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Newton's Laws of Motion Dr. Robert MacKay Clark College, Physics

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**Introduction Newtons 3 laws of motion 1. Law of inertia**

2. Net Force = mass x acceleration ( F = M A ) 3. Action Reaction Newton’s Universal Law of Gravity

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**Isaac Newton 1642-1727 Isaac Newton 1689 Knighted by Queen Anne 1705**

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**Other topics Why do objects accelerate? Why do objects not accelerate?**

Forces in Balance (Equilibrium) Forces out of Balance Friction Air resistance Terminal Velocity

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Law of inertia (1st Law) Every object continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it. acceleration = 0.0 unless the objected is acted on by an unbalanced force

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Law of inertia (1st Law) Inertia (The intrinsic tendency of an object to resist changes in motion) Mass is a measure of an object’s inertia Mass is also a measure of the amount of an object’s matter content. (i.e. protons, neutrons, and electrons) Weight is the force upon an object due to gravity

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Newton’s 2nd Law Net Force = Mass x Acceleration F = M A

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**Newton’s Law of Action Reaction (3rd Law)**

You can not touch without being touched For every action force there is and equal and oppositely directed reaction force

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A M=2.0 kg F=?**

= 2.0 kg x 6.0 m/s2 =12.0 Newtons = 12.0 N A= 6.0 m/s2

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An object experiences a net force and exhibits an acceleration in response. Which of the following statements is always true? (a) The object moves in the direction of the force. (b) The acceleration is in the same direction as the velocity. (c) The acceleration is in the same direction as the force. (d) The velocity of the object increases.

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An object experiences a net force and exhibits an acceleration in response. Which of the following statements is always true? (a) The object moves in the direction of the force. (b) The acceleration is in the same direction as the velocity. X(c) The acceleration is in the same direction as the force. (d) The velocity of the object increases.

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A M=2.0 kg**

F=6.0 N A=F / M = ? m/s2 A= ?

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A M=2.0 kg**

F=6.0 N A=F / M = m/s2 A= m/s2

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A M=? F= 10.0 N**

M= F/A = ? kg A= 20.0 m/s2

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A M=? F= 10.0 N**

M= F/A = kg A= 20.0 m/s2

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A M= 8.0 kg**

F= ? N F=M A = N A= 10.0 m/s2

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A A= 10.0 m/s2**

M= 8.0 kg F=M A = ? N F= ? N F= weight

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A A= 10.0 m/s2**

M= 8.0 kg F=M A = 80 N F= 80 N F= weight

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**Weight W = m g g = 10 m/s2 weight is the force due to the**

m= 6.0 kg W = m g g = 10 m/s2 weight is the force due to the gravitational attraction between a body and its planet W= ?

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**Weight W = m g g = 10 m/s2 weight is the force due to the**

m= 6.0 kg W = m g g = 10 m/s2 weight is the force due to the gravitational attraction between a body and its planet W= 60N

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**Question 1: A force of 45 N pushes horizontally on a 15 kg**

crate resting on a level frictionless surface. (Actually the crate has real good tiny wheels) What is the acceleration of the crate? 3.0 m/s/s 30.0 m/s/s 60.0 m/s/s 0. 33 m/s/s

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**Question 1: A force of 45 N pushes horizontally on a 15 kg**

crate resting on a level frictionless surface. (Actually the crate has real good tiny wheels) What is the acceleration of the crate? X m/s/s 30.0 m/s/s 60.0 m/s/s 0. 33 m/s/s

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**Question 2: A 15.0 kg crate is in contact with a 30.0 kg**

crate on a level frictionless surface as shown. If the 15.0 kg mass is pushed with a force of 45.0 N what is the acceleration of the two masses? A m/s/s B m/s/s C m/s/s D m/s/s

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**Question 2: A 15.0 kg crate is in contact with a 30.0 kg**

crate on a level frictionless surface as shown. If the 15.0 kg mass is pushed with a force of 45.0 N what is the acceleration of the two masses? A m/s/s B m/s/s C m/s/s D m/s/s

