# Lesson 04: Newton s laws of motion

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Lesson 04: Newton s laws of motion If you are not familiar with the basics of calculus and vectors, please read our freely available lessons on these topics, before reading this lesson. SCIMS Academy 1

2 Dynamics an introduction In kinematics, we described motion (of matter) how is a body moving? In dynamics, we relate the motion to its cause, that is forces. Why is the body moving in a particular way? Three laws of motion described by the scientist Newton in the 17 th century is the foundation of dynamics. These laws are based on experimental observations they are not mathematically derived from something more fundamental. SCIMS Academy 2

3 What is force? We associate force with a push or a pull when we push or pull a body, we say that we apply a force to the body. Note there is an interaction between two bodies the body applying the force (called body A), and the body on which the force is applied (called body B). For a force to act on a body, some other body must be present. We must be able to answer the question which body is applying the force? The two bodies may be in physical contact or not so we have contact forces and long range forces. Contact forces occur between bodies in contact (e.g. the force of friction or tension in a string). Contact forces, at the molecular level, are actually electrical forces. Long range forces refer to gravitational and electrical forces where the bodies exerting (& hence experiencing) these forces can be separated from each other, even by empty space. SCIMS Academy 3

4 Newton s first law of motion First law states that in the absence of external interactions (forces), a body P remains in a state of uniform motion (i.e. v = constant, which may be 0 (body P at rest); and a = 0). This property of a body to remain in uniform motion in the absence of external forces is called inertia. Note v is measured with respect to a coordinate system (frame of reference). a) We know that v PA = v PB + v BA. b) If body P has no external interactions and v PA is constant, then Newton s first law is valid from A s frame of reference. But if v BA is not constant, then v PB is also not constant (since their sum has to be the constant v PA ). This means that the first law is invalid in B s frame of reference. c) We call A as an inertial frame of reference it is a frame of reference in which Newton s first law holds. The earth is a good approximation to an inertial frame of reference. d) In point b), if v BA is constant, i.e. B is moving at a constant velocity relative to A, then v PB is also constant. That is Newton s first law holds in B as well. So any frame of reference moving at a constant velocity with respect to an inertial frame, is also inertial. e) A frame of reference which is not inertial is called an accelerated (or non-inertial) frame of reference. e.g. B in point b) above. Note v BA is not constant, which means B is accelerating with respect to the inertial frame A. SCIMS Academy 4

5 Newton s second law of motion In an inertial frame of reference, the following is experimentally observed. Magnitude of acceleration of a body P is directly proportional to the magnitude of force applied (i.e. acceleration a is proportional to force F). Different bodies when subject to the same force have different accelerations. For two bodies A and B subjected to the same force (of any magnitude), the ratio of their accelerations is a constant. So presumably the acceleration is dependent on some property of the body that we call mass. Since we haven t yet defined how to measure force, we can think of the same force being exerted, when we pull a body using a specific spring stretched to a certain length. Using two similar springs side by side, stretched to the same extent is equivalent to doubling the force. If m A and m B is the mass of body A and B respectively, we define their ratio m A /m B = a B /a A where a A and a B is the magnitude of the acceleration of A and B respectively, when acted on by the same force. So we now know, how to compare masses. Acceleration a is inversely proportional to mass m greater the mass, the lesser is its acceleration (i.e. it shows greater resistance to change its state of uniform motion). Therefore, mass is a measure of inertia. We can pick any object as representing unit mass in the SI unit, a mass of 1kg is defined as that of a cylinder maintained by the standards body. With unit mass defined, we can assign a mass to any other body by comparing accelerations as given above. Tying body A and B together, the mass found based on comparing accelerations is m A + m B. Thus mass represents the quantity of matter in a body. SCIMS Academy 5

6 Newton s second law (continued) Based on the experimental observation of acceleration proportional to force, and the definition of mass; we can write a = cf/m, where c is a constant. We take c as 1, so that one unit of force produces one unit of acceleration, when acting on a unit mass. Thus F = ma. In the SI system, the unit of force 1 N (newton) is defined as the force which produces an acceleration of 1 m/s 2, when acting on a mass of 1kg. It is also experimentally verified that The acceleration that a force F produces, has the same direction as F. Hence we can write F = ma. Forces obey the principle of superposition: The acceleration a from several forces acting on a body is the vector sum of the accelerations produced by each force acting alone. F a F ma n n n i ai i 1 i 1 m i 1 i (applies only in an inertial frame of reference) This is Newton s 2nd law. The mass times the acceleration of a body is equal to the vector sum of the forces (total force) acting on the body. This confirms the vector nature of force. It is usually applied in component form to solve problems, i.e. F ma F ma F ma x x y y z z SCIMS Academy 6

