* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Solutions to Tutorial Problem Bab
Survey
Document related concepts
Jerk (physics) wikipedia , lookup
Coriolis force wikipedia , lookup
Nuclear force wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Fictitious force wikipedia , lookup
Centrifugal force wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Transcript
Bab 4 Hukum-hukum Gerakan SOALAN-SOALAN Q5.4 In the motion picture It Happened One Night (Columbia Pictures, 1934), Clark Gable is standing inside a stationary bus in front of Claudette Colbert, who is seated. The bus suddenly starts moving forward and Clark falls into Claudette's lap. Why did this happen? Solution When the bus starts moving, the mass of Claudette is accelerated by the force of the back of the seat on her body. Clark is standing, however, and the only force on him is the friction between his shoes and the floor of the bus. Thus, when the bus starts moving, his feet start accelerating forward, but the rest of his body experiences almost no accelerating force (only that due to his being attached to his accelerating feet!). As a consequence, his body tends to stay almost at rest, according to Newton’s first law, relative to the ground. Relative to Claudette, however, he is moving toward her and falls into her lap. (Both performers won Academy Awards.) Q5.7 A rubber ball is dropped onto the floor. What force causes the ball to bounce? Solution The molecules of the floor resist the ball on impact and push the ball back, upward. The actual force acting is due to the forces between molecules that allow the floor to keep its integrity and to prevent the ball from passing through. Notice that for a ball passing through a window, the molecular forces weren’t strong enough. the larger upward force that the floor exerts on them together. Around the top of the weight’s motion, the scale reads less than average. If the iron is moving upward, the lifter can declare that she has thrown it, just by letting go of it for a moment, so our answer applies also to this case. Q5.14 Identify the action-reaction pairs in the following situation: a man takes a step; a snowball hits a girl in the back; a baseball player catches a ball; a gust of wind strikes a window Solution As a man takes a step, the action is the force his foot exerts on the Earth; the reaction is the force of the Earth on his foot. In the second case, the action is the force exerted on the girl’s back by the snowball; the reaction is the force exerted on the snowball by the girl’s back. The third action is the force of the glove on the ball; the reaction is the force of the ball on the glove. The fourth action is the force exerted on the window by the air molecules; the reaction is the force on the air molecules exerted by the window. We could in each case interchange the terms ‘action’ and ‘reaction.’ Masalah-masalah 3. A 3.00-kg object undergoes an acceleration given by Solution As the barbell goes through the bottom of a cycle, the lifter exerts an upward force on it, and the scale reads the resultant force acting on it and the magnitude of the resultant force. Solution m 3.00 kg ˆ 5.00ˆ a 2.00i j m s2 F m a Q5.10 A weightlifter stands on a bathroom scale. He pumps a barbell up and down. What happens to the reading on the bathroom scale as this is done? What if he is strong enough to ctually throw the barbell upward? How does the reading on the scale vary now? 2 a 2.00ˆi 5.00ˆj m / s . Find 6.00iˆ 15.0ˆj N 2 2 F 6.00 15.0 N 16.2 N 5. To model a spacecraft, a toy rocket engine is securely fastened to a large puck, which can glide with negligible friction over a horizontal surface, taken as the xy plane. The 4.00-kg puck has a velocity of 3.00ˆi m / s at one instant. Eight seconds later, its