chap4 solutions - CHAPTER FOUR SOLUTIONS CHAPTER FOUR...

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Unformatted text preview: CHAPTER FOUR SOLUTIONS CHAPTER FOUR SOLUTIONS Chapter Four Readings Brancazio, P., "The Physics of Kicking a Football," The Physics Teacher, October 1985, p. 403 Cohen, B.I., "Isaac Newton," Scientific American, December 1955, p. 73. Crane, H.R., "Digital Electronic Balances: Mass or Weight?," The Physics Teacher, 29, 142, 1991 McCloskey, M., "Intuitive Physics," Scientific American, April 1983, p. 122. Palmer, F., "Friction," Scientific American, February 1051, p. 54. Zimmerer, R., "The Measurement of Mass," The Physics Teacher, September 1983, p. 354 4.1 (a) FR=ma -= (6.0 kg)(2.0 m/sz) = 12 N. (b) a=fl= 12 N 3.0m/32. m 4.0 kg‘ 4.2 The acceleration given to the football is found from v= V0 + aras V-Vo_10m/S-0_ 2 r ' 0.205 ‘50m/5° Then, from Newton's 211d law, we findF= ma = (0.50 kg)(SO m/sZ) = 25 N. a= 4.3 Onearth w=mg. If g changes to g', the new weight would be w'- mg‘. L“ in“ g: g. I f _. z w _ , here ore, w 1m g g , or g w On the moon, g'_— -6 g. Thus, w'z-IE w= "65 lb. This is equivalent to 3.71 N. On Jupiter, g' -= 2.64 g. The techniques above give the weight on Jupiter to be 58.7 N. The mass is the same at all three locations. The weight on w 22.2 N the earth is 22.2 N, and E=m= 2.27 kg, 4.4 33 L— OOOON =0.05 m/sZ, and a=m'”1.5 x 107 kg V=Vo+at=0+(0.05)t with v=80kmfh=22.22 m/sgives t=444.4s=7.4min 4.5 Summing the forces on the plane a = 2 m/sz shOWn gives '—'.'* ‘ E'FX=F_f=10N-f % = 2 (0.20 kg)(2.0 m/s ). f F=10N From which 1‘ - 9.6 N 4.6 First, find the acceleration of the bullet from the equations of motion with constant acceleration, as 38 CHAPTER FOUR SOLUTIONS 2 - 2 2 _V V9 5320 m/31 - 0_ 4 2 a— .= 2(0.82 ) —6.24x10 m/s . Then, F= ma = (5.0 x 10-3 kg)(6.24 x 104 m/sZ) = 310 N 4.7 (a) From the second law, the acceleration of the boat is PR 2000 N - 1800 N a=E=—-————"‘—1000 kg =0-20m/82 r=1soon F=2000N (b) The distance moved is found from x= vot+%at2 =0 +% (0.2 m/sz)(10.0 s)2 = 10111. (c) Its velocity is found as v= v0+ at= 0+ (0.2 m/sz)(10.0 s) = 2 m/s. ‘ - 4.8 (a) We resolve the forces shown into their components as man: 31.591119 400 N: 200. N 346. N 450 N: M m FR: 122 N 789 N u The magnitude of the resultant force is found from the gorean theorem as FR= (213,02 + (2Py)2 = (122 N)2 + (789 N)? = 798 N, 2 and tantLe2 =§—3* = 6.47, from which 9 = 81.2”. Thus, the resul- tant force is at an angle of 8.8“ to the right of the forward direction. (13) The acceleration is in the same direction as F3 and is given by a=PR/m= 798 N/ 3000 kg= 0.266 m/sz. 4.9 The resultant of the two forces is found from the Pythagorean theorem as FR=\I(390 N)2 + (180 N)?- =430 N, and tan 9=i% 2- 2.166, ,, fig 430 N or 8—652. Thus, a=m=270 kg = 1.59 m/s2 at 65.2” north of east. 4.10 Using v2 = v02 - 2gb with v0; 0, we find that at the moment the ball hits the ground, its velocity v1 is v1 =V(-2)(-9.80)(30 = 24.25 m/s in the downward direction. If the ball rebounds to a height of 20 m, then its velocity V2 as it leaves the ground is V2 =V(—2)(—9.80)(20) = 19.80 m/s in the upward direction. The average acceleration between when the ball hits the ground and when it leaves the ground is then V2 ' V1 19.8 — -24.25 2 a= At = 0.002 s = 22025 m/s upward. Therefore, the average force is F= ma = (0.50 kg)(22025 111/82) = 1.10 x 104 N upward. CHAPTER FOUR SOLUTIONS 4.17 The forces on the bucket are the tension in the rope and the weight, 49 N, of the bucket We call the positive direction upward, and use the second law. EFy = may T- 49 N = (5.0 kg)(3.0 m/sz) T= 64 N. 4.18 (a) From the second law, we find the acceleration as =55: :3 fig- 0.333 m/sZ. _ —) To find the distance moved, we use F = 10 N x= vot+ %at2 =0+ % (0.333 m/s2)(3.0 s)2= 1.5 m (b) If the shopper places her 30 N (3.06 kg) child in the cart, the new F 10 N __ , 2 Moral-33.06 kg — 0.302 m/s , and the new distance traveled in 3.0 s will be x' = 0 + %(0.302 m/sz)(3.0 s)2 = 1.4 m. acceleration will be a = 4.19 (a) The average acceleration is given by 3= Av/Ata (Vf - vi)/At= (5.00 m/s - 20.0 m/s)/4.000 s = - 3.75 m/sz. The average force is found from the second law as 1F= an; a (2.000): 103 kg)(3.75 m/sz) = 7500 N (b) The distance traveled is: x=tz r=;———15'00 m“ a 20'0 m/s (4.000 s)=50.0m. 4.20 Letting m1 = 10.0 kg and 1112 = 5.00 kg, we have mla=m1g - T. m2a= T- ngsine. Adding these gives (1111 + m2)aam1g - mggsinfl. _‘_2'_ Thus, a=(m1 m Sm m1 + 1112 I - I o o a= (__,._.§___)___10 0 51500 ““40 ) 9.80 = 4.43 m/sZ. Using this value of a'gives: T= m1(g - a) = 10.0(9.80 - 4.43) = 53.7 N. g,or 4.21 (a) The resultant external force acting on this system having a total mass of 6.0 kg is 42 N directed horizontally toward the right. Thus, the . . "I: 42 N 2 acceleration produced is a= m = 6.0 kg = 7.0 m/s . The acceleration of the system is 7.0 m/s2 horizontally toward the right. (b) Draw a free body diagram of the 3.0 kg object and apply Newton‘s second law to the horizontal forces acting on this object. This gives 41 4.27 4.28 4.29 4.30 CHAPTER FOUR SOLUTIONS From (2), we find: T2 = 118 N, and using this in (1), we find T1 = 236 N. -F= -2795N Apply the 2nd law to the trailer to get: T= (300 kg)(2.15 m/sz) = 645 N. For the car: F- r= 1000 kg(2.15 m/sz), which gives: F= 2795 N. (a) The net force on the car = F— T= mcara = 2150 N (directed forward). (b) From above, we know that T= 645 N in forward direction. (c) The force exerted on the trailer by the car is - T= 645 N ( rearward). (d) The force exerted mm by the car is the resultant of the re- action force -F and the weight of the car as show above. This force is found to be PR = 10,190 Nat e = 159" to the left of the vertical. First, consider the 3.00 kg rising mass. The . T T forces on it are the tension, T, and its weight, 29.4 N. With the upward direction . as positive, the second law hec0mes + T- 29.4 N = (3.00 kg)a. (1) The forces on the falling 5.00 kg mass are its weight and T, and its acceleration is the same as that of the rising mass. Calling the positive direction doWn for this mass, gives w1- 29.4 N 49 N — T= (5.00 kg)a. (2) Equations (1) and (2) can be solved simultaneously to give (a) the tension as T= 36.8 N, (b) and the acceleration as a = 2.45 m/sz. (c) Consider the 3.00 kg mass. We have y= vot+%at2 =0+ % (2.45 m/s2)(1.00 s)2 = 1.23 m. (a) To bring the block to the verge of motion, we exert a force of 75 N, but from the first law, we know: EFX = 0, which gives 75 N-f=0,or f=75 N wherefis the force of friction. We know, since motion is imminent, that f= usN where its is the coefficient of static friction and N is the normal force. For our case, N = mg. Thus, f=nsmg, and us =m%g= 17956 1; =0.38. (b) When there is motion, the analysis is virtually the same as that above. We find: pk = 16906 I; = 0.31. (a) The static friction force on M1 can have a maximum value of f= PsN = [1.51113 = 0.5(10)9.80 = 49 N. This is more than the weight of M2 (39.2 N). Hence, the system does not move. a = 0 (b) Once in motion the friction force on N11 becomes f= ukN =ukmg = 0.3(10)9.80 = 29.4 N. This is less than the gravitational force of 39.2 N on M2. 43 4.39 4.40 4.41 CHAPTER FOUR SOLUTIONS 2 0 = (13.9 m/s)2 +2(—ukg)x, or x= W. (1) With pk = 0.100, this gives a value for xof X= 98.6m (b) With ilk - 0.600, (1) above gives X= 16.4111 First, let us find the normal force acting on the box by use of 2B,, =0. We have N - WCOS30.0° = O, or N =mgcos30.0°. Now apply Newton's 2nd law to the motion parallel to the ramp. 213x = mgsin30° - f--— max. But, f=pkN = nkmgcosBOD". Thus, the 2nd law becomes mgsinSOD" - ukmgcosStLO“ = max, or sin30.0° _ ax "k“ cos30.0° gc0530.0° . _._ax__ = “30-0 ' gcos30.0° 1.20 m/s2 3 0'57? ‘ (9.80 m/sz)-(0.866) ‘ (1436’ ._________ Using the given distance and time, X= vot+ % at2 gives 1.00m=0 +%a(16.00 s2). or a = 0.125 111/52. (a) Since the acceleration of the 4.00 kg mass is then 0.125 m/sz. Newton's second law gives T— 39.2 N - (4.00 kg)(0.125 m/sZ) or T= 39.7 N. (b) Then the an law for the 9.00 kg mass gives: sz = may -= 0, or N = wcos40.0° = (9.00 kg)(9.80 m/sz)cos40.0° = 67.6 N, and 28x = max , or mgsin40.0° - T- f = (9.00 kg)a, which yields 1’2 (9.00 kg)(9.80 m/s2)sin40.0° -— 39.7 N — (9.00 kg)(0.125 m/sz) = 15.9 N. Therefore, :1 == §=%3_2—% = 0.235. ( c) First, apply the 211‘:1 law to the 12.0 kg block: 118N- T= (12.0kg)a (1) For the 7.00 kg block, we have N —.Wcos37.0° = 54.8 N, and f= ukN = 0.25(54.8 N) = 13.7 N Apply the second law no the 7.00 kg block. We have: T— f- (68.6 N)sin37“ = (7.00 kg)a. (2) Solving (1) and (2) simultaneously gives: a = 3.32 m/sz. ...
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