Question

Consider a particle that moves through the force field F(x, y) = (y − x)i + xyj from the point (0, 0) to the point (0, 1) along the curve x = kt(1 − t), y = t. Find the value of k such that the work done by the force field is 1.

Answer 1

The work done by $\vec F$ is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r$

where $C$ is the given curve and $\vec r(t)$ is the given parameterization of $C$. We have

$\mathrm d\vec r=\dfrac{\mathrm d\vec r}{\mathrm dt}\mathrm dt=k(1-2t)\,\vec\imath+\vec\jmath$

Then the work done by $\vec F$ is

$\displaystyle\int_0^1((t-kt(1-t))\,\vec\imath+kt^2(1-t)\,\vec\jmath)\cdot(k(1-2t)\,\vec\imath+\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1((k-k^2)t-(k-3k^2)t^2-(k+2k^2)t^3)\,\mathrm dt=-\frac k{12}$

In order for the work to be 1, we need to have $\boxed{k=-12}$.