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**Question 3: A 15.0 kg crate is in contact with a 30.0 kg**

crate on a level frictionless surface as shown. If the 15.0 kg mass is pushed with a force of 45.0 N what is the force that the 15.0 kg mass exerts on the 30.0 kg mass? 15 N 20 N 25 N 30 N

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**Question 3: A 15.0 kg crate is in contact with a 30.0 kg**

crate on a level frictionless surface as shown. If the 15.0 kg mass is pushed with a force of 45.0 N what is the force that the 15.0 kg mass exerts on the 30.0 kg mass? 15 N 20 N 25 N 30 N The contact force is what gives the 30 kg crate its acceleration of 1 m/s/s. Thus this contact force must be equal to 30kg(1 m/s/s)=30 N

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Weight W = m g 8.0 kg W= ?

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Weight W = m g 8.0 kg W= 80 N

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**A baseball of mass m is thrown upward with some initial speed**

A baseball of mass m is thrown upward with some initial speed. A gravitational force is exerted on the ball (a) at all points in its motion (b) at all points in its motion except at the highest point (c) at no points in its motion

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**A baseball of mass m is thrown upward with some initial speed**

A baseball of mass m is thrown upward with some initial speed. A gravitational force is exerted on the ball (a) at all points in its motion (b) at all points in its motion except at the highest point (c) at no points in its motion

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A F= 150.0 N**

M= 5.0 kg D =120.0 N Net Force = ? A= ? m/s2

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A F= 150.0 N**

M= 5.0 kg D =120.0 N 1. Net Force = 30.0 N 2. A= 6.0 m/s2

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A F= 150.0 N**

M= 5.0 kg 2. D=?? Net Force =?? A= 20.0 m/s2

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A F= 150.0 N**

M= 5.0 kg 2. D=50 N Net Force = 100N Using MA A= 20.0 m/s2

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A F= 150.0 N**

M= 5.0 kg D=? Net Force = ? A= 0.0 m/s2

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**Newton’s 2nd Law Net Force = Mass x Acceleration F = M A F= 150.0 N**

M= 5.0 kg D=150.0 N Net Force = 0.0 A= 0.0 m/s2

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**Which of the following statements is most correct?**

(a) It is possible for an object to have motion in the absence of forces on the object. (b) It is possible to have forces on an object in the absence of motion of the object. (c) Neither (a) nor (b) is correct. (d) Both (a) and (b) are correct.

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**Which of the following statements is most correct?**

(a) It is possible for an object to have motion in the absence of forces on the object. (b) It is possible to have forces on an object in the absence of motion of the object. (c) Neither (a) nor (b) is correct. (d) Both (a) and (b) are correct.

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**An object experiences no acceleration.**

Which of the following cannot be true for the object? (a) A single force acts on the object. (b) No forces act on the object. (c) Forces act on the object, but the forces cancel.

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**An object experiences no acceleration.**

Which of the following cannot be true for the object? (a) A single force acts on the object. (b) No forces act on the object. (c) Forces act on the object, but the forces cancel.

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**QUICK QUIZ 5.3 (end of section 5.5)**

On Earth, where gravity is present, an experiment is performed on a puck on an air hockey table, with negligible friction. A constant horizontal force is applied to the puck and its acceleration is measured. The experiment is performed on the same puck in the far reaches of outer space where both friction and gravity are negligible. The same constant force is applied to the puck and its acceleration is measured. The puck’s acceleration in outer space will be a) greater than its acceleration on Earth, b) less than its acceleration on Earth, c) exactly the same as its acceleration on Earth, d) infinite since neither friction nor gravity are holding it back?

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**QUICK QUIZ 5.3 (end of section 5.5)**

On Earth, where gravity is present, an experiment is performed on a puck on an air hockey table, with negligible friction. A constant horizontal force is applied to the puck and its acceleration is measured. The experiment is performed on the same puck in the far reaches of outer space where both friction and gravity are negligible. The same constant force is applied to the puck and its acceleration is measured. The puck’s acceleration in outer space will be a) greater than its acceleration on Earth, b) less than its acceleration on Earth, c) exactly the same as its acceleration on Earth, d) infinite since neither friction nor gravity are holding it back?

