Reading a Potential Energy Curve — Question 2

Question 2

Turning positions in a displaced potential well

Assume one-dimensional classical motion with $K=E-U(x)$ and $F_x=-dU/dx$, with no dissipation. A 1 kg particle has E=8 J in $U(x)=2(x-3)^2$ J, with x in meters.

1. Find the allowed interval and turning positions.

2. Find the position and value of the maximum speed.

3. If the particle is released from rest at the right turning point, determine its initial force direction.

Question 2 - Solution

Turning positions in a displaced potential well

(a) Require nonnegative kinetic energy.

Allowed motion requires nonnegative kinetic energy: $$2(x-3)^2\leq8\Rightarrow|x-3|\leq2\Rightarrow\boxed{1\leq x\leq5\ \mathrm m}.$$ At the endpoints $U=E$, so $\boxed{x_{\mathrm{turn}}=1,5\ \mathrm m}$.

(b) Find the bottom of the well.

The potential minimum is at $x=3$ m: $$U_{\min}=0,\quad K_{\max}=8\ \mathrm J\Rightarrow\boxed{v_{\max}=\sqrt{2(8)/1}=4\ \mathrm{m/s}}.$$

(c) Evaluate the slope at release.

The release force at the right turning point is $$F=-U'(x)=-4(x-3)\Rightarrow\boxed{F(5)=-8\ \mathrm N}.$$ It points left, into the allowed region. The equilibrium at the well center is not a turning point at this energy.