Path Independence of Conservative Forces — Question 8

Question 8

A one-dimensional force with a finite step

On a line, a stationary force is +2 N for x<0 and $-4$ N for x>0. Its value at x=0 is finite. A particle is moved from $-1$ m to +2 m and back along the line.

1. Find outward and return works.

2. Construct a continuous potential with U(0)=0 on each side.

3. Does the jump in force prevent an endpoint potential from existing?

Question 8 - Solution

A one-dimensional force with a finite step

(a) Split the integrals at the step.

Split at the force discontinuity: $$\boxed{W_{-1\to2}=\int_{-1}^0 2\,dx+\int_0^2(-4)dx=2-8=-6\ \mathrm J}.$$ $$\boxed{W_{2\to-1}=+6\ \mathrm J},\qquad W_{\mathrm{cycle}}=0.$$

(b) Integrate with a common joining value.

Integrating from the common reference gives $$\boxed{U(x)=\begin{cases}-2x,&x\leq0,\\4x,&x\geq0,\end{cases}}\quad\text{(SI units)}.$$ Thus $U(-1)=2$ J, $U(2)=8$ J and $U(-1)-U(2)=-6$ J.

(c) Distinguish a kink from a jump in potential.

A bounded force gives no finite potential jump: $$|U(\epsilon)-U(-\epsilon)|\leq\int_{-\epsilon}^{\epsilon}|F|dx\leq8\epsilon\longrightarrow0.$$ The force step creates a kink. Its finite value at one isolated point contributes no integral area.