How do I prove whether a force perpendicular to the motion is conservative and $\mathbf{F}=\mathbf{F_{0}}\sin(at)$ conservative, where $\mathbf{F_{0}}$ is a constant vector.
I knew that for a force to be conservative, it's $\nabla \times \mathbf{F}=0$ everywhere or the work done around a closed path without including origin should be zero.
Self-learning from Kleppner and Kolenkow's book.
Best Answer
If a force is perpendicular to the motion then $\mathbf{F}\cdot \dot{\mathbf{x}}=0 \quad\forall t$, then the work
$$ W=\oint_\ell \mathbf{F}\cdot d\mathbf{x}=0$$
So it's conservative.
If $t$ is time, then the force does not depend on the position, then all the derivatives are zero and its trivial to say that $\nabla \times \mathbf{F}=0$.