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 the1. 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