University Physics Al No. 3 Newton's Laws of motion Class Number ame 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 cent Solution: For uniform circular motion, the speed or the angula d of the particle is a constant According to F =ma= m-=mo r, the magnitude of the force toward the center should be constant 2. Suppose the net force F on an object is a nonzero constant Which of the following could also (A)Position. (B)Speed. ( C) Velocity. (D) Acceleration Solution: According to Newton's second law F=ma, 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=br? The dashed line shows the velocity graph if there was no air resistance 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->00, a>0,v>constant. Thinking about this, the right answer is(B) I Filling the blank
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 m r r v Fc mac m 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 F ma 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 → ∞ , a → 0 , v → constant . 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.A2.50 kg system has an acceleration a=(4. 00 m/s2)i One of the forces acting on the system is (3.00N)i-(6.00N)j. Is this the total force on the system? No_ If not, the other force acting on the system is 7.001+6.00J(N) Solution: The total force is F=ma=2.5x.00i=10.0i(N) The other force is F= Fotol-(3.001-600)=7.001+600j(N) 2. Two masses, my and m2, 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 of the system of two masses is 28 m1+m2 The magnitude of the tension in the cord is n1+m2 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 m2g-T=m2a T-m,g=m,a Solving it, we can get T=2m,m, 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 0 with the vertical direction. (a) Make 100m 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 toO.(c)Apply Newtons second law to Fig 2 the horizontal and vertical direction to calculate the
1. A 2.50 kg system has an acceleration a i ˆ (4.00m/s ) 2 = r . One of the forces acting on the system is i j ˆ (6.00N) ˆ (3.00N) − . Is this the total force on the system? No . If not, the other force acting on the system is (N) 00 ˆ 6. 00 ˆ 7. i + j . Solution: The total force is (N) 0 ˆ 10. 00 ˆ F = ma = 2.5× 4. i = i v r The other force is (N) 00 ˆ 6. 00 ˆ ) 7. 00 ˆ 6. 00 ˆ F F (3. i j i j = total − − = + v v 2. 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 Atwood’s machine. The magnitude of the acceleration a r of the system of two masses is g m m m m 1 2 2 1 + − . The magnitude of the tension in the cord is g m m m m 1 2 2 1 2 + . Solution: The second law force diagrams of two mass are shown in figure. Assume the mass m2 moves downward. Apply Newton’s second law of motion, we have ⎩ ⎨ ⎧ − = − = T m g m a m g T m a 1 1 2 2 Solving it, we can get g m m m m a 1 2 2 1 + − = g m m m m T 1 2 2 1 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 mg v O y O x ′ m2 m1 Fig.1 T v m g v 2 a v T v m g v 1 m1 m2
angle 0is 8=arctan-(d)If the angle 0=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 217N.(f) The string breaks unexpectedly when the mass is moving exactly eastward. The cato he 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 Tcos 6-mg=0 Tsin e Solving the equations, we have 0=arctan (d=acan"→=grtn6=√98×15×tn474°=40ms (e)Tcos6-mg=0→T 1.5×98 =21.7N cOs 6 cOs 47.4 1, in which y=1.Om Then x=1 1. 8 m V9.8 I. 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 us=0.20 and 4=0. 15. See M=50kg 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 1g 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
angle θ is gr v 2 θ = arctan .(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 T mg 2 sin cos 0 θ θ Solving the equations, we have gr v 2 θ = arctan . (d) arctan tan 9.8 1.5 tan 47.4 4.0 m/s 2 = ⇒ = = × × = o θ v gr θ gr v (e) 21.7 N cos 47.4 1.5 9.8 cos cos 0 = × − = ⇒ = = o θ θ mg T mg T (f) Using ⎪ ⎩ ⎪ ⎨ ⎧ = = 2 2 1 y gt x vt , in which y=1.0m. Then 1.8 m 9.8 2 1 4 2 = × = = × g y x v 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 µs=0.20 and µk=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 m=H,M=0.2×50=10(kg) ∫,=0 f=MgH2=50×9.8×0.2=980(N f= Nu 10×9.8=980(N) IMg-N=0 (c)If m=m=5kg, assume the system does not move,then T=0 f T=f 10×9.8=49.0(N) As the maximum static friction force fm=Mg2=50×9.8×0.2=980(N) From the above, we know T<f 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=9.5kg)slides on a frictionless plane inclined at an angle 6=34. The block is attached by a string to a second block(of mass m2=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 Fig 4 block I and the plane, what is the acceleration of the tension in the string? Use the values 八=0.24and=0.15
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 Mg N f N T f mg T s s s µ ⎪ ⎩ ⎪ ⎨ ⎧ = = = × = = = × × = = = × = ⇒ 10 9.8 98.0(N) 50 9.8 0.2 98.0(N) 0.2 50 10(kg) or f T mg f Mg m M s s s s µ µ (c) If 5kg 2 1 m′ = m = , assume the system does not move, then ⎩ ⎨ ⎧ − ′ = ′ − = 0 0 s T f m g T 10 9.8 49.0 (N) 2 1 2 1 ⇒ T = fs ′ = mg = × × = As the maximum static friction force is 50 9.8 0.2 98.0(N) fs max = Mgµ s = × × = From the above, we know s max T < f 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 m1=9.5kg) slides on a frictionless plane inclined at an angle o θ = 34 . The block is attached by a string to a second block (of mass m2=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 µs=0.24 and µk=0.15 N v Mg v T v mg v x y T v f v m1 Fig.4 m2 o 34
Solution: (a)The second law force diagrams of two mass are shown in figure. Set the coordinate system as in pply Newton,'s second law of motion, we have re Assuming the acceleration is a shown in fig m1g m2g N-m,g cos 8=0 m2-m, sing 2.6-95×sin34 =-22m/s2 m1+m2 5+ m1m28 9.5×2.6×sin34 (+sin 8) (1+sin34)=31 9.5+26 (b The second law force diagrams of two mass are shown in figure. Set the coordinate system as in Assuming the acceleration is a shown in figure. N Apply Newton's second law of motion, we have m,gsin 8-T-f=m,a m1g g N-m,g cos0=0 m,g=m2 f=urN Solving the equations, we get n, ( sin Ak 9.5×(sin34+0.15c0s34)-26 =1.2m/s m,+ 9.5+26 仍+m(1+sinO-C0:95×26×sin34 m,m2g 95+)×(1+sin34-0.15×cos34)=29N 3. An object is drop from rest. Find the terminal speed assuming that the drag D force is given by D= by Solution The forces acting on the object and the coordinate system are shown in figure
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 ⎪ ⎩ ⎪ ⎨ ⎧ − = − = − = m g T m a N m g T m g m a 2 2 1 1 1 cos 0 sin θ θ Solving it, we get 2 1 2 2 1 2.2 m/s 9.5 2.6 sin 2.6 9.5 sin 34 = − + − × = + − = o g m m m m a θ (1 sin 34 ) 31 N 9.5 2.6 9.5 2.6 sin 34 (1 sin ) 1 2 1 2 × + = + × × + = + = o o θ m m m m g 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 ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = − = − = − − = f N T m g m a N m g m g T f m a µ k θ θ 2 2 1 1 1 cos 0 sin Solving the equations, we get 2 1 2 1 2 1.2 m/s 9.5 2.6 (sin cos ) 9.5 (sin 34 0.15cos34 ) 2.6 = + × + − = + − − = o o g m m m m a θ µ k θ (1 sin34 0.15 cos34 ) 29 N 9.5 2.6 9.5 2.6 sin34 (1 sin cos ) 1 2 1 2 × + − × = + × × + − = + = o o o θ µk θ m m m m g T 3. An object is drop from rest. Find the terminal speed assuming that the drag force is given by 2 D = bv . Solution: The forces acting on the object and the coordinate system are shown in figure. Apply Newton’s second law of motion, we have mg v D v j ˆ m1 m2 x y m g v 1 T v m g v 2 N v T v a v m1 m2 x y m g v 1 T v m g v 2 N v T v a v f v a v
dv dt dt du b dt In( b +,/"e When t→>∞, the terminal speed of the object is v= 4. Assume the kinetic frictional force on a falling mass m is proportional to its speed v, with a proportionality constant B. Choose j to be vertically downward (a)Show that Newton's second law of motion yields dv,=mg-pvy (b)When the mass reaches its terminal sped, what is (c)Show that the terminal speed is v (d)If the mass is dropped from rest, show that v, (0=vem(l-e-pnm)
D − mg = −ma dt dv ⇒ bv − mg = −m 2 dt m b dv b mg v = − − ⇒ 2 1 ∫ ∫ ∫ ∫ = − − ⇒ = − − ⇒ v t v t dt m b dv b mg v dt m b dv b mg v y 0 0 2 0 0 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 ˆ to be vertically downward. (a) Show that Newton’s second law of motion yields y y mg v t v m = − β d d . (b)When the mass reaches its terminal sped, what is t vy d d ? (c) Show that the terminal speed is β mg vterm = . (d) If the mass is dropped from rest, show that ( ) (1 ) t / m y term v t v e−β = −
Solution: The forces acting on the mass are shown in figure. Apply Newtons second law of motion we have (a)Apply Newton's second law of motion mg-f= ma, we have dt That is =mg- (b)When the mass reaches its terminal speed, the acceleration does not change mg-f=ma=0, thus mg -Bv,=m-=0,that is (c)according to the answer in(b), we know 0. So mg -Bv =0 The terminal speed is (d) If the mass is dropped from rest, then dt d(mg-Bv, B ng-Bi >In(mg-B,)-In(mg)=-BILZ →加、B,)=-B B,1%z fern (1
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 mg − f = ma , we have t v mg v ma m y y y d d − β = = That is y y mg v t v m = − β d d (b) When the mass reaches its terminal speed, the acceleration does not change, so mg − f = ma = 0 , thus 0 d d − = = t v mg v m y β y , that is 0 d d = t vy . (c) according to the answer in (b), we know 0 d d = t vy . So mg − βvy = 0 . The terminal speed is β mg vterm = . (d) If the mass is dropped from rest, then (1 ) (1 ) 1 ln(1 ) ln( - ) ln( ) d - d( - ) dt d d d vy 0 0 m t term m t y m t y y y t y y y y y y e v e mg v v e mg m t v mg m t mg v mg t mg v m mg v mg v m v mg v t v m β β β β β β β β β β β β β β − − − ⇒ = − = − ⇒ − = ⇒ − = − ⇒ − = − ⇒ = − = − = − ⇒ ∫ ∫ mg v f v j ˆ
