need it to go on to part b.

w = f x d
f = (-Ax + Bx4)
d = x

PE = (-Ax + Bx^4)x = -Ax^2 + Bx^5
 
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Since the force is continuously variable over the range of x isn't your U(x) the integral of the Force F(x) and not just the product of x and F(x)?
 
oh yeah.. so should it be

(-1/2)Ax^2 + (1/4)Bx^5 ?

i'm not sure what form my final answer should be.
 
ramenmeal said:
oh yeah.. so should it be

(-1/2)Ax^2 + (1/4)Bx^5 ?

i'm not sure what form my final answer should be.

How about -1/2Ax2 + 1/5Bx5 instead?
 

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