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Unformatted text preview: Husain, Zeena – Exam 2 – Due: Oct 15 2003, 9:00 pm – Inst: H L Berk 1 Thisprintoutshouldhave14questions, check that it is complete. Multiplechoice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . A 5 . 5 kg mass on a frictionless inclined surface is connected to a 2 . 4 kg mass. The pulley is massless and frictionless, and the connecting string is massless and does not stretch. The 2 . 4 kg mass is acted upon by an upward force of 4 . 1 N, and has a downward acceleration of 4 . 7 m / s 2 . 5 . 5 k g μ = θ 2 . 4 kg 4 . 1 N 4 . 7m / s 2 What is the tension in the connecting string? Correct answer: 8 . 14 N. Explanation: Given : m 1 = 5 . 5 kg , m 2 = 2 . 4 kg , F = 4 . 1 N , and a = 4 . 7 m / s 2 . Basic Concept: X i ~ F i = m~ a Solution: It is easiest to analyze the forces on each block separately. m 2 F m 2 g T a m 1 a T m 1 g s i n θ N m 1 g c o s θ Note: T is the tension in the string, F is the external force, and the weight of m 2 is m 2 g . From the freebody diagram on the 2 . 4 kg block m 2 a = m 2 g F T , so T = m 2 g m 2 a F = (2 . 4 kg)(9 . 8 m / s 2 4 . 7 m / s 2 ) 4 . 1 N = 8 . 14 N . 002 (part 1 of 1) 10 points Given: The surface is frictionless, Two blocks in contact with each other move to the right across a horizontal surface by the two forces shown. 1 . 9 kg 5 . 8 kg 76 N 13 N μ = 0 Determine the magnitude of the force ex erted on the block with mass 1 . 9 kg by the block with mass 5 . 8 kg. Correct answer: 60 . 4545 N. Explanation: Given : F 1 = 76 N , F 2 = 13 N , M 1 = 1 . 9 kg , M 2 = 5 . 8 kg , and μ = 0 . Husain, Zeena – Exam 2 – Due: Oct 15 2003, 9:00 pm – Inst: H L Berk 2 M 1 M 2 F 1 F 2 μ = 0 a The common acceleration of the system is F 1 F 2 = ( M 1 + M 2 ) a a = F 1 F 2 M 1 + M 2 . (1) This result is obtained by treating the two masses as a single object. M 1 + M 2 F 2 F 1 ( M 1 + M 2 ) g N To determine the force exerted on M 1 by M 2 , we need to analyze M 1 separately. M 1 F 1 F 12 M 1 g N From the free body diagram of M 1 we see that the only two horizontal forces acting on M 1 are the external force F 1 and the force F 12 exerted by M 2 , resulting in the acceleration of the mass F 1 F 12 = M 1 a, so F 12 = F 1 M 1 a F 12 = F 1 M 1 • F 1 F 2 M 1 + M 2 ‚ F 12 = (76 N) (1 . 9 kg) × • (76 N) (13 N) (1 . 9 kg) + (5 . 8 kg) ‚ F 12 = (76 N) (15 . 5455 N) = 60 . 4545 N . 003 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . A curve of radius r is banked at angle θ so that a car traveling with uniform speed v can round the curve without relying on friction to keep it from slipping to its left or right....
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This note was uploaded on 03/23/2008 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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