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A 85 kg sky diver jumps out of a plane at an altitude of 700 m....


A 85 kg sky diver jumps out of a plane at an altit
A 85 kg sky diver jumps out of a plane at an altitude of 700 m. After the skydiver falls for a period the parachute opens and lands with a speed of 3.6 m/s. What is the energy dissipated by the parachute during the dive? 75 kJ 14 kJ 114 kJ 1.8 kJ
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Answer #1

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If the parachute had not been used, landing speed would have been,

v = (2gh)1/2 = (2 * 9.81 * 700)1/2 = 117.2 m/s

Energy dissipated, E = 85 * (117.22 - 3.62) / 2 = 583 kJ

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Answer #2

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SOLUTION :


P. E. lost on reaching the ground 

= m g h 

= 85 * 9.8 * 700 

= 583100 J 


K. E. of the driver at  the time of landing 

= 1/2 m v^2

= 1/2 * 85 * (3.6)^2

= 550.8 J


Energy dissipated by the parachute during dive 

= 583100 - 550.8

= 582549.2 J

= 583  kJ  approx. (ANSWER)

answered by: Tulsiram Garg
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