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1、University Physics AI No. 3 Newtons Laws of Motion Class Number Name IChoose the Correct Answer 1. Which statement is most correct? ( C ) (A) Uniform circular motion causes a constant force toward the center. (B) Uniform circular motion is caused by a constant force toward the center. (C) Uniform ci
2、rcular motion is caused by a constant magnitude net force toward the center. (D) Uniform circular motion is caused by a constant magnitude net force away the center. Solution: For uniform circular motion, the speed or the angular speed of the particle is a constant. According to rmrvmmaFcc22 =, the
3、magnitude of the force toward the center should be constant. 2. Suppose the net force Fron an object is a nonzero constant. Which of the following could also be constant? ( D ) (A) Position. (B) Speed. (C) Velocity. (D) Acceleration. Solution: According to Newtons second lawamFrr=, the answer is (D)
4、. 3. Which of the graph in Figure 1 best shows the velocity-time graph for an object launched vertically into the air when air resistance is given by D=bv? The dashed line shows the velocity graph if there was no air resistance. ( B ) Solution: During the course of the object moves up, the accelerat
5、ion of the object by thinking about air resistance is larger than that no air resistance, it implies that the tangent of the graph should be larger than that no air resistance. Thus we remove the answer (D). During the course of the object moves down, when t, 0a, constantv. Thinking about this, the
6、right answer is (B) II. Filling the Blanks tvtvtvtv(A) (B) (C) (D) 1. A 2.50 kg system has an accelerationia)m/s00. 4(2=r. One of the forces acting on the system is ji)N00. 6()N00. 3(. Is this the total force on the system? No . If not, the other force acting on the system is )N(00. 600. 7ji +. Solu
7、tion: The total force is )N(0 .1000. 45 . 2iiamF=rvThe other force is )N(00. 600. 7)00. 600. 3(jijiFFtotal+=vv2. Two masses, m1 and m2, hang over an ideal pulley and the system is free to move (see Figure 1). Such an arrangement is called an Atwoods machine. The magnitude of the acceleration arof th
8、e system of two masses is gmmmm2112 +. The magnitude of the tension in the cord is gmmmm21212 +. Solution: The second law force diagrams of two mass are shown in figure. Assume the mass m2 moves downward. Apply Newtons second law of motion, we have =amgmTamTgm1122Solving the equations, we can get gm
9、mmma2112 += gmmmmT21212 += 3. Wahoo! You are swinging a mass m at speed v around on a string in circle of radius r whose plane is 1.00 m above the ground (see Figure 2). The string makes an angle with the vertical direction. (a) Make a second law force diagram about the mass and indicate the directi
10、on to the center of its circular path. (b) The direction of the acceleration of the mass is pointing to O . (c) Apply Newtons second law to the horizontal and vertical direction to calculate the r m 1.00m Fig. 2 TvgmvO y xOm2m1 Fig.1 Tvgmv 2avTvgmv 1m1m2 angle is grv2arctan=.(d) If the angle = 47.4
11、and the radius of the circle is 1.50 m, the speed of the mass is 4.0m/s . (e) If the mass is 1.50 kg, the magnitude of the tension in the string is 21.7N . (f) The string breaks unexpectedly when the mass is moving exactly eastward. The location the mass will hit the ground is 1.8m to the point O .
12、Solution: The second law force diagram is shown in figure. (c) Apply Newtons second law of motion, we have =rmvTmgT2 sin0cosSolve the equations, we have grv2arctan=. (d) m/s0 . 44 .47tan5 . 18 . 9tanarctan2 =ogrvgrv(e) N7 .214 .47cos8 . 95 . 1cos0cos=omgTmgT (f) Using =2 21gtyvtx, in which y=1.0m. T
13、hen m8 . 18 . 9 1242=gyvx III. Give the Solutions of the Following Problems 1. The static and kinetic coefficients of friction for a 50 kg mass on a table surface are s=0.20 and k=0.15. See Figure 3. The pulley is ideal. The system is at rest in equilibrium. (a) For each mass, sketch a second law fo
14、rce diagram indicating schematically all forces that are acting on each mass. (b) How large can the mass m be and not move the system? What the magnitude of the frictional force on the 50 kg mass when this situation exists? (c) If m is only half the maximum mass calculated in part (b), what is the m
15、agnitude of the frictional force on the 50 kg block? Indicate whether this is a force of static friction or a force kinetic friction. m M=50kg Fig.3 Solution: (a) The free body diagrams of two masses are shown in figure. (b) Assume that the system does not move, so we have =000NMgNffTTmgsss =)N(0 .988 . 910)N(0 .982 . 08 . 950)kg(10502 . 0mgTforMgfMmssss (c) If kg521=mm, assume the sys