velocity is to be 8.00 ˆi 10.0ˆjm / s . Assuming the rocket 11. Two forces F1 and F2 act on a 5.00-kg object. If F1 = 20.0 N and F2 = 15.0 N, find the accelerations in (a) and (b) of Figure P5.11. engine exerts a constant horizontal force, find (a) the components of the force and (b) its magnitude. Solution ˆ m s, m 4.00 kg , vi 3.00i ˆ 10.0ˆ v8 8.00i j m s , t 8.00 s ˆ 10.0ˆ j v 5.00i m s2 t 8.00 F m a 2.50ˆ i 5.00ˆ j N a F 2.50 2 5.00 2 Figure P5.11 5.59 N Solution F F1 F2 20.0iˆ 15.0ˆj (a) 7. An electron of mass 9.11 10–31 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to 7.00 105 m/s in a distance of 5.00 cm. Assuming its acceleration is constant, (a) determine the force exerted on the electron and (b) compare this force with the weight of the electron, which we neglected. F m a: F m a and a v2f vi2 2x f (b) a 5.00 m s2 at 36.9 ˆ 13.0ˆ F2 7.50i j N F F1 F2 27.5iˆ 13.0ˆj vi2 a 2x f F 9.111031 2 5 2 3.00 105 m s 2 7.00 10 m s kg 2 0.050 0 m 3.64 1018 N . (b) or v2f vi2 2ax f or . 2 f The weight of the electron is Fg m g 9.11 1031 kg 9.80 m s2 8.93 1030 N The accelerating force is 4.08 1011 tim es the w eightofthe electron. F2x 15.0cos60.0 7.50 N F2y 15.0sin 60.0 13.0 N Therefore v F m ˆ 3.00ˆ a 4.00i j m s2 Solution (a) ˆ 15.0ˆ 20.0i j 5.00a N 2 5.50iˆ 2.60ˆj m N m a 5.00a s2 6.08 m s2 at25.3 2 1 Fg T1 1 2 T2 T3 23. A 1.00-kg object is observed to accelerate at 10.0 m/s2 in a direction 30.0° north of east (Fig. P5.23). The force F2 acting on the object has a magnitude of 5.00 N and is directed north. Determine the magnitude and direction of the force F1 acting on the object. Figure P5.18 Problems 18 and 19 19. A bag of cement of weight Fg hangs from three wires as shown in Figure P5.18. Two of the wires make angles 1 and 2 with the horizontal. If the system is in equilibrium, show that the tension in the left-hand wire is T1 = Fg cos 2 /sin( 1 + 2) Solution T3 Fg (1) T1 sin1 T2 sin2 Fg (2) T1 cos1 T2 cos 2 Eliminate T2 and solve for T1 (3) T1 Fg cos 2 sin 1 cos 2 cos1 sin 2 Figure P5.23 Fg cos 2 sin 1 2 T3 Fg 325 N Solution cos25.0 T1 Fg 296 N sin 85.0 Choose a coordinate system with î East and ĵ North. cos1 cos60.0 T2 T1 296 N 163 N cos25.0 cos 2 F m a 1.00 kg 10.0 m s2 at 30.0 5.00 N ˆj F1 10.0 N 30.0 5.00 N ˆj 8.66 N iˆ F1 8.66 N East 25. A block is given an initial velocity of 5.00 m/s up a frictionless 20.0° incline (Fig. P5.22). How far up the incline does the block slide before coming to rest? Solution After it leaves your hand, the block’s speed changes only because of one component of its weight: Fx m ax m g sin 20.0 m a Figure P5.31 Solution Forces acting on 2.00 kg block: v2f vi2 2a xf xi . T m 1g m 1a Taking vf 0 , vi 5.00 m s, and a g sin 20.0 gives Forces acting on 8.00 kg block: 0 5.00 2 9.80 sin 20.0 xf 0 2 Fx T m 2a or xf 25.0 3.73 m 2 9.80 sin 20.0 . (1) (a) (2) Eliminate T and solve for a: a Fx m 1g m1 m2 a 0 forFx m 1g 19.6 N . (b) Eliminate a and solve for T: T m1 Fx m 2g m1 m2 T 0 forFx m 2g 78.4 N . 31. In the system shown in Figure P5.31, a horizontal force Fx acts on the 8.00-kg object. The horizontal surface is frictionless. (a) For what values of Fx does the 2.00-kg object accelerate upward? (b) For what values of Fx is the tension in the cord zero? (c) Plot the acceleration of the 8.00kg object versus Fx. Include values of Fx from –100 N to +100 N. Newton’s second law is: Fy m ay S 72.0 kg 9.80 m s2 72.0 kg ay 33. A 72.0-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.20 m/s in 0.800 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.50 s and comes to rest. What does the spring scale register (a) before the elevator starts to move? (b) during the first 0.800 s? (c) while the elevator is traveling at constant speed? (d) during the time it is slowing down? Solution First, we will compute the needed accelerations: (1) Before it starts to move: ay 0 (2) During the first 0.800 s: ay (3) While moving at constant velocity: 4 During the last 1.50 s: v yf v yi t 1.50 m s 2 ay 0 1.20 m s 0 0.800 s (a) When ay 0 , S 706 N . (b) When ay 1.50 m s2 , S 814 N . (c) When ay 0 , S 706 N . (d) When ay 0.800 m s2 , S 648 N . 