PHY303Kquiz1soln (2005)

# PHY303Kquiz1soln (2005) - Johnson, Matthew Quiz 1 Due: Feb...

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Unformatted text preview: Johnson, Matthew Quiz 1 Due: Feb 16 2005, 10:00 pm Inst: Kleinman 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A string going over a massless frictionless pul- ley connects two blocks of masses 4 . 2 kg and 9 . 3 kg. As shown on the picture below, the 4 . 2 kg block lies on a 32 incline; the coeffi- cient of kinetic friction between the block and the incline is = 0 . 22. The 9 . 3 kg block is hanging in the air. The 9 . 3 kg block accelerates downward while the 4 . 2 kg block goes up the incline with the same acceleration. 9 . 3 kg T 4 . 2 k g = . 2 2 T a 32 Given g = 9 . 8 m / s 2 , what is the accelera- tion of the system? Correct answer: 4 . 56661 m / s 2 . Explanation: Let : m 1 = 4 . 2 kg , m 2 = 9 . 3 kg , = 32 , and = 0 . 22 . Consider the Free Body Diagram for the mass m 1 : T F N W where W = m 1 g , T is the string tension, N is the normal force, and F is the kinetic friction force. Let the x axis point in the direction of the m 1 motion upward along the incline while the y axis is perpendicular to the incline. Then according to Newtons Second Law X F x = T -F -W sin = m 1 a, (1) X F y = N -W cos = 0 . (2) Substituting W = m 1 and the kinetic friction law F = N (3) into Eqs. 12 and solving for the string tension T , we find N = m 1 g cos = F = m 1 g cos T = m 1 a + m 1 g (sin + cos ) . (4) Now consider the secomd mass m 2 . There are only two forces acting on it: T pulling ver- tically up, and m 2 g pulling vertically down. Taking the downward acceleration of the m 2 as positive, we have m 2 a = m 2 g-T (5) and hence, in light of Eq. 4, we have m 2 ( g- a ) = m 1 a + m 1 g (sin + cos ) . (6) Johnson, Matthew Quiz 1 Due: Feb 16 2005, 10:00 pm Inst: Kleinman 2 This is a linear equation for the acceleration a ; its solution is a = m 2- m 1 (sin + cos ) m 2 + m 1 g = (9 . 8 m / s 2 ) (9 . 3 kg) + (4 . 2 kg) n (9 . 3 kg)- (4 . 2 kg) [ sin 32 + (0 . 22) cos 32 ] o = 4 . 56661 m / s 2 . (7) 002 (part 1 of 3) 10 points Consider a force pulling 3 blocks along a rough horizontal surface, where the masses are a multiple of a given mass m , as shown in the figure below. The coefficient of the kinetic friction is . The blocks are pulled by a force of 90 mg . 6 m 9 m 3 m T 1 T 2 F Determine the acceleration. 1. a = 8 g 2. a = 5 g 3. a = 6 g 4. a = 3 g 5. a = 4 g correct 6. a = 2 g 7. a = g 8. a = 7 g 9. a = 9 g Explanation: m 1 m 2 m 3 T 1 T 2 F Given : m 1 = 6 m, m 2 = 9 m, m 3 = 3 m, m 1 + m 2 + m 3 = 18 m, and F = 90 mg ....
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## This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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PHY303Kquiz1soln (2005) - Johnson, Matthew Quiz 1 Due: Feb...

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