Reading a Potential Energy Curve — Question 10

Question 10

A cusp minimum and the limits of a derivative test

Assume one-dimensional classical motion with $K=E-U(x)$ and $F_x=-dU/dx$, with no dissipation. An ideal potential is $U(x)=2|x|$ J with x in meters. A 1 kg particle is released from rest at x=2 m. Away from zero, use the ordinary one-sided slopes; no additional impulse acts at zero.

1. Find force for x<0 and x>0.

2. Find speed as the particle crosses x=0.

3. Explain why the cusp is a stable minimum even though $dU/dx$ is not defined there.

Question 10 - Solution

A cusp minimum and the limits of a derivative test

(a) Take slopes on the open sides.

The one-sided slopes give $$\boxed{F(x)=\begin{cases}+2\ \mathrm N,&x<0,\\-2\ \mathrm N,&x>0.\end{cases}}$$ Both forces point toward the cusp.

(b) Use continuity of potential and kinetic energy.

The release energy is $E=U(2)=4$ J. At the origin, $$\tfrac12(1)v^2=4-0\Rightarrow\boxed{v=2\sqrt2\ \mathrm{m/s}}.$$ The finite force switch supplies no extra impulse, so velocity remains continuous.

(c) Use the minimum and restoring directions.

The energy minimum is strict: $$U(x)-U(0)=2|x|>0\quad(x\ne0).$$ Small displacements increase U and produce a restoring force. The smooth condition $U'=0$ is inapplicable at a cusp; assigning zero static force exactly at the origin does not alter the crossing work.