The Laws of Motion ( ) N SOLUTIONS TO PROBLEMS ! F = ( 6.00) 2 + ( 15.0) 2 N = 16.2 N. Section 4.4. Newton s Second Law The Particle Under a Net Force
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1 SOLUTIONS TO PROBLEMS The Laws of Motion Section 4.3 Mass P4. Since the ca is moving with constant speed and in a staight line, the esultant foce on it must be zeo egadless of whethe it is moving (a) towad the ight o the left. Section 4.4 Newton s Second Law The Paticle Unde a Net Foce P4.3 m = 3.00 kg a = (.00î ĵ ) m s F = ma = 6.00î ĵ ( ) N!! F = ( 6.00) + ( 15.0) N = 16. N 107
2 108 The Laws of Motion ( ) N P4.4 (a)! F = F 1 + F = "9.00î ĵ Acceleation Velocity a = a x î + a y ĵ =! F m = v x î + v y ĵ = v i + at = at ( = "9.00î ĵ ) N = ("4.50î.00 kg ĵ ) m s = (!4.50î ĵ )m s ( ) 10 s ( ) m s ( ) =!45.0î ĵ The diection of motion makes angle θ with the x-diection. v! = tan "1 # y & % ( $ ' = # tan"1 " 15.0 m s & $ % 45.0 m s' ( v x! = " = 16 fom + x-axis (c) Displacement: x-displacement = x f! x i = v xi t + 1 a x t = 1 (!4.50 m s ) 10.0 s ( ) =!5 m (d) Position: y-displacement = y f! y i = v yi t + 1 a y t = 1 ( m s ) 10.0 s! = ("5î ĵ ) m f = i +! f = (!.00î ĵ ) + (!5î ĵ ) = (!7î ĵ ) m ( ) = m P4.7 (a) F! = F 1 + F = 0.0î ĵ ( ) N! F = ma : 0.0î ĵ = 5.00 a a = 4.00î ĵ ( ) m s o a = 5.00 m s at! = 36.9 F x = 15.0cos 60.0 = 7.50 N F y = 15.0sin 60.0 = 13.0 N F = ( 7.50î ĵ ) N! F = F 1 + F = 7.5î ĵ a = 5.50î +.60ĵ ( ) N = ma = 5.00a ( ) m s = 6.08 m s at 5.3 FIG. P4.7
3 P4.8 F! = m a eads Chapte whee â epesents the diection of a (!.00î +.00ĵ î! 3.00ĵ! 45.0î) N = m( 3.75 m s )â (!4.0î! 1.00ĵ ) N = m( 3.75 m s )â! F = ( 4.0) + ( 1.00) N at tan!1 " 1.00 % # $ 4.0& ' below the x-axis! F = 4.0 N at 181 = m( 3.75 m s )â. Fo the vectos to be equal, thei magnitudes and thei diections must be equal. (a)! â is at 181 counteclockwise fom the x-axis 4.0 N m = = 11. kg 3.75 m s (d) = v i + at = 0 + ( 3.75 m s at 181 )10.0 s so v f = 37.5 m s at 181 = 37.5 m s cos181 î m s sin181 ĵ so v f = (!37.5î! 0.893ĵ ) m s (c) Section 4.5 = m s = 37.5 m s The Gavitational Foce and Weight P4.9 (a) F g = mg = 10 lb = ( N lb) ( 10 lb) = 534 N m = F g g = 534 N = 54.5 kg 9.80 m s P4.11 Imagine a quick tip by jet, on which you do not visit the est oom and you pespiation is just canceled out by a glass of tomato juice. By subtaction, F g ( ) = mg p p and ( F g ) = mg C C give!f g = m( g p " g C ). Fo a peson whose mass is 88.7 kg, the change in weight is!f g = 88.7 kg( " ) =.55 N. A pecise balance scale, as in a docto s office, eads the same in diffeent locations because it compaes you with the standad masses on its beams. A typical bathoom scale is not pecise enough to eveal this diffeence.
