(690) Problem 17: Mike has a mass of mM= 85 kg. He jumps out of a perfectly good
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(690) Problem 17: Mike has a mass of mM= 85 kg. He jumps out of a perfectly good airplane that is 2000 m above the ground. After he falls 1000 m, when his downward speed is v1 = 42 ms, Mike opens his parachute. The positive y-direction is downward Randomized Variables mM= 85 kg v,-42 m/s > 3396 Part (a) Calculate the average magnitude of the upward force, Fal in Newtons, of the air resistance on Mike during his initial descent. Grade Summary Deductions 0% Potential 100% sin Submissions Attempts remaining:- (0% per attempt) detailed view tan( ) ??? cosO cotanasin acosO atan acotan sinho cosh tanh cotanhO END Degrees Radians No DELI CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 3 Feedback: deduction per feedback. 3396 Part (b) After Mike opens his parachute, he continues to descend, eventually reaching the ground with a speed of 4.0 m/s. Calculate the average upward force, Fa2 in Newtons, during this part of Mike's descent 33% Part (c) At the same time Mike jumps out of the airplane, his wallet (mass of mw-.3 kg) falls out of his pocket. Calculate the wallet's downward speed, Vwfin m/s, when it reaches the ground. For this calculation, assume that air resistance is negligibleExplanation / Answer
(i) Relative to where he opens his chute,
GPE = mgh = 85kg * 9.8m/s² * 1000m = 833 kJ
When he opens his chute,
KE = ½mv² = ½ * 85kg * (42m/s)² = 74.97 kJ
So the friction work done = (833-74.97) kJ = 758.03 kJ = F * d = F * 1000m,
and average F = 758.03 N
(ii) Relative to the ground, when he opens his chute,
TME = KE + GPE = 74.97kJ + 833kJ = 907.97 kJ
and at the ground,
TME = ½ * 85kg * (4m/s)² = 680 J
friction work = 907.97 kJ = F * 1000m,
and average F = 907.97 N.
(iii) Without friction, v = ?(2gh) = ?(2 * 9.8m/s² * 2000m) = 198 m/s
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