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

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 15

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online