Reading a Potential Energy Curve — Question 6

Question 6

A tangent gives force while the curve gives kinetic energy

Assume one-dimensional classical motion with $K=E-U(x)$ and $F_x=-dU/dx$, with no dissipation. Near x=2 m, $U(x)=(x-3)^2+2$ J. The tangent at x=2 passes through (1 m,5 J) and (3 m,1 J). A 2 kg particle there has E=7 J.

1. Find force at x=2 from the tangent.

2. Find speed at x=2 from the actual curve.

3. Would substituting the tangent value at x=3 into a later energy calculation be exact?

Question 6 - Solution

A tangent gives force while the curve gives kinetic energy

(a) Use the tangent slope.

The tangent gives the local derivative: $$\left.\frac{dU}{dx}\right|_2=\frac{1-5}{3-1}=-2\ \mathrm{J/m}\Rightarrow\boxed{F(2)=+2\ \mathrm N}.$$

(b) Read the potential itself.

Use the actual potential value for energy: $$U(2)=(2-3)^2+2=3\ \mathrm J,\quad K=7-3=4\ \mathrm J.$$ $$\boxed{v(2)=\sqrt{2(4)/2}=2\ \mathrm{m/s}}.$$

(c) Keep a local linearization local.

The tangent model and exact function differ away from the contact point: $$U_{\mathrm{tan}}(x)=3-2(x-2),\qquad U_{\mathrm{exact}}(x)-U_{\mathrm{tan}}(x)=(x-2)^2.$$ At $x=3$, they give 1 J and 2 J respectively. A tangent supplies the local slope, not exact finite-interval energy values.