Asked by Helen
A 71 kg skydiver jumps out of a plane at an altitude of 1370m. She reaches terminal velocity of 68m/s. She opens the parachute at an altitude of 328m, and lands with a speed of 2.8m/s.
a)How much (negative) work is done by air resistance before she opens the chute?
b)How much is done by the parachute?
a)How much (negative) work is done by air resistance before she opens the chute?
b)How much is done by the parachute?
Answers
Answered by
Damon
How much Ke should she have at 328 m?
m g h = 71(9.81)(1370-328) Joules
How much Ke does she actually have?
(1/2) m v^2 = (1/2)(71)(68)^2
The difference is how much work is done by air friction.
==========================
At 328 meters she has a total energy of
m g 328 + (1/2) m (68)^2
At the ground she has a total energy of
0 + (1/2)m (2.8)^2
The difference is work done by the chute.
m g h = 71(9.81)(1370-328) Joules
How much Ke does she actually have?
(1/2) m v^2 = (1/2)(71)(68)^2
The difference is how much work is done by air friction.
==========================
At 328 meters she has a total energy of
m g 328 + (1/2) m (68)^2
At the ground she has a total energy of
0 + (1/2)m (2.8)^2
The difference is work done by the chute.
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