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5 Solution Set 5 on chapters 10&amp;16&amp;17

# 5 Solution Set 5 on chapters 10&amp;16&amp;17 - 1...

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Chapter 10: Simple and Harmonic Motion and Elasticity. Selected Conceptual Questions, pp.312. Problems pp. 313: # 10, 13, 24, 34, 39, 49, 57, 66 Chapter 16: Waves and Sound. Selected Conceptual Questions, pp.499. Problems pp. 500: # 6, 16, 20, 23, 27, 42, 60, 70, 88, 104 Chapter 17: The Principle of Linear Superposition and Interference Phenomena . Selected Conceptual Questions, pp.527. Problems pp. 528: # 21, 37, 39, 43, 51, 52 Chapter 10: Simple and Harmonic Motion and Elasticity 10. REASONING The free-body diagram shows the magnitudes and directions of the forces acting on the block. The weight mg acts downward. The maximum force of static friction f s max acts upward just before the block begins to slip. The force from the spring F x Applied = kx (Equation 10.1) is directed to the right. The normal force F N from the wall points to the left. The magnitude of the maximum force of static friction is related to the magnitude of the normal force according to Equation 4.7 ( 29 max s s N f F μ = , where μ s is the coefficient of static friction. Since the block is at equilibrium just before it begins to slip, the forces in the x direction must balance and the forces in the y direction must balance. The balance of forces in the two directions will provide two equations, from which we will determine the coefficient of static friction. SOLUTION Since the forces in the x direction and in the y direction must balance, we have N s N and F kx mg F μ = = Substituting the first equation into the second equation gives ( 29 ( 29 ( 29 ( 29 ( 29 2 s N s s 1.6 kg 9.80 m/s or 0.79 510 N/m 0.039 m mg mg F kx kx μ μ μ = = = = = + y + x F N F s max = μ s F N mg F x Applied = kx 1

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13. REASONING AND SOLUTION From the drawing given with the problem statement, we see that the kinetic frictional force on the bottom block (#1) is given by f k1 = µ k ( m 1 + m 2 ) g (1) and the maximum static frictional force on the top block (#2) is MAX s2 s 2 f m g μ = (2) Newton’s second law applied to the bottom block gives F f k1 kx = 0 (3) Newton’s second law applied to the top block gives MAX s2 0 f kx - = (4) a. To find the compression x , we have from Equation (4) that x = MAX s2 f / k = µ s m 2 g / k = (0.900)(15.0 kg)(9.80 m/s 2 )/(325 N/m) = 0.407 m b. Solving Equation (3) for F and then using Equation (1) to substitute for f k1 , we find that F = kx + f k1 = kx + µ k ( m 1 + m 2 ) g F = (325 N/m)(0.407 m) + (0.600)(45.0 kg)(9.80 m/s 2 ) = 397 N ______________________________________________________________________________ 24. REASONING AND SOLUTION The cup slips when the force of static friction is overcome. So F = ma = µ s mg , where the acceleration is the maximum value for the simple harmonic motion, i.e., so that a max = A ϖ 2 = A (2 π f ) 2 (10.10) μ s = A (2 π f ) 2 / g = (0.0500 m)4 π 2 (2.00 Hz) 2 /(9.80 m/s 2 ) = 0.806 ______________________________________________________________________________ 34. REASONING It is assumed in Example 16 that the only forces acting on the jumper are the gravitational force (his weight) and, for the latter part of his descent, the elastic force of the bungee cord. Therefore, only conservative forces are present, and we may use energy conservation to guide our solution. He possesses gravitational potential energy with respect to the water and elastic potential energy. At the lowest point in his fall, he has no kinetic energy since he comes to a momentary halt and has zero speed.
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5 Solution Set 5 on chapters 10&amp;16&amp;17 - 1...

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