7 Newton s third law of motion The third law is a statement of the fact that forces arise because of interactions between bodies. It states that if body B exerts a force F A on body A, then body A exerts a force F B on body B such that F B = -F A. Forces always appear in pairs. Sometimes, you call them as action and reaction, but this does not mean that the reaction force is a result of the action force (there is no cause effect relationship between the forces). Either force can be called the action, and the other one is the reaction. Note, the forces equated above act on different bodies. When we apply Newton s 2 nd law to a body, only one of the forces in an action-reaction pair (namely that which acts on the body) contributes to F. SCIMS Academy 7

8 Weight of a body We know from the kinematics lesson, that a body falling under the force of earth s gravity, has a constant acceleration of magnitude g = 9.8 m/s 2. From Newton s 2 nd law, the gravitational force on the body has a magnitude mg, where m is the mass of the body. We call this force as the weight of the body, and its magnitude W = mg. When the weight of body A and body B are equal, it implies their mass is equal. The familiar equal-arm balance determines when two weights are equal. SCIMS Academy 8

9 Contact forces normal force and friction When a body is in contact with a surface, the surface exerts a force on the body. This force consists of two components one perpendicular to the surface called the normal force and another parallel (tangential) to the surface called friction. Friction arises only when a body A moves or tries to move relative to the surface of a second body B. The magnitude of the frictional force F k on body A when it is moving relative to B, is given approximately by F k = μ k N, where N is the normal force and μ k is called the coefficient of kinetic friction. μ k is determined experimentally and depends on the nature of the surfaces in contact. F k is in the direction opposite to the motion of body A and is called kinetic friction. Body A moving with velocity v F k = μ k N A N v Body B W SCIMS Academy 9

10 Normal force and friction (continued) When we apply a small force F to a body at rest, it doesn t move. This is because of the frictional force F s opposing its motion. F s is called static friction. As we increase the applied force F, static friction F s also increases, keeping the body at rest. But F s has a certain maximum value, and if F increases beyond this, the body starts to move. This maximum value of F s = μ s N, where N is the normal force and μ s is called the coefficient of static friction. μ s is also determined experimentally and depends on the nature of the surfaces in contact. Note F s μ s N where the equality holds only when the body is about to move (that is F s has attained its maximum value). For a given pair of surfaces, μ s > μ k hence the maximum force of static friction is greater than the kinetic friction. The normal force is what balances W, the weight of body A. The normal force is due to inter-molecular repulsions (electrical forces) between the surfaces in contact. Frictional force is also electrical in nature (at the molecular level). Body A at rest A F s = F μ s N N F Body B SCIMS Academy 10 W

11 Applying Newton s laws Based on the forces applied to a body, we can use Newton s laws to determine how the body moves. The problems where we apply the laws, consist of a collection of bodies that interact with each other. a) We take each body in turn, and draw a force diagram (also called a free body diagram) showing the magnitude and direction of each force acting on it. For each force, we must be able to answer the question which body exerts this force? b) In the force diagram of body A, when we show a force F exerted on it by body B; then in the force diagram of body B, we can show the force F in the opposite direction, which is exerted on it by body A (Newton s third law). c) We setup an inertial frame of reference and for each body, we write Newton s 2 nd law in component form. e.g. using the force diagram for body A, we find what is the x component of each force, based on the indicated magnitude and direction. We sum up the x components of all the forces on body A, and equate it to Ma x, where M is the mass of body A, and a x is the x component of its acceleration. d) The bodies may be constrained in some way as implied by the problem statement write down the kinematical equations for this. e.g. the magnitude of acceleration for two bodies connected by an inextensible taut string is the same. e) Keep track of the known and unknown variables. Newton s 2 nd law along with the constraint equations should provide enough relations to find the unknowns. SCIMS Academy 11