41. A 3.00-kg block starts from rest at the top of a 30.0° incline and slides a distance of 2.00 m down the incline in 1.50 s. Find (a) the magnitude of the acceleration of the block, (b) the coefficient of kinetic friction between block and plane, (c) the friction force acting on the block, and (d) the speed of the block after it has slid 2.00 m. Solution m 3.00 kg , 30.0 , x 2.00 m , t 1.50 s 1 (a) x at2 : 2 1 2 2.00 m a 1.50 s 2 4.00 a 1.78 m s2 2 1.50 v yf v yi 0 1.20 m s t 1.50 s 0.800 m s 2 ay S 706 N 72.0 kg ay . F n f m g m a : A long x: 0 f m g sin 30.0 m a f m g sin 30.0 a A long y: n 0 m g cos30.0 0 n m g cos30.0 f m g sin 30.0 a , n m g cos30.0 a k tan 30.0 0.368 gcos30.0 (b) k (c) f m g sin 30.0 a , f 3.00 9.80sin30.0 1.78 9.37 N (d) v2f vi2 2a xf xi where xf xi 2.00 m v2f 0 2 1.78 2.00 7.11 m vf 7.11 m 2 2 s2 s2 2.67 m s 45. Two blocks connected by a rope of negligible mass are being dragged by a horizontal force F (Fig. P5.45). Suppose that F = 68.0 N, m1 = 12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100. (a) Draw a free-body diagram for each block. (b) Determine the tension T and the magnitude of the acceleration of the system. over a frictionless pulley (Fig. P5.51), Pat pulls on the loose end of the rope with such a force that the spring scale reads 250 N. Pat's true weight is 320 N, and the chair weighs 160 N. (a) Draw freebody diagrams for Pat and the chair considered as separate systems, and another diagram for Pat and the chair considered as one system. (b) Show that the acceleration of the system is upward and find its magnitude. (c) Find the force Pat exerts on the chair. Figure P5.45 Solution (a) See Figure T m1 m2 F Figure P5.51 n1 m1 f1 = nk n2 T T f2 = nk 1 m1g = 118 N (b) m2 F Solution (a) 2 m2g = 176 N 68.0 T m 2g m 2a (Block #2) T m 1g m 1a (Block #1) Adding, 68.0 m 1 m 2 g m 1 m 2 a a (b) First consider Pat and the chair as the system. Note that two ropes support the system, and T 250 N in each rope. Applying F m a 68.0 g 1.29 m s2 m m 1 2 T m 1a m 1g 27.2 N 51. An inventive child named Pat wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes 2T 480 m a , where m 480 49.0 kg . 9.80 Solving for a gives 500 480 0.408 m s2 . 49.0 F m a on Pat: a (c) F n T 320 m a, where 320 32.7 kg 9.80 n m a 320 T 32.7 0.408 320 250 83.3 N m 55. An object of mass M is held in place by an applied force F and a pulley system as shown in Figure P5.55. The pulleys are massless and frictionless. Find (a) the tension in each section of rope, T1, T2, T3, T4, and T5 and (b) the magnitude of F. Suggestion: Draw a free-body diagram for each pulley. 61. What horizontal force must be applied to the cart shown in Figure P5.61 in order that the blocks remain stationary relative to the cart? Assume all surfaces, wheels, and pulley are frictionless. (Hint: Note that the force exerted by the string accelerates m1.) Solution (a) First, we note that F T1 . Next, we focus on the mass M and write T5 M g . Next, we focus on the bottom pulley and write T5 T2 T3 . Finally, we focus on the top pulley and write T4 T1 T2 T3 . Since the pulleys are not starting to rotate and are frictionless, T1 T3 , and T2 T3 . From this Mg . 2 Mg 3M g Then T1 T2 T3 , and T4 , and 2 2 information, we have T5 2T2 , so T2 T5 M g . (b) Since F T1 , we have F Mg . 2 Figure P5.61 Problems 61 and 63 Solution F ma For m 1 : T m 1a For m 2 : T m 2g 0 Eliminating T, a m 2g . For all 3 blocks: m1 F M m 1 m 2 a M m g m 1 m 2 2 m1