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F10P1AChap5Sol

F10P1AChap5Sol - More Applications of Newtons Laws...

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147 More Applications of Newton’s Laws SOLUTIONS TO PROBLEMS Section 5.1 Forces of Friction P5.1 For equilibrium: f = F and n = F g . Also, f = μ n i.e., μ = f n = F F g μ s = 75.0 N 25.0 9.80 ( ) N = 0.306 and μ k = 60.0 N 25.0 9.80 ( ) N = 0.245 . FIG. P5.1 P5.2 F y ! = ma y : + n " mg = 0 f s # μ s n = μ s mg This maximum magnitude of static friction acts so long as the tires roll without skidding. F x ! = ma x : " f s = ma The maximum acceleration is a = ! μ s g . The initial and final conditions are: x i = 0 , v i = 50.0 mi h = 22.4 m s , v f = 0 , v f 2 = v i 2 + 2 a x f ! x i ( ) : ! v i 2 = ! 2 μ s gx f (a) x f = v i 2 2 μ g x f = 22.4 m s ( ) 2 2 0.100 ( ) 9.80 m s 2 ( ) = 256 m (b) x f = v i 2 2 μ g x f = 22.4 m s ( ) 2 2 0.600 ( ) 9.80 m s 2 ( ) = 42.7 m
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148 More Applications of Newton’s Laws P5.8 m suitcase = 20.0 kg , F = 35.0 N F x ! = ma x : " 20.0 N + F cos # = 0 F y ! = ma y : + n + F sin # " F g = 0 (a) F cos ! = 20.0 N cos ! = 20.0 N 35.0 N = 0.571 ! = 55.2 ° FIG. P5.8 (b) n = F g ! F sin " = 196 ! 35.0 0.821 ( ) [ ] N n = 167 N P5.9 m = 3.00 kg , ! = 30.0 ° , x f = 2.00 m , t = 1.50 s , x i = 0 , v xi = 0 (a) x f = 1 2 at 2 : 2.00 m = 1 2 a 1.50 s ( ) 2 a = 4.00 1.50 ( ) 2 = 1.78 m s 2 FIG. P5.9 r F ! = r n + r f + m r g = m r a : Along x : 0 ! f + mg sin 30.0 ° = ma f = m g sin 30.0 ° ! a ( ) Along y : n + 0 ! mg cos30.0 ° = 0 n = mg cos30.0 ° (b) μ k = f n = m g sin 30.0 ° ! a ( ) mg cos30.0 ° , μ k = tan 30.0 ° !
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