anschp07

Chapter 7 Newton’s Second LawPhysics, 6th Edition Chapter 7. Newton’s Second Law Newton’s Second Law 7-1. A 4-kg mass is acted on by a resultant force of (a) 4 N, (b) 8 N, and (c) 12 N. What are the resulting accelerations? (a) a = 4N = 1 m/s2 4 kg (b) a = 8N = 2 m/s2 4 kg (c) a = 12N = 3 m/s2 4 kg 7-2. A constant force of 20 N acts on a mass of (a) 2 kg, (b) 4 kg, and (c) 6 kg. What are the resulting accelerations? (a) a = 20N = 10 m/s2 2 kg (b) a = 20N = 5 m/s2 4 kg (c) a = 20N = 3.33 m/s2 6 kg 7-3. A constant force of 60 lb acts on each of three objects, producing accelerations of 4, 8, and 12 N. What are the masses? m= 60 lb = 15 slugs 4 ft / s2 m= 60 lb = 7.5 slugs 8 ft / s2 m= 60 lb = 5 slugs 12 ft / s2 7-4. What resultant force is necessary to give a 4-kg hammer an acceleration of 6 m/s2? F = ma = (4 kg)(6 m/s2); F = 24 N 7-5. It is determined that a resultant force of 60 N will give a wagon an acceleration of 10 m/s2. What force is required to give the wagon an acceleration of only 2 m/s2? m= 60 N = 6 slugs ; 10 m / s2 F = ma = (6 slugs)(2 m/s2); F = 12 N 7-6. A 1000-kg car moving north at 100 km/h brakes to a stop in 50 m. What are the magnitude and direction of the force? 2 2as = v 2 − vo ; f Convert to SI units: 100 km/h = 27.8 m/s 2 v 2 − vo f a= 2s = (0) 2 − (27.8 m / s) 2 ; 2(50 m) a = 7.72 m / s2 F = ma = (1000 kg)(7.72 m/s2); 65 F = 772 N, South. 8 m/s2) = 31. 32 ft / s2 W = (7 slugs)(32 ft/s2) = 224 lb 7-9.8 kg mailbox? What is the mass of a 40-N tank? W = (4.4 N 66 .8 m/s2) = 47.0 m / s2 W = (3. Find the mass and the weight of a body if a resultant force of 16 N will give it an acceleration of 5 m/s2.08 kg 9. m= Physics. 32 ft / s2 gm = mm = me = 5.33 ft/s2 7-10.625 slugs.8 m / s2 7-8. 7-11. What is the weight of a 4. 5. What is the acceleration due to gravity on the moon and what is her mass on the moon? On the Earth? Her mass is the same on the moon as it is on the earth. What is the weight of a 70-kg astronaut on the surface of the earth.8 kg)(9. What is the mass of a 60-lb child? What is the weight of a 7-slug man? m= 60 lb = 1. Wm = mmgm 30 lb .8 m/s2) = 686 N . A woman weighs 180 lb on earth.Chapter 7 Newton’s Second Law The Relationship Between Weight and Mass 7-7. so we first find the constant mass: me = 180 lb = 5.20 kg . she weighs only 30 lb. Compare the resultant force required to give him or her an acceleration of 4 m/s2 on the earth with the resultant force required to give the same acceleration in space where gravity is negligible? On earth: W = (70 kg)(9. When she walks on the moon.20 kg)(9. m= 16 N = 3. FR = (70 kg)(4 m/s2) = 280 N Anywhere: FR = 280 N The mass doesn’t change. 6th Edition 40 N = 4.88 slugs .62 slugs .625 slugs gm = 5. 5.0 N . 7 m/s) m= 2500 lb = 78. µ k = 1270 lb . W = 2940 N m = 300 kg W = mg = (300 kg)(9.16. a= ∆v −4 m / s = . t 3s m= −400 N . 