Chapter 13 - CHAPTER 13 Quick Quizzes 1. 2. 3. 4. 5. (d)....

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C HAPTER 13 Quick Quizzes 1. (d). 2. (c). 3. (b). 4. (d). 5. (b) and (c). An accelerating elevator is equivalent to a gravitational field. Thus, if the elevator is accelerating upward, this is equivalent to an increased effective gravitational field magnitude g , and the period will decrease. Similarly, if the elevator is accelerating downward, the effective value of g is reduced and the period increases. If the elevator moves with constant velocity, the period of the pendulum is the same as that in the stationary elevator. 6. (a). The clock will run slow . With a longer length, the period of the pendulum will increase. Thus, it will take longer to execute each swing, so that each second according to the clock will take longer than an actual second. 7. (b). Greater. The value of g on the Moon is about one-sixth the value of g on Earth, so the period of the pendulum on the moon will be greater than the period on Earth. 405
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CHAPTER 13 Problem Solutions 13.1 (a) The force exerted on the block by the spring is ( )( ) 160 N m 0.15 m 24 N s Fk x =− =+ , or s F = 24 N directed toward equilibrium position (b) From Newton's second law, the acceleration is 2 24 N m 60 0.40 kg s s F a m + == = + = 2 m 60 toward equilibrium position s 13.2 (a) The spring constant is 3 -2 50 N 1.0 10 N m 5.0 10 m s F mg k xx === × . () 3 1.0 0.11 m s FF k x === × = 2 1.1 10 N × (b) The graph will be a straight line passing through the origin with a slope equal to 3 1.0 k . 13.3 (a) Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m before coming to rest momentarily. It will then repeat this motion over and over again with a regular period. (b) From 2 1 2 yi y yv t a t ∆= + , with 0 yi v = , the time required for the ball to reach the ground is 2 2 2 4.00 m 0.904 s 9.80 m s y y t a ∆− = . This is one-half of the time for a complete cycle of the motion. Thus, the period is T = 1.81 s . (c) No . The net force acting on the mass is a constant given by F mg (except when it is contact with the ground). This is not in the form of Hooke’s law. 406
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CHAPTER 13 13.4 The force the hand exerts on the handle is equal in magnitude and opposite in direction to the sum of the forces exerted on the handle by the springs, or ( ) 1234 4 h =− + + + FF F F F 1 F . Thus, ( ) 1 44 h FF k x == and if ( ) 0.800 0.435 m 0.365 m max x = −= , then ) () 4 65.0 N m 0.365 m h max F 94.9 N 13.5 Since object A is in equilibrium, the net force acting on it must be zero, giving Hence, 121 2 80.0 N FFk xk x += + = 12 80.0 N 80.0 N = 40.0 N cm 25.0 N cm x kk ++ 1.23 cm 13.6 (a) The sketch at the right is a free-body diagram of the upper end of the spring shown in Figure P13.6. This point is in equilibrium, so 11 2 sin 32.5 0 y k x Σ = . If 4 1 210 N and 5.60 10 N m Fk × , the elongation of the spring is ( ) 3 1 4 1 2 210 N sin 32.5 2 sin 32.5 4.03 10 m 5.60 F x k ° ° = × = 4.03 mm × 32.5° 32.5° F 1 = k 1 x 1 (b) The rope is now replaced by a pair of identical springs, lying along the original line of the rope. If the force constant of each of these springs is k and the elongation of each is 1 2 x x = , then gives 0 y F Σ= ( ) 2 sin 32.5 kx k x °= , or 1 2sin 32.5 2sin 32.5 2 k x x °° 4 5.60 4sin32.5 × ° 4 2.61 × 407
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CHAPTER 13 13.7 (a) The spring constant of each band is 3 -2 15 N 1.5 10 N m 1.0 10 m s F k x == × Thus, when both bands are stretched 0.20 m, the total elastic potential energy is
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Chapter 13 - CHAPTER 13 Quick Quizzes 1. 2. 3. 4. 5. (d)....

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