The document discusses Newton's second law of motion which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It provides examples showing that applying equal forces to masses of different sizes produces different accelerations, in accordance with this law. It also defines key terms like force, mass, weight, and acceleration and establishes relationships between them using equations.

2. Newton’s 2 nd Law of Motion Applying a constant force of 12 N in succession to 1-, 2- and 3-kg masses will produce accelerations of 12 m/s 2 , 6 m/s 2 and 4m/s 2 , respectively. a = — — = —— = 12 m/s 2 F 12N m 1 1kg a = — — = —— = 6 m/s 2 a = — — = —— = 4 m/s 2 F 12N m 2 2 kg F 12N m 3 3kg

3. Newton’s Second Law of Motion : “Whenever an unbalanced force acts on a body, it produces in the direction of the force an acceleration that is directly proportional to the force and inversely proportional to the mass of the body.” Σ F = + F 1 + F 2 + F 3 + … Force (N) = mass (kg) x acceleration (m/s 2 ) Force (lb) = mass (slug) x acceleration (ft/s 2 ) 1 lb = 4.448 N 1 slug = 14.59 kg F net = ma

6. F = ma = (1000kg)(- 7.72 m/s 2 ) F = 7720 N, South Therefore, we can summarize as follows: SI: W(N) = mkg) x g(9.8m/s 2 ) English: W(lb) = m(slug) x g(32 ft/s 2 )

14. Solution: Given: m = 15.0 kg; F 1 = +15.0 N; F 2 = +18.0 N Find: a = ? a = — — = ———————— F net m 15.0 N + 18.0 N 15.0 kg F net = F 1 + F 2 = ma 2. A 15.0kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box? a = 2.20 m/s 2

15. Solution: Given: m = 15.0 kg; F 1 = +15.0 N; F 2 = +18.0 N Find: a = ? a = — — = ———————— F net m 15.0 N + 18.0 N 15.0 kg F net = F 1 + F 2 = ma 2. A 15.0kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box? a = 1.53 m/s 2 to the right

16. Solution: Given: m = 15.0 kg; F 1 = +15.0 N; F 2 = -18.0 N Find: a = ? a = — — = ———————— F net m 15.0 N - 18.0 N 15.0 kg F net = F 1 + F 2 = ma 3. If the forces in number 2 are applied opposite each other, what is the magnitude and direction of the acceleration? a = 0.2 m/s 2 to the left

17. The Atwood Machine m M T T M m T T Mg mg a a

18. The Atwood Machine M m T T Mg mg a a For mass M: T – Mg = - Ma Mg - T = Ma Eq. 1 For mass m: T – mg = ma Eq. 2 Conbining eq.1 & eq. 2: Mg - T = Ma T – mg = ma + Mg - mg = (M + m)a a = — —— g M – m M + m

20. Vertical and horizontal problems m 1 m 2 A 2.00-kg hanging block pulls a 3.00-kg block along a frictionless table. Calculate for the acceleration of the system and the tension on the cord.

21. Vertical and horizontal problems m 1 m 2 m 1 m 2 T a w 2 a T F net = T = m 1 a - eq.1 F net = T- w 2 = -m 2 a or w 2 – T = m 2 a - eq.2 T T

23. a = — —————— w 2 (m 1 + m 2 ) a = — —————— (2.00kg)(9.80 m/s 2 ) (3.0 kg + 2.0 kg) a = 3.92 m/s 2 For acceleration: For the tension: from eq.1 T = m 1 a = (3.0 kg)(3.92m/s 2 )= 11.8 N Another Example

26. m 1 m 2 m 3 T 1 T 2 T 2 T 1 Vertical and horizontal problems If a 100-g counter weight is attached on the left side of m 1 which is 300 g, and a 200-g mass on the right, what would be the acceleration of the system and the tensions on the left and right cords?

27. m 1 T 1 T 2 Free-body Diagrams m 2 m 3 T 2 T 1 w 3 w 3 F net = T 1 – T 2 = m 1 a F net = T 2 – W 2 = m 2 a F net = T 1 – W 3 = -m 3 a a a m 1 = 300 g m 2 = 100g m 3 = 200 g

29. a = — ———— = ————————————— W 3 – W 2 m 1 + m 2 + m 3 a = 1.63 m/s 2 (0.2kg)(9.8m/s 2 ) – (0.1kg)(9.8m/s 2 ) 0.3 kg + 0.2 kg + 0.1 kg m 1 = 300 g m 2 = 100g m 3 = 200 g

30. m 1 = 300 g m 2 = 100g m 3 = 200 g T 1 – T 2 = m 1 a –Eq.1 T 2 – W 2 = m 2 a –Eq.2 W 3 – T 1 = m 3 a –Eq.3 For Tensions, T 1 and T 2 W 3 – T 1 = m 3 a m 3 g – T 1 = m 3 a T 1 = m 3 g – m 3 a T 1 = m 3 (g – a) T 1 = 0.2kg(9.8m/s 2 – 1.63m/s 2 ) Using Equation 3 T 1 = 1.63 N

31. Using Equation 2 T 2 – W 2 = m 2 a T 2 – m 2 g = m 2 a T 2 = m 2 a + m 2 g T 2 = m 2 (g – a) T 2 = 0.1kg(9.8m/s 2 + 1.63m/s 2 ) T 2 = 1.14 N

32. Using Equation 1 T 1 – T 2 = m 1 a 1.63N – T 2 = 0.3kg(1.63m/s 2 ) 1.63N – T 2 = 0.489 N T 2 = 1.63N – 0.489 N T 2 = 1.14 N T 1 = 1.63 N

33. Solution a = — —— g M – m M + m a = — ———— (9.8m/s 2 ) 5kg – 2kg 5kg + 2kg a = 4.2 m/s 2

34. Another Solution 5.0 kg 2.0 kg a a F = ma Unbalanced force F : F = W 2 + W 1 = 5kg(9.8m/s 2 ) – 2kg(9.8m/s 2 ) = 49 N – 19.6 N = 29.4N The moving masses : m = 2kg + 5kg = 7kg T T The acceleration: F = ma 29.4N = 7kg (a) a = 29.4 N / 7kg = 4.2 m/s 2 w 1 w 2

35. Solution: The tension on the cord Consider the 2.0-kg body as the moving part of the system. We can isolate it as shown in the free-body diagram on the right . Let T be the tension on the cord. The unbalanced force is T – W 1 . The acceleration is 4.2 m/s 2 . F = ma T – 19.6N = (2.0kg)(4.2m/s 2 ) 2.0 kg T a T = 8.4 N + 19.6 N = 28.0 N