Show that a force that does no work must be a velocity-dependent force.

(N/A) The work done by a force $\overrightarrow{F}$ over a displacement $d\overrightarrow{l}$ is given by $dW = \overrightarrow{F} \cdot d\overrightarrow{l}$.
Given that the work done is zero,we have $\overrightarrow{F} \cdot d\overrightarrow{l} = 0$.
Since the velocity is defined as $\vec{v} = \frac{d\overrightarrow{l}}{dt}$,we can write $d\overrightarrow{l} = \vec{v} dt$.
Substituting this into the work equation: $\overrightarrow{F} \cdot (\vec{v} dt) = 0$.
Since $dt \neq 0$,it follows that $\overrightarrow{F} \cdot \vec{v} = 0$.
This implies that the force $\overrightarrow{F}$ must always be perpendicular to the velocity vector $\vec{v}$ of the particle.
If the particle changes its direction of motion,the velocity vector $\vec{v}$ changes.
To maintain the condition $\overrightarrow{F} \cdot \vec{v} = 0$ (i.e.,the angle $\theta$ between $\overrightarrow{F}$ and $\vec{v}$ remains $90^{\circ}$),the direction of the force $\overrightarrow{F}$ must also change in accordance with the change in the direction of $\vec{v}$.
Therefore,the force must be velocity-dependent.