2 Solution Set 2 on chapters 4&amp;5

# 2 Solution Set 2 on chapters 4&amp;5 - 1 Chapter 4:...

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Chapter 4: Forces and Newton’s Laws of Motion Selected Conceptual Questions, pp.125. Problems pp. 126: # 9, 10, 17, 28, 36, 48, 55, 76, 84, 93, 112, 115 Chapter 5: Dynamics of Uniform Circular Motion Selected Conceptual Questions, pp.155. Problems pp. 156: # 4, 12, 23, 30, 38, 48 Chapter 4: Forces and Newton’s Laws of Motion 9. SSM WWW REASONING Let due east be chosen as the positive direction. Then, when both forces point due east, Newton's second law gives A B 1 F F ma F + = Σ 142 43 (1) where 2 1 0.50 m/s a = . When A F points due east and B F points due west, Newton's second law gives A B 2 F F ma F = Σ 142 43 (2) where 2 2 0.40 m/s a = . These two equations can be used to find the magnitude of each force. SOLUTION a. Adding Equations 1 and 2 gives ( 29 ( 29 ( 29 2 2 1 2 A 8.0 kg 0.50 m /s 0.40 m / s 3.6 N 2 2 m a a F + + = = = b. Subtracting Equation 2 from Equation 1 gives ( 29 ( 29 ( 29 2 2 1 2 B 8.0 kg 0.50 m /s 0.40 m /s 0.40 N 2 2 m a a F - - = = = ________________________________________________________________________________ 1

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10. REASONING From Newton’s second law, we know that the net force F Σ acting on the electron is directly proportional to its acceleration, so in part a we will first find the electron’s acceleration. The problem text gives the electron’s initial velocity ( v 0 = +5.40×10 5 m/s) and final velocity ( v = +2.10×10 6 m/s), as well its displacement ( x = +0.038 m) during the interval of acceleration. The elapsed time is not known, so we will use Equation 2.9 ( 29 2 2 0 2 v v ax = + to calculate the electron’s acceleration. Then we will find the net force acting on the electron from Equation 4.1 ( 29 F ma Σ = and the electron’s mass. Because F 1 points in the + x direction and F 2 points in the − x direction, the net force acting on the electron is 1 2 F F F Σ = - . In part b of the problem, we will rearrange this expression to obtain the magnitude of the second electric force. SOLUTION a. Solving Equation 2.9 for the electron’s acceleration, we find that ( 29 ( 29 ( 29 2 2 6 5 2 2 13 2 0 2.10 10 m/s 5.40 10 m/s 5.42 10 m/s 2 2 +0.038 m v v a x + × - + × - = = = + × Newton’s 2 nd law of motion then gives the net force causing the acceleration of the electron: ( 29 ( 29 31 13 2 17 9.11 10 kg 4.94 10 N F ma - - = = × + × = + × b. The net force acting on the electron is 1 2 F F F Σ = - , so the magnitude of the second electric force is 2 1 F F F = - Σ , where F Σ is the net force found in part a : 17 17 17 2 1 7.50 10 N 4.94 10 N 2.56 10 N F F F - - - = - Σ = × - × = × 17. REASONING Equations 3.5a ( 29 2 1 0 2 x x x v t a t = + and 3.5b ( 29 2 1 0 2 y y y v t a t = + give the displacements of an object under the influence of constant accelerations a x and a y . We can add these displacements as vectors to find the magnitude and direction of the resultant displacement. To use Equations 3.5a and 3.5b, however, we must have values for a x and a y . We can obtain these values from Newton’s second law, provided that we combine the given forces to calculate the x and y components of the net force acting on the duck, and it is here that our solution begins.
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## This note was uploaded on 04/01/2011 for the course PHY 135 taught by Professor Wagihghobriel during the Spring '11 term at University of Toronto- Toronto.

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2 Solution Set 2 on chapters 4&amp;5 - 1 Chapter 4:...

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