(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Think that we have only conservative force in this question and no nonconservative force like friction exists. Also## U ##means potential energy (eg Gravitational Potential Energy),##K## means Kinetic Energy (##1/2 mv^2##) and ##E = K + U = constant##

We have a arbitrary Potential energy (U) in terms of position (x) diagram like the picture. The object is at first, in a position like ##x_0## and wants to get to point ##x_1##. At ##x_1## we have ##dU / dx = 0## and Also the initial Mechanical Energy (E = K + U) of the object at first is equal to the U at point ##x_1##. We want to prove that it takes infinite time -theoretically- for this object to get to the point ##x_1##. (The movement is 1 Dimensional)

2. Relevant Equations

##E = K + U = Constant##

##K = 1/2 m v^2##

##F = m a##

##F = -dU/dx##

Taylor Series at ##x = x_1## for function ##f(x)##: ##f(x) = f(x_1) + f'(x) (x - x_1) + f''(x) (x - x_1)^2 / 2! + ...##

All forces are conservative in this question.

3. The attempt at a solution

At first we know that the K at ##x_1## equals to zero. Intuitively, I can understand the at the end of the path, the acceleration goes to zero and also the velocity goes to zero (because ## K -> 0 ==> V -> 0##). But I don't know how to prove this. I know that F = -dU/dx.

For the start, I get this from taylor series but I don't know what to do with it:

##U(x) = U(x1) + dU/dx (x - x1) + O(x^2)## -> ##U(x) = U(x1) + (-m a) (x - x1)

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Infinite time for an object to get K=0 conservative Force

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**