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I wanted to show that a force such as drag is not a conservative force but cannot apply my method of approach. Suppose we took the situation with drag at high speeds, so $\vec D=c|\vec v|\vec v$ for some $c$.

Since work is defined as $\int_C \vec F \cdot d\vec s$ along a path $C$, if I could show that this integral differs depending on the path taken, then I would have shown that drag is non-conservative.

So the integral I would have is: $\int_C -(c|\vec v|\vec v)\cdot d\vec s$ and we can write $\vec v$ as $|v|\hat{e_t}$ which is a vector in the tangential direction to the path. Since $d\vec s$ is defined as $\hat{e_t}\cdot ds$ then the integral becomes $$\int_C -(c|\vec v|\vec v)\cdot d\vec s=\int_C -(c|\vec v|^2)ds$$

How would I proceed from here?

user258521
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1 Answers1

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Consider the work done by the drag when the motion is in a closed loop.

Because the drag is always directed opposite to the motion, the infinitesimal work is always negative. The total work, which is the integral, must therefore also be negative.

But for a conservative force, the work done in a closed loop is always zero.

We conclude that drag cannot be conservative.

Robin Ekman
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