2(200 ft) and a = .0 slugs)(32 ft/s2). Fk = µ k N .0 slugs 8 ft / s2 W = 800 lb W = mg = (25. Find the mass and weight of a body if a resultant force of 400 N causes it to decrease its velocity by 4 m/s in 3 s.7 ft / s) 2 = . Applications for Single-Body Problems: 7-14. a= 60 ft / s . Find the mass and weight of a body if a resultant force of 200 lb causes its speed to increase from 20 ft/s to 60 ft/s in a time of 5 s.508 F = ma = (78. What horizontal pull is required to drag a 6-kg sled with an acceleration of 4 m/s2 if a friction force of 20 N opposes the motion? P – 20 N = (6 kg)(4 m/s2). A 2500-lb automobile is speeding at 55 mi/h.20 ft / s = 8 ft / s2 . (55 mi/h = 80. P = 44.3 m / s2 F = -1270 lb µk = 0. 5s m= 200 lb = 25.8 m/s2). What must be the coefficient of kinetic friction? We first find the mass and then the acceleration.0 N 20 N 6 kg P 7-15. 6th Edition 7-12. What resultant force is required to stop the car in 200 ft on a level road. a = −133 m / s2 .Chapter 7 Newton’s Second Law Physics.3 ft/s2). 2500 lb 67 . 32 ft / s2 2 v 2 − v0 f 2 Now recall that: 2as = v 2 − v0 f a= 2s ( 0) − (80.1 slugs)(-16. −133 m / s2 .1 slugs. 7-13. . (a) 7840 N – (800 kg)(9. 100 N – 15.5 m/s2 2 N F mg 100 N 100 N – F = ma.0 N W = mg and T = .60 N + 98 N or 7-17. (b) 6 m/s2 upward.30 m/s2 m mg + T 7-19.Chapter 7 Newton’s Second Law Physics.7 N = (8 kg) a. g  32 ft/s  W  64 lb  a. g  32 ft/s  a=0 + T m = W/g W (b) T − W = a = -12. Find the acceleration of the cabinet if µk = 0.8 m/s2) = 98 N. What is the tension in the cable if the + acceleration is (a) zero. Find the acceleration of the load if the tension in the cable is (a) 64 lb. An 800-kg elevator is lifted vertically by a strong rope.2. A 10-kg mass is lifted upward by a light cable. (b) 7840 N. (a) 2000 N – (800 kg)(9. Find the acceleration of the elevator if the rope tension is (a) 9000 N. Newton’s law for the problem is: T – mg = ma (up is positive) (a) 9000 N – (800 kg)(9. A horizontal force of 100 N pulls an 8-kg cabinet across a level floor. 64 lb − 64 lb =  a 2  . (a) T – 98 N = (10 kg)(0 m/s ) (b) T – 98 N = (10 kg)(6 m/s ) (c) T – 98 N = (10 kg)(-6 m/s ) 2 2 2 10 kg and and T = 98 N T = 60 N + 98 N or T = 158 N T = 38. 68 . A 64-lb load hangs at the end of a rope. 40 lb − 64 lb =  a 2  .0 ft/s2 7-18. 6th Edition 7-16. F = µkN = µk mg F = 0.2(8 kg)(9. 96 lb − 64 lb =  a 2  . (a) T − W = W  64 lb  a. (b) 40 lb. a = 1. g  32 ft/s  W  64 lb  a.0 ft/s2 (b) T − W = a = 16.45 m/s2 a=0 a = -7. and (c) 96 lb.8 m/s ) = 15.8 m/s2) = (800 kg)a .8 m/s2) = (800 kg)a .7 Ν a = 10.8 m/s2) = (800 kg)a . and (c) 2000 N. and (c) 6 m/s2 downward? T Note that up is positive and that W = (10 kg)(9. F + mg sin 300 = ma . mg sin 30 = ma . 7-22: F is up the plane now. Still.8 m/s2)(0. P = 114 N + F N P 300 300 mg *7-23. + P 300 N F 300 Σ Fx = ma.8 m/s2) sin 300 = 4. What is the acceleration in the absence of friction? Physics.20 m/s2.0 N = 40 N. See Prob.2(9. What is the acceleration? Why did you not need to know the mass of the block? ΣFx = max. a = 3. 3.90 m/s2. 