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Unformatted text preview: Version 105 – Quiz 3 – coker – (58245) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This quiz covers topics from class lectures over Chs. 10, 11, 12, 13 and 14 in the text book. 001 10.0 points A 80 . 1 kg ice skater, moving at 8 . 4 m / s, crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at 4 . 2 m / s. Suppose the average force a skater can experience without breaking a bone is 4424 N. If the impact time is 0 . 11 s, what is the magnitude of the average force each skater experiences? 1. 4862.41 2. 3496.96 3. 7131.25 4. 5807.08 5. 2884.72 6. 1375.86 7. 3659.15 8. 5118.75 9. 3891.77 10. 3058.36 Correct answer: 3058 . 36 N. Explanation: Let : m = 80 . 1 kg , v i = 8 . 4 m / s , v f = 4 . 2 m / s , F = 4424 N , and Δ t = 0 . 11 s . From the impulsemomentum equation, F av Δ t = Δ p = m ( v f v i ) . F av = m ( v f v i ) Δ t = (80 . 1 kg) (4 . 2 m / s 8 . 4 m / s) . 11 s = 3058 . 36 N , which has a magnitude of 3058 . 36 N . The average force on skater 2 has the same magni tude but opposite direction by Newton’s third law. This force is not great enough to break a bone. 002 (part 1 of 2) 10.0 points Consider the rectangular block of mass m = 43 kg height h = 0 . 58 m, length l = 0 . 7 m. A force F is applied horizontally at the upper edge. The acceleration of gravity is 9 . 8 m / s 2 . l m h F What is the minimum force required to start to tip the block? 1. 86.45 2. 81.0133 3. 238.378 4. 123.284 5. 129.392 6. 112.531 7. 109.564 8. 75.5491 9. 155.018 10. 254.293 Correct answer: 254 . 293 N. Explanation: Basic Concepts: In equilibrium summationdisplay vector F = 0 summationdisplay vector τ = 0 Part 1: Let the right lower corner of the block be denoted as A . Then summationdisplay τ A = hF l 2 mg = 0 Therefore F = mg l 2 h = (43 kg) (9 . 8 m / s 2 ) . 7 m 2 (0 . 58 m) = 254 . 293 N . Version 105 – Quiz 3 – coker – (58245) 2 003 (part 2 of 2) 10.0 points What is the minimum coefficient of static fric tion required for the block to tip with the application of a force of this magnitude? 1. 0.603448 2. 0.243333 3. 0.254167 4. 0.427083 5. 0.290909 6. 0.211538 7. 0.396907 8. 0.555556 9. 0.414894 10. 0.5 Correct answer: 0 . 603448. Explanation: summationdisplay F y = N A mg = 0 Therefore N A = mg and summationdisplay F x = F f = F μN A = 0 . Solving for μ , μ = F N A = F mg = 254 . 293 N (43 kg) (9 . 8 m / s 2 ) = 0 . 603448 004 10.0 points A sticky blob strikes and sticks to a free rod, which is initially at rest, as shown. Let E be the mechanical energy of the sys tem, vector P the linear momentum of the system, and vector L the angular momentum of the system....
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This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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