Kaplunovsky - practice 05

# Kaplunovsky - practice 05 - practice 05 Due noon Question 1...

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practice 05 – – Due: Feb 20 2007, noon 1 Question 1, chap 5, sect 2. part 1 of 1 5 points A 4100 kg helicopter accelerates upward at 1 . 8 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . What lift force is exerted by the air on the propellers? Correct answer: 47560 N (tolerance ± 1 %). Explanation: The propellers exert a force F upward and gravity acts downward. The net force (di- rected upward) on the helicopter is F = m a = Σ F up Σ F down = F m g Thus F = m a + m g = m ( a + g ) Note: Here a and g are absolute values of the helicopter’s acceleration and the acceleration of gravity, respectively. Question 2, chap 5, sect 5. part 1 of 1 5 points Two students sit in identical office chairs facing each other. Bob has a mass of 95 kg, while Jim has a mass of 77 kg. Bob places his bare feet on Jim’s knees, as shown to the right. Bob then suddenly pushes outward with his feet, causing both chairs to move. Bob Jim In this situation, while Bob’s feet are in con- tact with Jim’s knees, 1. Each student exerts the same amount of force on the other. correct 2. Neither student exerts a force on the other. 3. Bob exerts a force on Jim, but Jim doesn’t exert a force on Bob. 4. None of these answers is correct. 5. Each student exerts a force on the other, but Bob exerts a larger force. 6. Each student exerts a force on the other, but Jim exerts a larger force. Explanation: By Newton’s third law, action and reaction are of the same magnitude and in the opposite direction. Question 3, chap 5, sect 6. part 1 of 3 5 points Note: a and g are magnitudes; the direc- tions of vectora and vectorg are shown in the figure below (the acceleration is pointing up ). Consider a force of 14 m g pulling three blocks connected by inextensible strings (which can be considered massless) in the ver- tical direction, where the masses are a multi- ple of a given mass m . 2 m 7 m 3 m F = 14 m g T 1 T 2 a g Find the acceleration a of this system of blocks. 1. a = 1 2 g 2. a = 1 6 g correct 3. a = 1 14 g 4. a = 1 3 g 5. a = 1 4 g

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practice 05 – – Due: Feb 20 2007, noon 2 6. a = 1 12 g 7. a = 1 8 g 8. a = g 9. a = 1 5 g Explanation: m 1 m 2 m 3 F T 1 T 2 a g Given : m 1 = 2 m , m 2 = 7 m , m 3 = 3 m , m 1 + m 2 + m 3 = 12 m , and F = 14 m g . Basic Concept: Newton’s 2nd law summationdisplay vector F = M vectora . Solution: The system is accelerating as if there were only one block whose mass is m 1 + m 2 + m 3 = 12 m . m 1 + m 2 + m 3 F a g ( m 1 + m 2 + m 3 ) g Newton’s Second Law for the entire system gives summationdisplay vector F = M vectora F ( m 1 + m 2 + m 3 ) g = +( m 1 + m 2 + m 3 ) a a = + F ( m 1 + m 2 + m 3 ) g m 1 + m 2 + m 3 = + 14 m g 12 m g 12 m = 2 m g +12 m = g 6 . Question 4, chap 5, sect 6. part 2 of 3 5 points What is the equation of motion for the block with mass 7 m ? 1. T 1 + 7 m g T 2 = 7 m a 2. T 1 + 7 m g + T 2 2 m g = 7 m a 3. T 1 7 m g + T 2 2 m g = 7 m a 4. T 1 7 m g T 2 = 7 m a correct 5. T 1 7 m g T 2 2 m g = 7 m a 6. T 1 + 7 m g + T 2 + 2 m g = 7 m a 7. T 1 + 7 m g + T 2 = 7 m a 8. T 1 7 m g + T 2 = 7 m a 9. T 1 7 m g + T 2 + 2 m g = 7 m a 10. T 1 + 7 m g T 2 2 m g = 7 m a Explanation: m 2 T 1 a g T 2 m 2 g Applying Newton’s Second Law for m 3 and
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