7-10 is required to cause the acceleration DOWN the plane to be 4 m/s2? Assume that in = IO kg and µk = 0. P is down plane (+). P = 16. Assume that µk = 0.5 N mg 69 . 0 300 mg N F 300 300 mg sin 300 . In Fig. 0 N = mg cos 300 mg 0 0 mg sin 30 .µkN = ma .5 N P .7-10 will produce an acceleration of 4 m/s2 also up the incline? F = µkN = µkmg cos 300.5 N ΣFx = ma. What push P directed up and along the incline in Fig. down the plane.2 in Fig 7-10.5 N – (10 kg)(9. 0 a = g sin 30 0 a = (9. *7-22. Assume that m = I 0 kg and µk = 0. 6th Edition N 300 ΣFx = max. a = g sin 30 . an unknown mass slides down the 300 inclined plane.8 m/s2)(0.5) – 0.25. P .5 N – 49.8 m/s2)cos 300 = 25.5 N + (10 kg)(9. P – F – mg sin 300 = ma P – 25. 7-10. 7.866).5) = (10 kg)(4 m/s2) P – 25.8 m/s2)(0.Chapter 7 Newton’s Second Law 7-20.8 m/s2)(0.5) = (10 kg)(4 m/s2) P .3(10 kg)(9.0 N = 40 N.25.5 N + 49.µk g cos 30 a = (9. F = 0. F = 25. 3 in Fig. down the plane 7-21. What force P down the incline in Fig.10.µk mg cos 30 = ma . 6th Edition 7-24.88 m/s2 2 Now.8 m / s2 ) a= = . 7-12? 10 kg 45 N A Τ = 20 Ν 5 kg B ΣF = mTa. For total system: m2g = (m1 + m2)a m2 g (6 kg)(9.8 m / s2 ) a= 2 = m1 + m2 4 kg + 6 kg 70 .5 Ν Fig. a = 10 m/s2 2 2 kg 6 kg T 80 N To find T. 7-25. to find T. 45 N = (15 kg) a. 7-11. *7-27.88 m/s ).µkm1g = (m1 + m2)a m g − µ k m1 g (6 kg)(9.8 m / s2 ) − (0. What force does block A exert on block B in Fig. apply F = ma to 6-kg block only: 80 N – T = (6 kg)(10 m/s ).2 in 7-13. F = µkN = µkm1g Fk N T m1 g +a T m2 g For total system: m2g .Chapter 7 Newton’s Second Law Applications for Multi-Body Problems Physics.2)(4 kg)(9. m1 + m2 4 kg + 6 kg (m1g is balanced by N) N T m1 g +a T m2 g a = 5. If the coefficient of kinetic friction between the table and the 4 kg block is 0. N = m1g. Τ = 2 3. Assume zero friction in Fig. what is the acceleration of the system. 7-13? Assume zero friction and draw free-body diagrams. consider only m1 ΣF = m1a T = m1a = (4 kg)(5. What are the acceleration of the system and the tension in the connecting cord for the arrangement shown in Fig. What is the acceleration of the system? What is the tension T in the connecting cord? Resultant force = total mass x acceleration 80 N = (2 kg + 6 kg)a. a = 3 m/s2 F = 15 N Force ON B = mB a = (5 kg)(3 m/s2). *7-26. What is the tension in the cord? ΣFy = 0. 2 N *7-28.2). T = m1(g + a) = (2 kg)(9. What is the acceleration in Fig.8 m / s2 ) 2 kg + 8 kg Now look at m1 T . *7-29. Assume that the masses m1 = 2 kg and m2 = 8 kg are connected by a cord that passes over a light frictionless pulley as in Fig. Add friction force F up plane in figure for previous problem.69 m/s2 m2g *7-30.88 m/s 2 +a T T m2g m1g a= alone: (8 kg)(9. 7-15 starts from rest.8 N − 7. 7-15 as the 10-kg block moves down the plane against friction (µk = 0.4 N +a N T T 320 320 acceleration assuming zero friction? (assume motion down plane) Σ Fx = mT a. T = m2g – m2a Τ = (6 kg)(9.10 m/s2). What is the acceleration of the system and the tension in the cord? Resultant force = total mass of system x acceleration m2g – m1g = (m1 + m2)a a= m2 g − m1 g m1 + m2 a = 5. m1g sin 320 – m2g = (m1 + m2) a m1g (10 kg)(9.10 m/s2 T m1 g +a T m2 g To find T.8 m/s2 – 5. m2g – T = m2a.Chapter 7 Newton’s Second Law a= 58.8 m/s2) – (6 kg)(5.84 N 10 kg Fk N Physics. consider only m2 and make down positive: ΣFy = m2a . 71 .) or a = 5.(2 kg)(9.19. (Cont. The system described in Fig. T = 28.m1g = m1 a.8 m/s2)sin 320 – (2 kg)(9. 12 kg a = 2.8 m / s2 ) .8 m/s2) = (10 kg + 2 kg)a a= 519 N .88 m/s2). 6th Edition *7-27. What is the T = 31. 7-14.6 N . T = m2(g + a) = (2 kg)(9.5 slugs)(32 ft/s2 + 8 ft/s2). T = m(g + a).5 slugs)(32 ft/s2 + 8 ft/s2). T = (62. 6th Edition ΣFy = 0 . N = m(g + a). A 200-lb worker stands on weighing scales in the elevator of Problem 7-32. What is the maximum acceleration? Tmax – mg = ma a= Tmax − mg 200 N . F = µkN = µk m1g cos 320 m1g sin 320 – m2g – µk m1g cos 320 = (m1 + m2) a . T = 22. Find the minimum breaking strength of the cable pulling the elevator? ΣFy = ma. m= W 2000 lb = .2 m/s2 8 kg mg T = 200 N +a 72 .31 m/s2). m = 62. N = m1g cos 320 *7-30.2 N Challenge Problems 7-32. A 2000-lb elevator is lifted vertically with an acceleration of 8 ft/s2. N +a 200 lb T = 2500 lb 7-34. N – mg = ma. Physics. (Cont. What is the reading of the scales as he is lifted at 8 m/s? The scale reading will be equal to the normal force N on worker. a = 1. T = (62.(8 kg)(9. T = 2500 lb 7-33.Chapter 7 Newton’s Second Law m1g sin 320 – m2g – F = (m1 + m2) a .8 m / s2 ) = m 8 kg a = 15.) m1g sin 320 – m2g – F = (m1 + m2) a .5 slugs g 32 ft / s2 T +a 2000 lb T – mg = ma.31 m/s2 *7-31 What is the tension in the cord for Problem 7-30? Apply F = ma to mass m2 only: T – m2g = m2 a. A 8-kg load is accelerated upward with a cord whose breaking strength is 200 N.8 m/s2 + 1. If mass A is twice that of mass B. Consider two masses A and B connected by a cord and hung over a single pulley. F = 0. will cause the block to accelerate at 4 m/s2? F = µkN = µkmg cos 340. 7-13 are switched so that the larger mass is on the table.5 Ν *7-37.(0. Τ = 2 3.7)(9.7. 2 a = -6. what is the acceleration of the system? mA = 2mB . What push up the incline.vf . A 5-kg mass rests on a 340 inclined plane where µk = 0. m1 + m2 6 kg + 4 kg 2 N T m1 g +a T m2 g a = 3. where N = mg: F = -µkmg = ma. m1 = 6 kg.8 m / s ) = . What would be the acceleration and tension in the cord neglecting friction? For total system: m2g = (m1 + m2).4 N 73 . m2 = 4 kg a= m2 g (4 kg)(9. 6th Edition 7-35.92 m/s2 2 ΣF = m1a T = m1a = (6 kg)(5. 2 a = -µkg = .8 m / s2 = 3 3 mg = 3ma a = 3.8 m/s2)cos 340 = 8. the right mass A will be 2m.12 N mg ΣFx = ma.2.Chapter 7 Newton’s Second Law Physics. What is the horizontal stopping distance for a 1600-kg truck traveling at 20 m/s? The stopping distance is determined by the acceleration from a resultant friction force F = µkN.86 m / s2 ) *7-36.88 m/s ).2 km Recall that: 2as = vo . s= 2 v 2 − v0 f 2a 0 − (20 m / s) 2 = . 