Kaplunovsky - practice 05

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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern