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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 ί૿ The Particle Under a Net Force P4.3 m = 3.00 kg r a = 2.00 ˆ i + 5.00 ˆ j ( ) m s 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 ( ) m s 2 ( ) 10 s ( ) = ! 45.0 ˆ i + 15.0 ˆ j ( ) m s (b) The direction of motion makes angle θ with the x -direction. ! = tan " 1 v y v x # $ % & ' ( = tan " 1 " 15.0 m s 45.0 m s # $ % & ' ( ! = " 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 ! 4.50 m s 2 ( ) 10.0 s ( ) 2 = ! 225 m y -displacement = y f ! y i = v yi t + 1 2 a y t 2 = 1 2 + 1.50 m s 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 ( ) m s 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 ( ) m s 2 = 6.08 m s 2 at 25.3 ° FIG. P4.7
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