2mg – mg = (2m + m)a a= g 9. Suppose the 4 and 6-kg masses in Fig.2(5 kg)(9.86m/s2 s = 29.8 m/s2) sin 340 = (5 kg)(4 m/s2) P = 47. For rubber tires on a concrete road µk = 0. P – F – mg sin 34 = ma 0 P – 8. If the left mass B is m.27 m/s 2 +a T m mg N + F 340 B T 2m A 2mg P 340 *7-38. 2(-6.12 N – (5 kg)(9.8 m/s2). 4 4 2.42 kg ∆F ∆a 4 5 6 2 7 8 9 Acceleration. Critical Thinking Questions W = 35. m/s2 2 1. In a laboratory experiment.1 8 5.2 lb = 12 lb + 0. A 96-lb block rests on a table where µk = 0.1 12 8. m/s 74 .4. What is the significance of the slope of this curve? What is the mass? Force. A cord tied to this block passes over a light frictionless pulley.6 10 7. N 14 12 10 8 6 4 2 0 0 1 2 3 The slope is the change in Force over the change in acceleration.4 Plot a graph of weight (force) versus acceleration.2 lb=   4 ft/s 32 ft/s 2   F N T 96 lb W +a T ( ) W – 19. What weight must be attached to the free end if the system is to accelerated at 4 ft/s2? F = µkN = 0. the mass is found to be: m = 1.2 lb  96 lb + W  2 W . Larger and larger weights are transferred from the car to a hanger at the end of a tape that passes over a light frictionless pulley. 6th Edition *7-39. which is the mass of the system. Since the car moves on a horizontal air track with negligible friction. N Acceleration.7 lb 7-40. In this manner. The following data are recorded: Weight.2 (96 lb).125 W. F = 19.9 6 4. the mass of the entire system is kept constant. the resultant force is equal to the weights at the end of the tape. Thus. the acceleration of a small car is measured by the separation of spots burned at regular intervals in a paraffin-coated tape.Chapter 7 Newton’s Second Law Physics.19. what must be the tension in the top string in order that the system accelerate upward at 4 m/s2? In the latter case what are the tensions in the strings that connect masses? 7-43. 6th Edition 7-41. In the above experiment. Several runs are made. What is the acceleration of the system if the suspended mass is three times that of the mass on the table and µk = 0. What happens to the acceleration as the mass of the system is increased? What should the value of the product of mass and acceleration be for each run? Is it necessary to include the mass of the constant 4 N weight in these experiments? The acceleration increases with increasing mass. (3 . F = µkN = µkmg For total system: 3mg . N = mg. and 6 kg. What is the tension in each cord? If they are then detached from the ceiling. increasing the mass of the car each time by adding weights.3)(9. 75 A 2 kg B 4 kg C 6 kg . Thus.µkmg = (3m + m)a . are connected (in order) by strings and hung from the ceiling with another string so that the largest mass is in the lowest position.8 m / s2 ) = 4 4 a = 6. 2 kg. 4 kg.0. Three masses.62 m/s2 3mg 7-43. According to Newton’s second law. the product of the total mass of the system and the acceleration must always be equal to the resultant force of 4 N for each run. a student places a constant weight of 4 N at the free end of the tape.