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### HW4-solutions

Course: PHY 302k, Spring 2008
School: University of Texas
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Word Count: 415

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(avs294) seunsom HW4 bohm (59970) This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A soccer ball kicked with a force of 14.2 N accelerates at 6.9 m/s2 to the right. What is the mass of the ball? Correct answer: 2.05797 kg. Explanation: Basic Concept: Fnet = F = ma. 1 003 (part 2 of 2) 10.0 points What...

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(avs294) seunsom HW4 bohm (59970) This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A soccer ball kicked with a force of 14.2 N accelerates at 6.9 m/s2 to the right. What is the mass of the ball? Correct answer: 2.05797 kg. Explanation: Basic Concept: Fnet = F = ma. 1 003 (part 2 of 2) 10.0 points What should it weight on Jupiter, where gravity is 2.64 times of that on Earth? Correct answer: 67.316 N. Explanation: Given : gJ = 2.64 gE WJ = m gJ = m (2.64 gE ) = 2.64 WE = (2.64) (5.73 lb) 4.45 N 1 lb Let : Fnet = 14.2 N to the right a = 6.9 m/s2 to the right . Solution: Fnet a (14.2 N) m= (6.9 m/s2 ) = 2.05797 kg . m= 002 (part 1 of 2) 10.0 points A bag of produce weighs 5.73 lb on Earth. What should it weigh on the Moon, where 1 the free-fall acceleration is that on Earth? 6 Correct answer: 4.24975 N. Explanation: Given : WE = 5.73 lb and gE . gM = 6 The weight of the bag of sugar on the Moon is WM = m gM m gE = 6 WE = 6 5.73 lb 4.45 N = 6 1 lb = 4.24975 N . = 67.316 N 004 10.0 points A(n) 7.1 g bullet leaves the muzzle of a rie with a speed of 520 m/s. What constant force is exerted on the bullet while it is traveling down the 0.5 length m of the barrel of the rie? Correct answer: 1919.84 N. Explanation: Average acceleration can be found from 2 2 vf = vo + 2 a Since vo = 0, we have v2 a= 2 Thus v2 F = ma = m 2 (7.1 g)(520 m/s)2 1 kg = 2 (0.5 m) 1000 g = 1919.84 N 005 10.0 points A(n) 1492.2 kg car is coasting along a level road at 31.7 m/s. A constant braking force is applied, such that the car is stopped in a distance of 60.9 m. seunsom (avs294) HW4 bohm (59970) What is the magnitude of the braking force? Correct answer: 12311.1 N. Explanation: Basic Concepts: 2 2 vf = v0 + 2a S 2 Solution: FT = ma + Fg = ma + mg = (4 kg) 2.1 m/s2 + (4 kg) 9.81 m/s2 = 47.64 N upward. Newtons Second Law of Motion Solution: Since vf = 0 a= 2 v0 2S (31.7 m/s)2 = 2 (60.9 m) = 8.25033 m/s2 . So the retarding force should be F = ma = (1492.2 kg)(8.25033 m/s2 ) = 12311.1 N F = 12311.1 N . 006 10.0 points A 4.0 kg bucket of water is raised from a well by a rope. The acceleration of gravity is 9.81 m/s2 . If the upward acceleration of the bucket is 2.1 m/s2 , nd the force exerted by the rope on the bucket of water. Correct answer: 47.64 N. Explanation: FT Fg Note: Figure is not drawn to scale. Basic Concept: Fnet = ma = FT Fg Given: m = 4.0 kg a = 2.1 m/s2 g = 9.81 m/s2 2.1 m/s2
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Maryland - ECON - 441
Maryland - ECON - 441
Maryland - ECON - 441
Maryland - ECON - 441
Maryland - ECON - 441
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