Easy. P4.2 Since the car is moving with constant speed and in a straight line, the. resultant force on it must be regardless of whether it is moving
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1 Chapte 4 Homewok Solutions Easy P4. Since the ca is moving with constant speed and in a staight line, the zeo esultant foce on it must be egadless of whethe it is moving (a) towad the ight o the left. P4.5 m = 3.00 kg, (a) a.00î 5.00ĵ m s F ma 6.00î 15.0ĵ N F N 16. N P4.7 We use the paticle unde a net foce model. (a) F F 1 F (0.0î 15.0ĵ) N Newton s second law gives, with m = 5.00 kg, F a 4.00î 3.00ĵ m m/ s o a 5.00 m s at 36.9 F x 15.0cos N F y 15.0sin N F 7.50î 13.0ĵ N F F 1 F 0.0î 7.50î 13.0ĵ a (7.5î 13.0ĵ) N F m N 5.50î.60ĵ m/ s 6.08 m/ s at 5.3 ANS. FIG. P4.7
2 P4.9 (a) F g mg 10 lb N lb10 lb 534 N m F g g 534 N 54.5 kg 9.80 m s P4.11 (a) Fom v f v i ax and F ma, we can solve fo the acceleation and then the foce: a v f v i x Substituting to eliminate a, F Substituting the given infomation, F F kg N m v f v i x m/ s ( m) m/ s The Eath exets on the electon the foce called weight, F g mg ( kg)(9.80 m/ s ) N The acceleating foce is times the weight of the electon. P4.13 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 p mg p and F g mg p g C F g C mg C give Fo a peson whose mass is 90.0 kg, the change in weight is F g 90.0 kg 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.
3 P4.14 We find acceleation: f i v i t 1 at 4.0 mî 3.30 mĵ 0 1 a1.0 s 0.70 s a Now F g F ma F m a becomes a 5.83î 4.58ĵ m s F.80 kg5.83î 4.58ĵ m s.80 kg F 16.3î 14.6ĵ N 9.80 m s ĵ P4.31 Afte it leaves you hand, the block s speed changes only because of one component of its weight: F x ma x mg sin 0.0 ma v f v i ax f x i Taking v f 0, v i 5.00 m/ s, and a gsin 0.0 gives o sin 0.0 x f x f 9.80sin m
4 Medium P4.4 (a) F F 1 F 6.00î 4.00ĵ 3.00î 7.00ĵ 9.00î 3.00ĵ N Acceleation: a a x î a y ĵ F 9.00î 3.00ĵ N m.00 kg Velocity: 4.50î 1.50ĵ m s v f v x î v y ĵ v i at at m/ s v f 4.50î 1.50ĵ 10.0 s 45.0î 15.0ĵ m/ s The diection of motion makes angle with the x diection. tan 1 v y v x 1 tan 15.0 m s 45.0 m s fom the x axis (c) Displacement: x-displacement x f x i v xi t 1 a x t m s 10.0 s 5 m y-displacement y f y i v yi t 1 a y t m s 10.0 s 75.0 m (d) 5î 75.0ĵ m Position: f i f.00î 4.00ĵ 5î 75.0ĵ 7î 79.0ĵ m
5 P4.6 F ma eads.00î.00ĵ 5.00î 3.00ĵ 45.0î N m3.75 m s â whee â epesents the diection of a 4.0î 1.00ĵ N m3.75 m s â F N at tan 1 F 4.0 N at 181 m 3.75 m s 4.0 below the x axis Fo the vectos to be equal, thei magnitudes and thei diections must be equal. â (a) Theefoe â is at 181 counte-clockwise fom the x axis m 4.0 N 11. kg 3.75 m s (c) v (d) v v i v v 0 a t (3.75 m/ s )(10.00 s) 37.5 m/ s F m t a t 0 4.0î 1.00ĵ 11. kg So, v f 37.5î 0.893ĵ m s 10.0 s37.5î 0.893ĵ m/ s P4.17 (a) 15.0 lb up, to countebalance the Eath s foce on the block. (c) 5.00 lb up, the foces on the block ae now the Eath pulling down with 15.0 lb and the ope pulling up with 10.0 lb. The foces fom the floo and ope togethe balance the weight. 0, the block now acceleates up away fom the floo.
6 P4. (a) Isolate eithe mass: T mg ma 0 T mg The scale eads the tension T, so (c) (d) T mg 5.00 kg9.80 m s 49.0 N The solution to pat (a) is also the solution to. Isolate the pulley: T T 1 0 T T 1 mg 98.0 N F n T mg 0 Take the component along the incline o n x T x mg x 0 ANS. FIG. P4. (a) and ANS. FIG. P4. (c) 0 T mg sin T mg sin 30.0 mg 4.5 N ANS. FIG. P4. (d) P4.6 (a) Fist constuct a fee-body diagam fo the 5-kg mass as shown in the Figue 4.6a. 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 4.6a. Again, since the system is moving at constant velocity, a 0, and applying Newton s second law in component fom gives ANS. FIG. 4.6 (a) ANS. FIG. 4.6
7 F x T cos 50 T 1 cos 40 0 T sin 50 T 1 sin N 0 F y Solving the above equations simultaneously fo T 1 and T gives T N and T 37.5 N and above we found T N. Poceed as in pat (a) and constuct a fee-body diagam fo the mass and fo the knot as shown in Figue 4.6b. 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 N, T 56.6 N and T N P4.8 (a) Diagams showing the foces on the objects ae below. and (c) Fist, conside m1, 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 m, the block that moves vetically. The foces on it ae tension T and its weight, 88. N. We have F y ma 88. N T = (9 kg)a [] ANS. FIG. P4.8 the 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 P4.9 Choose a coodinate system with î East and ĵ Noth. The acceleation is a [(10.0 cos 30.0 )î (10.0 sin 30.0 )ĵ] m/ s (8.66î 5.00ĵ) m/ s Fom Newton s second law, F ma (1.00 kg)(8.66î m/ s 5.00ĵ m/ s ) (8.66î 5.00ĵ) N and F F 1 F ANS. FIG. P4.9 so the foce we want is F 1 F F (8.66î 5.00ĵ 5.00ĵ) N 8.66î N 8.66 N east P4.30 m 1 =.00 kg, m = 6.00 kg, = 55.0 (a) Diagams showing the foces on the objects ae below. F x m gsin 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 (c) T m 1 a g 6.7 N ANS. FIG. P4.30 (d) Since v i 0, v f at 3.57 m s.00 s 7.14 m s. Had P4.51 Apply F ma. Fo m 1 : T m 1 g = 0 Fo m : T = m a Eliminating T, ANS. FIG. P4.51
9 a m 1 g m Fo all 3 blocks: F M m 1 m m 1g m P4.53 (a) Following the in-chapte example about a block on a fictionless incline, we have a gsin 9.80 m s sin 30.0 a 4.90 m s The block slides distance x on the incline, with sin m x x = 1.00 m: v f v i ax f x i m s 1.00 m v f 3.13 m s afte time t s x f v f 1.00 m 3.13 m s s. (c) Now in fee fall, y f y i v yi t 1 a yt : m s sin 30.0t m st 4.90 m s t 1.56 m st.00 m m s t 1.56 m s m s 9.80 m s Only the positive oot is physical:.00 m t s x f v x t 3.13 m scos s 1.35 m (d) total time t s t s s 1.14 s (e) The mass of the block makes no diffeence.
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