From the conservation of energy,

$\overline{){\mathbf{K}}{{\mathbf{E}}}_{{\mathbf{0}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{0}}}{\mathbf{=}}{\mathbf{K}}{{\mathbf{E}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{W}}}_{{\mathbf{d}}}}$

Drag has to be accounted for on the "final" side of the energy conservation equation because it is a non-conservative force.

Power,

$\overline{){\mathbf{P}}{\mathbf{o}}{\mathbf{w}}{\mathbf{e}}{\mathbf{r}}{\mathbf{=}}\frac{\mathbf{E}\mathbf{n}\mathbf{e}\mathbf{r}\mathbf{g}\mathbf{y}}{\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}}}$

a)

KE_{0} = 0 J

U_{0} = mgd

KE_{f} = (1/2)mv^{2}

A skydiver of mass m jumps from a hot air balloon and falls a distance d before reaching a terminal velocity of magnitude v. Assume that the magnitude of the acceleration due to gravity is g.

a) What is the work (W_{d}) done on the skydiver, over the distance, by the drag force of the air?

b) Find the power (P_{d}) supplied by the drag force after the skydiver has reached terminal velocity v.

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