µk)mg = 4 ma a= (3 − µ k ) g (3 . (Cont. An arrangement similar to that described by Fig.3.) The tension in each string is due only to the weights BELOW the string. Fk N T mg +a T Σ Fy = 0. 7-13 is set up except that the masses are replaced. It is necessary to add the mass of the 4-N weight to each of the runs because it is part of the total mass of the system. 7-42.Chapter 7 Newton’s Second Law Physics. TA = 48 N 2 a = + 4 m/s2 A 2 kg B C 4 kg 6 kg TB = (4 kg + 6 kg)(4 m/s2) = 40 N . TC = (6 kg)(4 m/s2) = 24 N 7-44.5 m/s2 Acceleration exists only while a force is applied. N = (400 lb)cos 600 = 200 lb . A 400-lb sled slides down a hill (µk = 0. What is the normal force on the sled? What is the force of kinetic friction? What is the resultant force down the hill? What is the acceleration? Is it necessary to know the weight of the sled to determine its acceleration? ΣFy = 0. F = 40 lb ΣFx = W sin 600 – F = (400 lb)sin 600 – 40 lb. Σ Fy = 0. both astronaut and solar panel move in opposite directions at the speeds obtained when contact is broken. 6th Edition TB = (6 kg + 4 kg)(9. Physics. 7-45.8 N .2) inclined at an angle of 600.8 m/s2) = 98. solve for aa: aa = − mp a p ma =− (200 kg)(2 m / s2 ) . TA = (2 kg + 4 kg + 6 kg)(4 m/s2). N – W cos 600 = 0. 80 kg a = . The force causes the panel to accelerate at 2 m/s2.8 m/s2) = 118 N Now consider the upward acceleration of 4 m/s . once the force is removed. An 80-kg astronaut on a space walk pushes against a 200-kg solar panel that has become dislodged from a spacecraft.2)(200 lb). Fp = mpap.0 N . TA = (6 kg + 4 kg + 2 kg)(9. Fa = maaa .8 m/s2) = 58. mpap = .maaa .Chapter 7 Newton’s Second Law TC = (6 kg)(9.. What acceleration does the astronaut receive? Do they continue to accelerate after the push? The force on the solar panel Fp is equal and opposite that on the astronaut Fa. Thus. FR = 306 lb +a 600 W = mg = 400 lb N F 600 76 . F = µkN = (0. Chapter 7 Newton’s Second Law Physics.) Since FR = ma.5 N 77 .63 m/s2 .7 N TB = m3g + m3a TB = 68.8 m/s2). we note that: W sin 600 .µkW = (W/g)a.63 m/s2) . F = 23. Thus.3(8 kg)(9. 7-16. 7-16. left is positive. Neglecting friction.63 m/s2).3 between the mass m2 and the table in Fig. The acceleration is not affected by m2. 2 2 TA = 81. *7-46. The masses m2 and F N m3 are 8 and 6 kg. TB = (6 kg)(9.8 m/s − 1. what is the acceleration of the system? What are the tensions in the cord on the left and in the cord on the right? Would the acceleration be the same if the middle mass m2 were removed? Total mass of system = (10 + 8 +6) = 24 kg Resultant Force on system = m1g – m3g The normal force N balances m2g.8 N.8 m/s2) – (6 kg)(9. What mass m1 is TB TA + required to cause the system to accelerate to the left at 2 m/s2? ( F = µkm2g acts to right. m3g 6 kg m1g – F – m3g = (m1 + m2 +m3)a . m1 = 10 kg. ΣF = mTa m1g – m3g = (m1 + m2 +m3)a . respectively. TA = m1g – m1a TA = m1(g – a) = (10 kg)(9. and m3 = 6 kg. the weight divides out and it is not necessary for determining the resultant acceleration.6 N *7-47.0 N – 58. Now apply to 6-kg mass: TB – m3g = m3a. TA 10 kg + TA N TB + TB 6 kg m2g 8 kg m1g m3g (10 kg)(9. Assume that µk = 0. F = µkm2g = 0.8 m/s + 1. 