Ch_4Word - Chapter 4 1 CHAPTER 4 Dynamics Newton's Laws of...

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Chapter  4 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: ? ma ;   255 N =  m (2.20 m/s 2 ), which gives        m  = 116 kg . 3. We apply Newton’s second  law to the object: ? 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: ? 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 ) =        0 . 6. The acceleration can be found  from the car’s one-dimensional motion: v  =  v 0  +  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 ? 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 0 2  + 2 a ( x  –  x 0 ); (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 ? 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  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 . Page 1
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Chapter  4 2 9. The required  average acceleration can be found  from the one-dimensional motion: v 2  =  v 0 2  + 2 a ( x  –  x 0 ); 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 ?
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