F10P1AChap4Sol

# F10P1AChap4Sol - The Laws of Motion SOLUTIONS TO PROBLEMS...

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107 The Laws of Motion SOLUTIONS TO PROBLEMS Section 4.3 Mass P4.2 Since the car is moving with constant speed and in a straight line, the resultant force on it must be zero regardless of whether it is moving (a) toward the right or (b) the left. Section 4.4 Newton’s Second Law ί The Particle Under a Net Force P4.3 m = 3.00 kg r a = 2.00 ˆ i + 5.00 ˆ j ( ) 2 r F = m r a = 6.00 ˆ i + 15.0 ˆ j ( ) N ! r F ! = 6.00 ( ) 2 + 15.0 ( ) 2 N = 16.2 N

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108 The Laws of Motion P4.4 (a) ! r F = r F 1 + r F 2 = " 9.00 ˆ i + 3.00 ˆ j ( ) N Acceleration r a = a x ˆ i + a y ˆ j = ! r F m = " 9.00 ˆ i + 3.00 ˆ j ( ) N 2.00 kg = " 4.50 ˆ i + 1.50 ˆ j ( ) m s 2 Velocity r v f = v x ˆ i + v y ˆ j = r v i + r a t = r a t r v f = ! 4.50 ˆ i + 1.50 ˆ j ( ) 2 ( ) 10 s ( ) = ! 45.0 ˆ i + 15.0 ˆ j ( ) (b) The direction of motion makes angle θ with the x -direction. ! = tan " 1 v y v x # \$ % & ' ( = tan " 1 " # \$ % & ' ( = " 18.4 ° + 180 ° = 162 ° from + x -axis (c) Displacement: x -displacement = x f ! x i = v xi t + 1 2 a x t 2 = 1 2 ! 2 ( ) 10.0 s ( ) 2 = ! 225 m y -displacement = y f ! y i = v yi t + 1 2 a y t 2 = 1 2 + 2 ( ) 10.0 s ( ) 2 = + 75.0 m ! r r = " 225 ˆ i + 75.0 ˆ j ( ) m (d) Position: r r f = r r i + ! r r r r f = ! 2.00 ˆ i + 4.00 ˆ j ( ) + ! 225 ˆ i + 75.0 ˆ j ( ) = ! 227 ˆ i + 79.0 ˆ j ( ) m P4.7 (a) r F ! = r F 1 + r F 2 = 20.0 ˆ i + 15.0 ˆ j ( ) N r F ! = m r a : 20.0 ˆ i + 15.0 ˆ j = 5.00 r a r a = 4.00 ˆ i + 3.00 ˆ j ( ) 2 or a = 5.00 m s 2 at = 36.9 ° (b) F 2 x = 15.0cos60.0 ° = 7.50 N F 2 y = 15.0sin 60.0 ° = 13.0 N r F 2 = 7.50 ˆ i + 13.0 ˆ j ( ) N r F ! = r F 1 + r F 2 = 27.5 ˆ i + 13.0 ˆ j ( ) N = m r a = 5.00 r a r a = 5.50 ˆ i + 2.60 ˆ j ( ) 2 = 2 at 25.3 ° FIG. P4.7