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Unformatted text preview: hanna (brh687) hk 4 Opyrchal (11105) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Joe pushes down the length of the handle of a(n) 12 . 5 kg lawn spreader. The handle makes an angle of 42 . 5 with the horizontal. Joe wishes to accelerate the spreader from rest to 1 . 32 m / s in 1 . 7 s. What force must Joe apply to the handle? Correct answer: 13 . 1645 N. Explanation: The horizontal component of the force is F h = F cos . Let v be the final velocity of the spreader. According to Newtons second law, F h = ma h so F cos = m v h t F = m v h t cos = mv t cos = (12 . 5 kg) (1 . 32 m / s) (1 . 7 s) cos 42 . 5 = 13 . 1645 N 002 10.0 points A person of mass 92 . 8 kg escapes from a burn ing building by jumping from a window situ ated 28 . 3 m above a catching net. The acceleration of gravity is 9 . 8 m / s 2 . If air resistance exerts a force of 119 . 1 N on him as he falls, determine his speed just before he hits the net. Correct answer: 21 . 9554 m / s. Explanation: The forces acting on him are the gravita tional force mg acting downward, and the air resistance, F a , acting upward. The accelera tion is downward and the net force is F net = ma = mg F a a = mg F a m The person is in free fall, so his final speed is defined by v f = 2 a h = radicalbigg 2 ( mg F a ) h m = radicalBigg 2 [(92 . 8 kg)(9 . 8 m / s 2 ) 119 . 1 N] (28 . 3 m) 92 . 8 kg = 21 . 9554 m / s . 003 (part 1 of 2) 10.0 points A 36900 kg sailboat experiences an eastward force 11400 N due to the tide pushing its hull while the wind pushes the sails with a force of 52200 N directed toward the northwest (45 westward of North or 45 northward of West). What is the magnitude of the resultant ac celeration of the sailboat? Correct answer: 1 . 21596 m / s 2 . Explanation: According to Newtons Second Law, mvectora = vector F net = vector F wind + vector F tide . To find the magnitude of the net force, we draw the parallelogram for vector addition vector F wind vector F tide vector F net NW W E Note that 135 is the angle between the tide and the wind forces. Use the Law of Cosines: F 2 net = F 2 wind + F 2 tide 2 F wind F tide cos = (52200 N) 2 + (11400 N) 2 2 (52200 N)(11400 N) cos135 = 2 . 01323 10 9 N 2 hanna (brh687) hk 4 Opyrchal (11105) 2 and the boats acceleration is a = F net m = 2 . 01323 10 9 N 2 36900 kg = 1 . 21596 m / s 2 . 004 (part 2 of 2) 10.0 points What is the direction of the boats accelera tion? Correct answer: 55 . 3498 (N of West)....
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 Spring '12
 James
 Physics

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