Path Independence of Conservative Forces — Question 6

Question 6

A separable force admits an endpoint calculation

A force field is $\vec F=(-2x\,\hat i-6y\,\hat j)$ N, with x,y in meters. Consider A=(0,0) and B=(2,1).

1. Find work along a horizontal-then-vertical path.

2. Find work along $x=2t$, $y=t$ for $0\le t\le1$.

3. Construct a potential and explain why it establishes more than agreement between these two paths.

Question 6 - Solution

A separable force admits an endpoint calculation

(a) Integrate the separate components.

Integrate along each coordinate leg: $$W=\int_0^2(-2x)dx+\int_0^1(-6y)dy=[-x^2]_0^2+[-3y^2]_0^1=\boxed{-7\ \mathrm J}.$$

(b) Evaluate the parametrized route.

With $\vec r(t)=(2t,t)$, $d\vec r=(2,1)dt$ and $\vec F=(-4t,-6t)$: $$\boxed{W=\int_0^1[(-4t)(2)+(-6t)(1)]dt=\int_0^1-14t\,dt=-7\ \mathrm J}.$$

(c) Exhibit a global potential.

A globally defined potential is $$\boxed{U=x^2+3y^2+C},\qquad-\nabla U=(-2x,-6y)=\vec F.$$ Therefore $W_{AB}=U_A-U_B$ on every route in the plane. This construction proves more than agreement between two examples.