Fundamental Theorem for Line Integrals — Question 9

Question 9 – Solution

Strategy. Apply the theorem to $-U$: work by $-\nabla U$ is the negative of the change in $U$.

Step 1: Potential-energy change $$U(A)=2^2+2(1^2)=6, \qquad U(B)=0^2+2(3^2)=18.$$ Therefore $\Delta U=U(B)-U(A)=12$.

Step 2: Work Since $\mathbf F=\nabla(-U)$, $$\begin{align*} W&=(-U)(B)-(-U)(A)\\ &=U(A)-U(B)=6-18=\boxed{-12}. \end{align*}$$ Equivalently, $W=-\Delta U=-12$.

Step 3: Interpret the sign The particle ends at a point with greater potential energy. The force derived from $-\nabla U$ therefore does negative net work during the motion.

Verification The identity $W+\Delta U=-12+12=0$ checks the sign. Reversing the trip would produce work $+12$.