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Unformatted text preview: Chapter 4 1 CHAPTER 4  Dynamics: Newtons 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 Newtons second law to find the mass: ? F = ma ; 255 N = m (2.20 m/s 2 ), which gives m = 116 kg . 3. We apply Newtons 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 Newtons 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 cars onedimensional 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 Newtons 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 onedimensional 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 Newtons 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 Newtons second law: ? 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 ....
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 Spring '08
 Hadley
 Mass

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