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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 θ. Let v be the final velocity of the spreader. According to Newton’s second law, F h = ma h so F cos θ = m Δ v h Δ t F = m Δ v h Δ t cos θ = mv t cos θ = (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 Newton’s 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 θ α 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 θ = (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 boat’s 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 boat’s accelera tion? Correct answer: 55 . 3498 ◦ (N of West)....
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This note was uploaded on 04/07/2012 for the course PHYSICS 121 taught by Professor James during the Spring '12 term at University of Texas.
 Spring '12
 James
 Physics

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