4 110 The Laws of Motion P4.13 (a)! F = ma and v f = v i + ax f o a = v f! v i. x f Theefoe, ( )! F = m v f " v i x f! F = 9.11 # 10 "31 kg $ ( 7.00 # 10 %& 5 m s ) " 3.00 # 10 5 m s m ( ) ( ) ' () = 3.64 # 10 "18 N. The weight of the electon is ( )( 9.80 m s ) = 8.93! 10 "30 N F g = mg = 9.11! 10 "31 kg The acceleating foce is 4.08! times the weight of the electon. P4.14 We find acceleation: f! i = v i t + 1 at Now 4.0 mî! 3.30 mĵ=0+ 1 a 1.0 s a = 5.83î! 4.58ĵ! F = ma becomes F g + F = ma F =.80 kg 5.83î! 4.58ĵ F = 16.3î ĵ ( ) = 0.70 s a ( ) m s. ( ) m s + (.80 kg) ( 9.80 m s ) ĵ ( ) N. Section 4.6 Newton s Thid Law P4.17 (a) 15.0 lb up 5.00 lb up (c) 0
5 Chapte Section 4.7 Applications of Newton s Laws P4. (a) Isolate eithe mass T! mg = ma = 0 T = mg. The scale eads the tension T, so T = mg = 5.00 kg( 9.80 m s ) = 49.0 N. FIG. P4.(a) Isolate the pulley T + T 1 = 0 T = T 1 = mg = 98.0 N. (c) F! = n + T + m g = 0 Take the component along the incline n x + T x + m g x = 0 FIG. P4. o 0 + T! mg sin 30.0 = 0 T = mg sin 30.0 = mg = 4.5 N = ( ) FIG. P4.(c)
6 11 The Laws of Motion P4.4 (a) Fist constuct a fee body diagam fo the 5 kg mass as shown in the Figue 4.4a. Since the mass is in equilibium, we can equie T 3! 49 N = 0 o T 3 = 49 N. Next, constuct a fee body diagam fo the knot as shown in Figue 4.4a. Again, since the system is moving at constant velocity, a = 0 and applying Newton s second law in component fom gives FIG. P4.4(a)! F x = T cos 50 " T 1 cos 40 = 0 F y! = T sin 50 + T 1 sin 40 " 49 N = 0 Solving the above equations simultaneously fo T 1 and T gives T 1 = 31.5 N and T = 37.5 N and above we found T 3 = 49.0 N. FIG. P4.4 Poceed as in pat (a) and constuct a fee body diagam fo the mass and fo the knot as shown in Figue 4.4b. Applying Newton s second law in each case (fo a constant-velocity system) we find: T 3! 98 N = 0 T! T 1 cos 60 = 0 T 1 sin 60! T 3 = 0 Solving this set of equations we find: T 1 = 113 N T = 56.6 N and T 3 = 98.0 N
7 Chapte P4.6 The two foces acting on the block ae the nomal foce, n, and the weight, mg. If the block is consideed to be a point mass and the x- axis is chosen to be paallel to the plane, then the fee body diagam will be as shown in the figue to the ight. The angle θ is the angle of inclination of the plane. Applying Newton s second law fo the acceleating system (and taking the diection up the plane as the positive x diection) we have! F y = n " mg cos# = 0 : n = mg cos!! F x = "mg sin# = ma : a =!g sin" FIG. P4.6 (a) When! = 15.0 a =!.54 m s Stating fom est = v i + a x f! x i ( ) = ax f ( )!.00 m = ax f =!.54 m s ( ) = 3.18 m s P4.7 Choose a coodinate system with î East and ĵ Noth. F! = m a = 1.00 kg 10.0 m s ( ) at 30.0 ( 5.00 N) ĵ + F 1 = ( 10.0 N)!30.0 = ( 5.00 N) ĵ + ( 8.66 N) î!f 1 = 8.66 N ( East) FIG. P4.7 P4.8 Fist, conside the block moving along the hoizontal. The only foce in the diection of movement is T. Thus,! F x = ma T = ( 5 kg) a (1) Next conside the block that moves vetically. The foces on it ae the tension T and its weight, 88. N. We have! F y = ma FIG. P N! T = ( 9 kg) a () Note that both blocks must have the same magnitude of acceleation. Equations (1) and () can be added to give 88. N = 14 kg ( ) a. Then a = 6.30 m s and T = 31.5 N.
8 114 The Laws of Motion P4.30 m 1 =.00 kg, m = 6.00 kg,! = 55.0 (a)! F x = m g sin" # T = m a and T! m 1 g = m 1 a a = m g sin"! m 1 g m 1 + m = 3.57 m s T = m 1 ( a + g) = 6.7 N FIG. P4.30 (c) Since v i = 0, = at = ( 3.57 m s )(.00 s) = 7.14 m s. Additional Poblems P4.44 (a) Following the in-chapte Example about a block on a fictionless incline, we have a = g sin! = ( 9.80 m s )sin 30.0 a = 4.90 m s The block slides distance x on the incline, with sin 30.0 = m x x = 1.00 m : = v f = v i + a( x f! x i ) = 0 + ( 4.90 m s )( 1.00 m) 3.13 m s afte time t s = x f = ( 1.00 m) 3.13 m s = s. (c) Now in fee fall y f! y i = v yi t + 1 a y t : ( 4.90 m s )t + ( 1.56 m s)t!.00 m = 0 Only one oot is physical!.00 = (!3.13 m s)sin 30.0 t! 1 ( 9.80 m s )t t =!1.56 m s ± t = s x f = v x t = ( 3.13 m s)cos 30.0 ( 1.56 m s)! 4( 4.90 m s )(!.00 m) 9.80 m s [ ]( s) = 1.35 m (d) (e) total time = t s + t = s s = 1.14 s The mass of the block makes no diffeence.
9 Chapte P4.47! F = ma Fo m 1 : T = m 1 a Fo m : T! m g = 0 Eliminating T, a = m g m 1 Fo all 3 blocks: FIG. P4.47 F = ( M + m 1 + m ) a = ( M + m 1 + m )! " # m g $ m 1 % &
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