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Unformatted text preview: mata (jpm2873) HW #5 Antoniewicz (57420) 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 . 2 kg block is placed on top of a 11 . 7 kg block. A horizontal force of F = 55 . 9 N is applied to the 11 . 7 kg block, and the 3 . 2 kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0 . 178. There is friction both between the masses and between the 11 . 7 kg block and the ground. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 2 kg 11 . 7 kg = 0 . 178 = 0 . 178 F T Determine the tension T in the string. Correct answer: 5 . 58208 N. Explanation: Given : m = 3 . 2 kg , M = 11 . 7 kg , m + M = 14 . 9 kg , = 0 . 178 , and F = 55 . 9 N . Basic Concept: Newtons second law. Solution: Consider the free body diagram for the situation. m N 1 mg mg T M N 1 = mg N 2 M g ( m + M ) g mg F Applying Newtons 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 = N 1 = mg = (0 . 178) (3 . 2 kg) (9 . 8 m / s 2 ) = 5 . 58208 N . 002 (part 2 of 2) 5.0 points Determine the magnitude of the acceleration of the 11 . 7 kg block. Correct answer: 2 . 07918 m / s 2 . Explanation: Applying Newtons 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 = N 2 = ( M + m ) g Then solving for a from (4), we have a = 1 M ( F f 1 f 2 ) = 1 M bracketleftBig F mg ( M + m ) g bracketrightBig mata (jpm2873) HW #5 Antoniewicz (57420) 2 = 1 M bracketleftBig F (2 m + M ) g bracketrightBig = 1 11 . 7 kg braceleftBig 55 . 9 N (0 . 178) bracketleftBig 2 (3 . 2 kg) + (11 . 7 kg) bracketrightBig (9 . 8 m / s 2 ) bracerightBig = 2 . 07918 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 1 . 4 m 5 . 6 kg 3 1 What is the speed of the ball when it is in circular motion? Correct answer: 2 . 06055 m / s....
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 Spring '08
 Turner
 Physics

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