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Unformatted text preview: rose (kmr2427) – HW #5 – Antoniewicz – (57380) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 5.0 points A 3 . 09 kg block is placed on top of a 10 . 8 kg block. A horizontal force of F = 55 N is applied to the 10 . 8 kg block, and the 3 . 09 kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is . 0963. There is friction both between the masses and between the 10 . 8 kg block and the ground. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 09 kg 10 . 8 kg μ = 0 . 0963 μ = 0 . 0963 F T Determine the tension T in the string. Correct answer: 2 . 91616 N. Explanation: Given : m = 3 . 09 kg , M = 10 . 8 kg , m + M = 13 . 89 kg , μ = 0 . 0963 , and F = 55 N . Basic Concept: Newton’s second law. Solution: Consider the free body diagram for the situation. m N 1 mg μ mg T M N 1 = mg N 2 M g μ ( m + M ) g μ mg F Applying Newton’s second law on block m yields m : summationdisplay F y = N 1 mg = 0 (1) m : summationdisplay F x = f 1 T = 0 , (2) where N 1 is the normal force exerted on block m by block M and f 1 is the frictional force between block m and block M . Using equa tion (2) to solve for T and noting that from equation (1), N 1 = mg , we obtain T = f 1 = μ N 1 = μ mg = (0 . 0963) (3 . 09 kg) (9 . 8 m / s 2 ) = 2 . 91616 N . 002 (part 2 of 2) 5.0 points Determine the magnitude of the acceleration of the 10 . 8 kg block. Correct answer: 3 . 60882 m / s 2 . Explanation: Applying Newton’s second law on block M yields M : summationdisplay F y = N 2 N 1 M g = 0 (3) M : summationdisplay F x = F f 1 f 2 = M a , (4) where N 2 is the normal force on block M by the ground and f 2 is the frictional force be tween block M and the ground. From equa tion (3), N 2 = M g + N 1 = M g + mg = ( M + m ) g . Thus f 2 = μ N 2 = μ ( M + m ) g Then solving for a from (4), we have a = 1 M ( F f 1 f 2 ) = 1 M bracketleftBig F μ mg μ ( M + m ) g bracketrightBig rose (kmr2427) – HW #5 – Antoniewicz – (57380) 2 = 1 M bracketleftBig F μ (2 m + M ) g bracketrightBig = 1 10 . 8 kg braceleftBig 55 N (0 . 0963) bracketleftBig 2 (3 . 09 kg) + (10 . 8 kg) bracketrightBig × (9 . 8 m / s 2 ) bracerightBig = 3 . 60882 m / s 2 . 003 (part 1 of 2) 5.0 points A small metal ball is suspended from the ceil ing by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9 . 8 m / s 2 . v 9 . 8 m / s 2 2 . 1 m 6 . 4 kg 2 7 ◦ What is the speed of the ball when it is in circular motion? Correct answer: 2 . 18187 m / s....
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This note was uploaded on 03/01/2011 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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