Newton's Second Law of Motion Explained Through Examples and Applications

Observations ,[object Object],— = — = — = constant F 1 F 2 F 3 a 1 a 2 a 3 The constant is a measure of how effective is the given force in producing acceleration. This ratio is the property of the body called mass . m = — F a A force of 1 newton (1N) is that resultant force that will give a 1-kg mass an acceleration of 1 m/s 2 . The newton (N) is the SI unit of force.

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

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

Illustrative Example 2 ,[object Object],F = ma = (6kg)(2 m/s 2 ) m = — — = ——— = 6 kg F 60N a 10m/s 2 F = 12 N

Illustrative Example 3 ,[object Object],Given: m = 1000kg; v i = 100 km/h = 27.8 m/s; v f = 0 Find: F net = ? Formula: F net = ma a = — —— v f – v i Δ t Δ X 50 m v 13.9 m/s Δ t = — — = ———— Δ t = 3.6 s a = — ——— 0 – 27.8 m/s 3.6 s a = – 7.72 m/s 2

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 )

Relationship Between Mass and Weight ,[object Object],[object Object],W = mg or m = — — W g ,[object Object],[object Object],[object Object]

Illustrative Examples ,[object Object],[object Object],[object Object],[object Object],W = mg = (4.8kg)(9.8 m/s 2 )= 47N m = F/g = (40N)/(9.8 m/s 2 )= 4.08 kg m = F/g = (60lb)/(32 ft/s 2 )= 1.9 slug W = mg = (7slug)(32 m/s 2 )= 224 lb

Illustrative Example 2 ,[object Object],Given: W E = 100N; g E = 9.80 m/s 2 g P = 2.0 m/s 2 Find: W P = ?

Solution ,[object Object],m = — — = ———— = 10.2 kg W E 100 N g E 9.80 m/s 2 W P = mg P = (10.2 kg)(2 m/s 2 ) Weight on the planet: W P = 20.4 N Ans.: a = 1.63 m/s 2 ; m = 81.6 kg on both places

[object Object],Practice Exercise (p.61-62) 2. A 15.0-kg 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? 3. If the forces in number 2 are applied opposite each other, what is the magnitude and direction of the acceleration? Solution Solution Solution

Solution: ,[object Object],Given: F net = 20.0 N; Find: m = ? a = 1.20m/s 2 m = — — = ————— = 16.7 kg F a 20.0 N 1.20 m/s 2 F = ma

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

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

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

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

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

Example: Atwood Machine ,[object Object],5.0 kg 2.0 kg a a T T

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.

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

[object Object],T = m 1 a w 2 – T = m 2 a Combining the 2 equations gives: w 2 = m 1 a + m 2 a w 2 = (m 1 + m 2 )a a = — —————— w 2 (m 1 + m 2 ) Working equation ---- Eq. 1 ---- Eq. 2

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

Example: Horizontal and Vertical motion ,[object Object],100 g 10 g

Solution a = — ————— w 2 (m 1 + m 2 ) a = — ——————— 10g (980cm/s 2 ) (100g + 10g) a = 89.1 cm/s 2 The acceleration: The tension: Isolating the 10-g mass, the tension T acts upward and the acceleration downward. The unbalanced force F = (9800 – T)dynes. Applying the second law: F = ma 9800-T = 10g(89.1cm/s 2 ) T = 9800dy – 891dy T = 8910 dynes ,[object Object]

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?

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

[object Object],T 1 – T 2 = m 1 a T 2 – W 2 = m 2 a W 3 – T 1 = m 3 a W 3 – W 2 = (m 1 + m 2 + m 3 )a a = — ————— W 3 – W 2 m 1 + m 2 + m 3 + + Working equation F net2 = T 2 – W 2 = m 2 a F net1 = T 1 – T 2 = m 1 a F net3 = T 1 – W 3 = -m 3 a

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

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

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

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

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

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

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

Newton's Second Law of Motion Explained Through Examples and Applications

Recommandé

Recommandé

Contenu connexe

Tendances

Tendances (20)

Similaire à Newton's Second Law of Motion Explained Through Examples and Applications

Similaire à Newton's Second Law of Motion Explained Through Examples and Applications (17)

Plus de Nestor Enriquez

Plus de Nestor Enriquez (6)

Newton's Second Law of Motion Explained Through Examples and Applications