1. Aim: How can we apply Newton’s 2 nd Law of Acceleration? Do Now: An object with mass m is moving with an initial velocity v o and speeds up to a final velocity of v in time t when an unbalanced force F is applied to it. From this information, derive Newton’s 2 nd Law, F = ma

2. Newton’s 2 nd Law Law of Acceleration Unbalanced forces produce acceleration ΣF = ma The key to solving these problems is in the free-body diagram

3. 1. Which of the following best indicates the direction of the net force, if any, on the ball at point Q ? (A) (B) (C) (D) (E) There is no net force on the ball at point Q. Horizontal motion is constant The only acceleration is vertical due to gravity Gravity only goes down No Calculator **1 minute**

4. 2. A ball falls straight down through the air under the influence of gravity. There is a retarding force F on the ball with magnitude given by F = bv, where v is the speed of the ball and b is a positive constant. The magnitude of the acceleration a of the ball at any time is equal to which of the following? (A) g ‑ b (B) g - bv/m (C) g + bv/m (D) g/b (E) bv/m F = bv F g = ma No Calculator **1 min 15 sec**

5. 3. Which of the following diagrams best represents the gravitational force W. the frictional force f, and the normal force N that act on the block? 4. The magnitude of the frictional force along the plane is most nearly (A)2.5 N (B)5 N (C)6 N (D)10 N (E)16 N A 2 ‑ kilogram block slides down a 30° incline as shown at the right with an acceleration of 2 meters per second squared. ΣF = ma F ║ - F F = ma mgsinθ – F F = ma (2 kg)(10 m/s 2 )sin30 – FF = (2 kg)(2 m/s 2 ) 10 – FF = 4 FF = 6 N No Calculator **2 min 30 sec**

6. No Calculator **2 min 30 sec** F Net = F ll F Net = mgsinθ F Net = (2 kg)(10 m/s 2 )(10/20) F Net = 10 N Sinθ = 10/20 v 2 = v 0 2 + 2ax v 2 = 2ax v 2 = 2(F Net /m)x v 2 = 2(10 N/2 kg)20 m v = 14 m/s 5. 6.

7. 7. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension in the string between the blocks is (A)2F (B)F (C)2/3 F (D)1/2 F (E)1/3 F 1 kg 2 kg ΣF = ma T = ma T = 1a ΣF = ma F - T = ma F - T = 2a F – T = 2a T = 1a F = 3a T = 1a F = 3a Therefore, T = 1/3 F Each block gets its own free-body TTF Add the two equations together to cancel out a variable No Calculator **1 min 15 sec**

8. 8. A helicopter holding a 70 ‑ kilogram package suspended from a rope 5.0 meters long accelerates upward at a rate of 5.2 m/s 2. Neglect air resistance on the package. a. On the diagram below, draw and label all of the forces acting on the package. b. Determine the tension in the rope. c. When the upward velocity of the helicopter is 30 meters per second, the rope is cut and the helicopter continues to accelerate upward at 5.2 m/s 2. Determine the distance between the helicopter and the package 2.0 seconds after the rope is cut. T FgFg Calculator **7.5 min**

9. b. ΣF = ma T – mg = ma T = ma + mg T = m(a + g) T = 70 kg(5.2 m/s 2 + 9.8 m/s 2 ) T = 1050 N c.y = v o t + ½ at 2 y Helicopter = (30 m/s)(2 s) + ½ (5.2 m/s 2 )(2 s) 2 = 70.4 m y Package = (30 m/s)(2 s) - ½ (9.8 m/s 2 )(2 s) 2 = 40.4 m y H – y P + 5 m = 30 m + 5 m = 35 m

10. 9. Part of the track of an amusement park roller coaster is shaped as shown above. A safety bar is oriented length­wise along the top of each car. In one roller coaster car, a small 0.10 ‑ kilogram ball is suspended from this bar by a short length of light, inextensible string. a. Initially, the car is at rest at point A. i. On the diagram to the right, draw and label all the forces acting on the 0.10 ‑ kilogram ball. ii. Calculate the tension in the string. T mg ΣF = 0 T – mg = 0 T = mg T = (0.1 kg)(9.8 m/s 2 ) T = 0.98 N Calculator **12 min**

11. The car is then accelerated horizontally, goes up a 30° incline, goes down a 30° incline, and then goes around a vertical circular loop of radius 25 meters. For each of the four situations described in parts (b) to (d), do all three of the following. In each situation, assume that the ball has stopped swinging back and forth. 1)Determine the horizontal component T h of the tension in the string in newtons and record your answer in the space provided. 2)Determine the vertical component T v of the tension in the string in newtons and record your answer in the space provided. 3)Show on the adjacent diagram the approximate direction of the string with respect to the vertical. The dashed line shows the vertical in each situation.

12. b. The car is at point B moving horizontally 2 m/s to the right with an acceleration of 5.0 m/s 2 T h = T v = ΣF x = ma T h = ma T h = (0.1 kg)(5 m/s 2 ) T h = 0.5 N ΣF y = 0 T v – mg = 0 T v = mg T v = (0.1 kg)(9.8 m/s 2 ) T v = 0.98 N 0.5 N 0.98 N

13. ΣF y = 0 T v – mg = 0 T v = mg T v = (0.1 kg)(9.8 m/s 2 ) T v = 0.98 N Same as aii c. The car is at point C and is being pulled up the 30° incline with a constant speed of 30 m/s. T h = T v = ΣF x = 0 T h = 0 N Velocity is constant 0 N 0.98 N

14. d. The car is at point D moving down the incline with an acceleration of 5.0 m/s 2. T h = T v = Fsinθ Fcosθ mg X-comp of tension is responsible for the x-comp of the acceleration ΣF x = ma T h = Fcosθ T h = macosθ T h = (0.1 kg)(5 m/s 2 )cos30 T h = 0.43 N 0.43 N y-comp of tension counters only part of the weight resulting in vertical acceleration ΣF y = ma T v = Fsinθ - mg T h = masinθ - mg T h = m(asinθ – g) T h = 0.1 kg[(5 m/s 2 )(sin30) – 9.8 m/s 2 ] T h = -0.75 N -0.75 N