University Physics AI No. 3 Newton’s Laws of Motion Class Number Name I．Choose 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 circular 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 rm r v mmaF cc 2 2 ω=== , the magnitude of the force toward the center should be constant. 2. Suppose the net force F r on 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 Newton’s second law amF r r = , the answer is (D). 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 acceleration of the object thinking about air resistance is more than that no air resistance, that is that the tangent of the graph should be more than that no air resistance. Thus we remove the answer (D). During the course of the object moves down, when ∞→t , 0→a , constant→v . Thinking about this, the right answer is (B) II. Filling the Blanks t v t v t v t v (A) (B) (C) (D) 1. A 2.50 kg system has an acceleration ia ? )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( ? 00.6 ? 00.7 ji + . Solution: The total force is )N( ? 0.10 ? 00.45.2 iiamF =×== r v The other force is )N( ? 00.6 ? 00.7) ? 00.6 ? 00.3( jijiFF total +=??= vv 2. Two masses, m 1 and m 2 , hang over an ideal pulley and the system is free to move (see Figure 1). Such an arrangement is called an Atwood’s machine. The magnitude of the acceleration a r of the system of two masses is g mm mm 21 12 + ? . The magnitude of the tension in the cord is g mm mm 21 21 2 + . Solution: The second law force diagrams of two mass are shown in figure. Assume the mass m 2 moves downward. Apply Newton’s second law of motion, we have ? ? ? =? =? amgmT amTgm 11 22 Solving it, we can get g mm mm a 21 12 + ? = g mm mm T 21 21 2 + = 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 θ with the vertical direction. (a) Make a second law force diagram about the mass and indicate the direction to the center of its circular path. (b) The direction of the acceleration of the mass is pointing to O . (c) Apply Newton’s second law to the horizontal and vertical direction to calculate the θ r m 1.00m Fig. 2 T v gm v O y x O′ m 2 m 1 Fig.1 T v gm v 2 a v T v gm v 1 m 1 m 2 angle θ is gr v 2 arctan=θ .(d) If the angle θ = 47.4° 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′ . Solution: The second law force diagram is shown in figure. (c) Apply Newton’s second law of motion, we have ? ? ? ? ? = =? r mv T mgT 2 sin 0cos θ θ Solving the equations, we have gr v 2 arctan=θ . (d) m/s0.44.47tan5.18.9tanarctan 2 =××==?= o θθ grv gr v (e) N7.21 4.47cos 8.95.1 cos 0cos = × ==?=? o θ θ mg TmgT (f) Using ? ? ? ? ? = = 2 2 1 gty vtx , in which y=1.0m. Then m8.1 8.9 12 4 2 = × ×== g y vx 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 μ s =0.20 and μ 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 force 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 magnitude 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 second law force diagrams of two mass are shown in figure. (b) Assume that the system does not move, so we have ? ? ? ? ? ? ? =? = =? =? 0 0 0 NMg Nf fT Tmg ss s μ ? ? ? ? ? =×=== =××== =×== ? )N(0.988.910 )N(0.982.08.950 )kg(10502.0 mgTfor Mgf Mm s ss s μ μ (c) If kg5 2 1 ==′ mm , assume the system does not move, then ? ? ? =′? =?′ 0 0 s fT Tgm N)(0.498.910 2 1 2 1 =××==′=? mgfT s As the maximum static friction force is )N(0.982.08.950 max =××== ss Mgf μ From the above, we know maxs fT < Thus the magnitude of the frictional force on the 50 kg block is 49N. The system does not move, so the friction force is a static friction force. 2. In the system shown in Fig.4, a block (of mass m 1 =9.5kg) slides on a frictionless plane inclined at an angle o 34=θ . The block is attached by a string to a second block (of mass m 2 =2.6kg). The system is released from rest. (a) Find the acceleration and the tension in the string. (b) If taking into account a frictional force between block 1 and the plane, what is the acceleration of the tension in the string? Use the values μ s =0.24 and μ k =0.15 N v gM v T v gm v x y T v f v m 1 Fig.4 m 2 o 34 Solution: (a) The second law force diagrams of two mass are shown in figure. Set the coordinate system as in figure. Assuming the acceleration is a v shown in figure. Apply Newton’s second law of motion, we have ? ? ? ? ? =? =? =? amTgm gmN amgmT 22 1 11 0cos sin θ θ Solving it, we get 2 21 12 m/s2.2 6.25.9 34sin5.96.2sin ?= + ×? = + ? = o g mm mm a θ N31)34sin1( 6.25.9 34sin6.25.9 )sin1( 21 21 =+× + ×× =+ + = o o θ mm gmm T (b) The second law force diagrams of two mass are shown in figure. Set the coordinate system as in figure. Assuming the acceleration is a v shown in figure. Apply Newton’s second law of motion, we have ? ? ? ? ? ? ? = =? =? =?? Nf amgmT gmN amfTgm k μ θ θ 22 1 11 0cos sin Solving the equations, we get 2 21 21 m/s2.1 6.25.9 6.2)34cos15.034(sin5.9)cos(sin = + ?+× = + ?? = oo g mm mm a k θμθ N29)34cos15.034sin1( 6.25.9 34sin6.25.9 )cossin1( 21 21 =×?+× + ×× =?+ + = oo o θμθ k mm gmm T 3. An object is drop from rest. Find the terminal speed assuming that the drag force is given by 2 bvD = . Solution: The forces acting on the object and the coordinate system are shown in figure. Apply Newton’s second law of motion, we have gm v D v j ? m 1 m 2 x y gm v 1 T v gm v 2 N v T v a v m 1 m 2 x y gm v 1 T v gm v 2 N v T v a v f v a v mamgD ?=? dt dv mmgbv ?=?? 2 dt m b dv b mg v ?= ? ? 2 1 ∫∫ ∫∫ ?= ? ? ?= ? ? tv tv dt m b dv b mg v dt m b dv b mg v y 00 2 00 2 1 1 t m bg e b mg v b mg v t m b b mg v b mg v b mg 2 )ln( 2 1 ? = + ? ? ?= + ? ? When ∞→t , the terminal speed of the object is b mg v = 4. Assume the kinetic frictional force on a falling mass m is proportional to its speed v, with a proportionality constant β. Choose j ? to be vertically downward. (a) Show that Newton’s second law of motion yields y y vmg t v m β?= d d . (b)When the mass reaches its terminal sped, what is t v y d d ? (c) Show that the terminal speed is β mg v term = . (d) If the mass is dropped from rest, show that )1()( / mt termy evtv β? ?= Solution: The forces acting on the mass are shown in figure. Apply Newton’s second law of motion, we have (a) Apply Newton’s second law of motion mafmg =? , we have t v mmavmg y yy d d ==?β That is y y vmg t v m β?= d d (b) When the mass reaches its terminal speed, the acceleration does not change, so 0==? mafmg , thus 0 d d ==? t v mvmg y y β , that is 0 d d = t v y . (c) according to the answer in (b), we know 0 d d = t v y . So 0=? y vmg β . The terminal speed is β mg v term = . (d) If the mass is dropped from rest, then )1()1( 1 )1ln( )ln()-ln( d - )-d( dt d d d y v 00 m t term m t y m t y y y t y y y y y y eve mg v ev mg m t v mg m t mgvmg t mvmg vmg vmg vm vmg t v m ββ β β β β β β β β β β β β ?? ? ?=?=? =?? ?=?? ?=?? ?=? = ? ??= ∫∫ gm v f v j ?