This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: rathbun (rar2954) 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 . 86 kg block is placed on top of a 9 . 87 kg block. A horizontal force of F = 54 . 6 N is applied to the 9 . 87 kg block, and the 3 . 86 kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is . 0958. There is friction both between the masses and between the 9 . 87 kg block and the ground. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 86 kg 9 . 87 kg = 0 . 0958 = 0 . 0958 F T Determine the tension T in the string. Correct answer: 3 . 62392 N. Explanation: Given : m = 3 . 86 kg , M = 9 . 87 kg , m + M = 13 . 73 kg , = 0 . 0958 , and F = 54 . 6 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 : F y = N 1 mg = 0 (1) m : 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 . 0958) (3 . 86 kg) (9 . 8 m / s 2 ) = 3 . 62392 N . 002 (part 2 of 2) 5.0 points Determine the magnitude of the acceleration of the 9 . 87 kg block. Correct answer: 3 . 85874 m / s 2 . Explanation: Applying Newtons second law on block M yields M : F y = N 2N 1 M g = 0 (3) M : 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 F mg ( M + m ) g rathbun (rar2954) HW #5 Antoniewicz (57380) 2 = 1 M F (2 m + M ) g = 1 9 . 87 kg 54 . 6 N (0 . 0958) 2 (3 . 86 kg) + (9 . 87 kg) (9 . 8 m / s 2 ) = 3 . 85874 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 . 3 m 5 . 8 kg 2 2 What is the speed of the ball when it is in circular motion? Correct answer: 1 . 3886 m / s. Explanation: Let : = 1 . 3 m , = 22 , g = 9 . 8 m / s 2 , and m = 5 . 8 kg . Use the free body diagram below. T mg The tension on the string can be decom posed into a vertical component which bal ances the weight of the ball and a horizontal component which causes the centripetal ac celeration, a centrip that keeps the ball on its horizontal circular path at radius r = sin ....
View
Full
Document
This note was uploaded on 03/08/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

Click to edit the document details