6th Edition 7-45 (Cont. Three masses. To find TA apply ΣF = m1a to 10-kg mass: m1g – TA = m1a .8 m/s2) = (24 kg) a a = 1. m2 = 8 kg. are connected as shown in Fig. (24 kg)a = 98. ) TA m1g m2g 8 kg + TB Apply ΣF = mTa to total system. Σ Fx = ma: mAg sin 600 . m1 = 14. ΣFx = ma: WB – WA sin 600 = (mA + mB) a N  W + WB  WB − WA sin 60 =  A a .g sin 400 .866) = 0.188(64 lb + WB) WB – 55.8 kg = 2m1 + 28 kg . F is directed down plane.Chapter 7 Newton’s Second Law Physics. WB = 83. It + continues to move up the plane (+) at an acceleration of –9 m/s2.mg sin 400 = ma 40 0 F 400 mg a = -µkg cos 400 .mg sin 400 = ma.mBg = (mA + mB) a (9.2 kg = 2 mA + 8 kg. F = µkN .(9. F What is the coefficient of kinetic friction? Since block is moving up plane. What must be the weight of block B if Block A moves up the plane with an acceleration of 6 ft/s2.0 lb + 0. Neglect friction. ΣFy = 0.360 *7-49.0 lb WA = 64 lb *7-50.(6 kg)(9.8 m/s2) = mA(2 m/s2) + (4 kg)(2 m/s2) 8.49 mA – 39.8 m/s2) cos 400 .8 m/s2)(0.5 N . 7-17 has a weight of 64 lb.4 lb = 12.8 m/s2) sin 400 Solving for µk we obtain: µk = 0. 7-17 is 4 kg. The mass of block B in Fig. What must be the mass of block A if it is to move down the plane at an acceleration of 2 m/s2? Neglect friction. -µkmg cos 400 .866)mA – (4 kg)(9.188 g 32 ft/s 2 600 T 600 +a T WB WB – (64 lb)(0.8 m1 – 23. 6th Edition m1(9. F = µkmg cos 400 ΣFx = ma. A block of unknown mass is given a push up a 400 inclined plane and then released. mA = 7.8 m/s2) – 23. 600 600 N T +a T mAg mBg 78 .5 kg – 58.1 kg *7-48.8 m/s2) = (m1 + 14 kg)(2 m/s2) 9. Block A in Fig.188WB . g   o a 6 ft/s 2 = = 0.28 kg. N = mg cos 400. -9 m/s2 = -µk(9. -F . 63 m/s2 The greater acceleration results from 60 0 the fact that the friction force is increasing the resultant force instead of decreasing it as was the case in part (a). F = µkmAg cos 600 Resultant force on entire system = total mass x acceleration mBg – mAg sin 600 . F = µkN .8 m/s2)(0. mBg – mAg sin 600 + µkmAg cos 600 = (mA + mB)a 98 N – 33. ΣFx = ma.53 m/s2 (b) If initial motion is down the plane. F is down the plane.9 N + 14. but the resultant force is still down the plane. and (b) if the system is initially moving down the plane? (a) With upward initial motion.9 N – 14. Find the acceleration if (a) the system is initially moving up the plane. mAg mBg 79 .8 m/s2) – (4 kg)(9.Chapter 7 Newton’s Second Law Physics. respectively.3(10 kg)(9. Assume that the masses A and B in Fig. 7-17 are 4 kg and 10 kg. 6th Edition *7-51. The block will side until it stops and then goes the other way. ΣFy = 0.7 N = (14 kg)a. then F is up the plane.3. N = mAg cos 600.7 N = (14 kg)a N F T 60 0 +a T a = 5. or a = 3. The coefficient of kinetic friction is 0.µkmAg cos 600 = (mA + mB)a N F 600 T 600 +a T mAg mBg (10 kg)(9.8 m/s2)(0.866) – 0.5) = (14 kg)a 98 N – 33.