Ch4Word - Chapter 4 1 Page 1 CHAPTER 4 Dynamics Newton’s Laws of Motion 1 We convert the units lb =(0.25 lb(4.45 N/lb ˜ 1 N 2 If we select the

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Unformatted text preview: Chapter 4 1 Page 1 CHAPTER 4 - Dynamics: Newton’s Laws of Motion 1. We convert the units: # lb = (0.25 lb)(4.45 N/lb) ˜ 1 N. 2. If we select the bike and rider as the object, we apply Newton’s second law to find the mass: ? F = ma ; 255 N = m (2.20 m/s 2 ), which gives m = 116 kg . 3. We apply Newton’s second law to the object: ? F = ma ; F = (7.0 × 10 –3 kg)(10,000)(9.80 m/s 2 ) = 6.9 × 10 2 N . 4. Without friction, the only horizontal force is the tension. We apply Newton’s second law to the car: ? F = ma ; F T = (1250 kg)(1.30 m/s 2 ) = 1.63 × 10 3 N . 5. We find the weight from the value of g . ( a ) Earth: F G = mg = (58 kg)(9.80 m/s 2 ) = 5.7 × 10 2 N . ( b ) Moon: F G = mg = (58 kg)(1.7 m/s 2 ) = 99 N . ( c ) Mars: F G = mg = (58 kg)(3.7 m/s 2 ) = 2.1 × 10 2 N . ( d ) Space: F G = mg = (58 kg)(0 m/s 2 ) = . 6. The acceleration can be found from the car’s one-dimensional motion: v = v + at ; 0 = [(90 km/h)/(3.6 ks/h)] + a (7.0 s), which gives a = – 3.57 m/s 2 . We apply Newton’s second law to find the required average force ? F = ma ; F = (1050 kg)(– 3.57 m/s 2 ) = – 3.8 × 10 3 N . The negative sign indicates that the force is opposite to the velocity. 7. The required average acceleration can be found from the one-dimensional motion: v 2 = v 2 + 2 a ( x – x ); (155 m/s) 2 = 0 + 2 a (0.700 m – 0), which gives a = 1.72 × 10 4 m/s 2 . We apply Newton’s second law to find the required average force ? F = ma ; F = (6.25 × 10 –3 kg)(1.72 × 10 4 m/s 2 ) = 107 N . 8. ( a ) The weight of the box depends on the value of g : F G = m 1 g = (30.0 kg)(9.80 m/s 2 ) = 294 N . We find the normal force from ? F y = ma y ; F N – m 1 g = 0, which gives F N = m 1 g = 294 N . ( b ) We select both blocks as the object and apply Newton’s s e c o n d l a w : ? F y = ma y ; F N1 – m 1 g – m 2 g = 0, which gives F N1 = ( m 1 + m 2 ) g = (30.0 kg + 20.0 kg)(9.80 m/s 2 ) = 490 N . If we select the top block as the object, we have ? F y = ma y ; F N2 – m 2 g = 0, which gives F N2 = m 2 g = (20.0 kg)(9.80 m/s 2 ) = 196 N . F N1 m 1 g x y m 2 g m 1 g F N2 F N m 1 g ( a ) ( b ) Chapter 4 2 Page 2 9. The required average acceleration can be found from the one-dimensional motion: v 2 = v 2 + 2 a ( x – x ); 0 = [(100 km/h)/(3.6 ks/h)] 2 + 2 a (150 m – 0), which gives a = – 2.57 m/s 2 . We apply Newton’s second law to find the required force ? F = ma ; F = (3.6 × 10 5 kg)(– 2.57 m/s 2 ) = – 9.3 × 10 5 N . The weight of the train is mg = (3.6 × 10 5 kg)(9.80 m/s 2 ) = 3.5 × 10 6 N, so Superman must apply a force that is 25% of the weight of the train. 10. The acceleration of the first box is a 1 = F / m 1 , so the speed after a time t is v 1 = v + a 1 t = a 1 t ....
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch4Word - Chapter 4 1 Page 1 CHAPTER 4 Dynamics Newton’s Laws of Motion 1 We convert the units lb =(0.25 lb(4.45 N/lb ˜ 1 N 2 If we select the

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