contestada

in the presence of air drag, an object of mass 0.126kg falls from height of 15.2m to 7.9m with the terminal speed v. How much work did the drag force do on the object

Respuesta :

Answer:

Explanation:

The distance traveled is 15.2 - 7.9 = 7.3 m

The potential energy converted that is not expressed as kinetic energy is the work of air friction

W = PE - KE

W = mgh - ½mv²

W = 0.126(9.8)(7.3) - ½(0.126)v²)

W = 9.01404 - 0.063v² J