Question

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.
What is the work (Wd) done on the skydiver, over the distance, by the drag force of the air?

Answer 1

Answer:

$W_{drag} = m\cdot g \cdot d - \frac{1}{2}\cdot m \cdot v^{2}$

Explanation:

Drag Force is the only nonconservative force that affects the system. According to the Principle of Energy Conservations and Work-Energy Theorem, the physical model for the skydiver is:

$U_{g,A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the skydriver by the drag is:

$W_{drag} = U_{g,A}-U_{g,B}-K_{B}$

$W_{drag} = m\cdot g \cdot d - \frac{1}{2}\cdot m \cdot v^{2}$