Question

A 45-kg skydiver jumps out of an airplane and falls 450 m, reaching a maximum speed of 51 m/s before opening her parachute. How much work, in joules, did air resistance do on the skydiver before she opened her parachute

Answer 1

Answer:

The work done by the friction force is - 139927.5 J.

Explanation:

mass of diver, m = 45 kg

distance falls, h = 450 m

initial speed, u = 0 m/s

final speed = 51 m/s

According to the work energy theorem,

Work done by the gravity + work done by the friction force = change in kinetic energy

$m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J$