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**Newton’s 2nd Law Friction depends on surfaces in contact (roughness)**

contact force pushing surfaces together Friction ? F= N M= 5.0 A= 0.0 m/s2 Net Force = ?

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**Fnet=P-fK P fs,max=msN fK=mKN f fs,max fK=mKN kinetic friction**

(sliding friction) Static friction F Applied Force=Static frictional force F=fs

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**Air ResistanceForce Depends on: velocity Air density**

Shape and aerodynamics of object

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**Terminal Velocity When air resistance force balances an objects weight**

Air Drag Acceleration= 0.0 ===> Terminal velocity w

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**Terminal Velocity Acceleration = 0.0 ===>Terminal velocity Air Drag**

80 kg 10 kg w w which has the greatest force of air resistance?

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**Not Terminal Velocity Acceleration = ? Air Drag = 240N W = ?**

80 kg W = ? which has the greatest force of air resistance?

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**Not Terminal Velocity Acceleration = ? Air Drag = 240N +x W = 800 N**

80 kg +x W = 800 N

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**Not Terminal Velocity Acceleration = ? Air Drag = 240N SF = m a**

+800N - 240N= 80kg a +560N=80kg a a=7.0 m/s2 Down 80 kg +x W = 800 N

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**Which encounters the greater force of air resistance—**

A falling elephant, or 2. A falling feather? Ch 4-6

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**1. A falling elephant, or 2. A falling feather?**

Which encounters the greater force of air resistance— 1. A falling elephant, or 2. A falling feather? Ch 4-6 Answer: 1 There is a greater force of air resistance on the falling elephant, which “plows through” more air than the feather in getting to the ground. The elephant encounters several newtons of air resistance, which compared to its huge weight has practically no effect on its rate of fall. Only a small fraction of a newton acts on the feather, but the effect is significant because the feather weighs only a fraction of a newton. Remember to distinguish between a force itself and the effect it produces!

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Two smooth balls of exactly the same size, one made of wood and the other of iron, are dropped from a high building to the ground below. The ball to encounter the greater force of air resistance on the way down is Ch 4-7 1. the wooden ball. 2. the iron ball. 3. Neither. The force is the same.

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Two smooth balls of exactly the same size, one made of wood and the other of iron, are dropped from a high building to the ground below. The ball to encounter the greater force of air resistance on the way down is Ch 4-7 Answer: 2 Air resistance depends on both the size and speed of a falling object. Both balls have the same size, but the heavier iron ball falls faster through the air and encounters greater air resistance in its fall. Be careful to distinguish between the amount of air drag and the effect of that air drag. If the greater air drag on the faster ball is small compared to the weight of the ball, it won’t be very effective in reducing acceleration. For example, 2 newtons of air drag on a 20-newton ball has less effect on fall than 1 newton of air drag on a 2-newton ball. 1. the wooden ball. 2. the iron ball. 3. Neither. The force is the same.

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**Just after jumping off a bridge, her acceleration is**

1. 10 m/s/s down. 2. zero 3. increasing Ch 4-8

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**Just after jumping off a bridge, her acceleration is**

1. 10 m/s/s down. 2. zero 3. increasing Ch 4-8

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**As she falls faster and faster through the air, her acceleration**

1. increases. 2. decreases. 3. remains the same. Ch 4-8

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**1. increases. 2. decreases. 3. remains the same.**

As she falls faster and faster through the air, her acceleration 1. increases. 2. decreases. 3. remains the same. Ch 4-8 Answer: 2 Acceleration decreases because the net force on her decreases. Net force is equal to her weight minus her air resistance, and since air resistance increases with increasing speed, net force and hence acceleration decreases. By Newton’s 2nd law: , where mg is her weight and R is the air resistance she encounters. As R increases, a decreases. Note that if she falls fast enough so that R = mg, a = 0, then with no acceleration she falls at constant velocity.

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**When she reaches terminal velocity her acceleration is**

1. zero. 2. decreasing 3. Equal to gravity. Ch 4-8

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**When she reaches terminal velocity her acceleration is**

1. zero. 2. decreasing 3. Equal to gravity. Ch 4-8

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Who’s falling faster? Heavy Light Both falling at same speed

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Who’s falling faster? Heavy Light Both falling at same speed

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