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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

User Nisetama
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1 Answer

6 votes

Answer:

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

Step-by-step 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* 9.8* 450 + W' = 0.5* 45* (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J

User Kat Cox
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