12 Example 1 Body A of mass m rests on a plank. The coefficient of static friction between the plank and the body is μ s. What is the maximum angle θ that the plank can make with the horizontal without the body slipping? Solution: The force diagram of body A is shown along with the positive x and y axis. The three forces acting on body A are its weight (mg), normal force N and the static friction F s. mg, N and F s are the magnitude of these forces in the direction indicated. At the maximum angle, the static friction force is at its maximum, i.e. F s = μ s N. a x and a y = 0 since the body is at rest. F s θ y N 1) mg sin N 0 (x components should add up to 0) 2) N mg cos 0 (y components should add up to 0) mg sin N mg cos So tan or tan s s 1 s s s Notes: mgsinθ is the component of weight along the x axis. -F s = -μ s N is the component of friction along the x axis. F x = 0 gives the first equation above. θ mg x N is the component of normal force along the y axis. -mgcos θ is the component of weight along the y axis. F y = 0 gives the second equation above. SCIMS Academy 12

13 T B Example 2 In the diagram shown, there is no friction. The pulleys are light and frictionless. The string is light and inextensible. Block A and Block B have mass m and M respectively. Find the acceleration of each block in terms of m, M and g. Solution: The force diagrams for body A and body B are shown. T is the uniform tension in the string. We choose a separate coordinate axis (fixed to the table) for each body, which is okay. Note when A moves a distance x to the right, B will move down by x/2. With our choice of axes, this means that a x for A is twice a y for B. Let us denote acceleration a y of B as a. m A Fixed Table T N y Fixed support B M For body B Mg 2 T Ma (Newton's 2nd law for y components) For body A T m(2 a) (Newton's 2nd law for x component) Mg Mg 4 ma Ma a (the acceleration of mass M) 4m M The acceleration of mass m is 2a. Mg y A mg T x Note: A taut string pulls the bodies to which it is connected, and the magnitude of this force is called tension. It is another example of a contact force. The tension at the two ends of a string is the same, when the string is light (its mass can be neglected). SCIMS Academy 13

14 Example 3 [IIT 1989]: Two blocks of masses 2.9 kg and 1.9 kg are suspended from a rigid support S by two inextensible wires each of length 1 meter. The upper wire has negligible mass and lower wire has a uniform mass of 0.2 kg/m. The whole system of blocks and support have an upward acceleration of 0.2 m/s 2. Find the tension at the mid point of a) the lower wire. b) the upper wire. Solution: Take the y axis to be vertically up (say fixed to the ground). Since the upper wire has negligible mass, the tension is uniform in it. Tension isn t uniform in the lower wire, since it has mass. To see why, consider its force diagram. If T u is the tension at its upper end, then it pulls down on m 1 with a force T u. Therefore, m 1 pulls the wire up with a force T u (3 rd law). Similarly, if T L is the tension at the lower end, m 2 pulls the wire down with a force T L. Therefore, the 2 nd law for the wire in the y direction gives: T u T L Mg = Ma (M is the mass of wire). For the wire to accelerate up (as given in the problem), T u has to be greater than (T L + Mg) by the amount Ma. Only when M is 0, T u = T L. S 1m 2.9 kg M 1m m 1 y Mg T u M T L Since we need to find the tension at the mid-point of the lower wire & since the whole system is moving up with the same acceleration, we can take one body as m 1 + top half of the lower wire (mass M/2). The other body is m 2 + the lower half of the lower wire. The force diagrams and the equations of motion are shown on the next page. 1.9 kg m 2 SCIMS Academy 14

15 Example 3 (continued) M M M T2 ( m2 ) g ( m2 ) a (2nd law in y direction for composite body m2 ) M T2 ( m2 )( g a) ( )( ) 20 N (tension at midpoint of lower wire). 2 Note half of the lower wire has mass 0.1 kg, since the whole wire is 1 m long and the mass per unit length = 0.2 kg/m. M M M T1 T2 ( m1 ) g ( m1 ) a (2nd law in y direction for composite body m1 ) M T1 T2 ( m1 )( g a) 20 ( )( ) 50 N (tension throughout the upper wire) 2 y y T 2 m 2 + M/2 T 1 m 1 + M/2 T 2 (m 2 + M/2)g (m 1 + M/2)g SCIMS Academy 15

16 Example 4 [IIT 1992]: A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plumb bob is suspended from the ceiling of the car by a light rigid rod of length 1 m. The angle made by the rod with the vertical is (take g = 10 m/s 2 ). a) zero b) 30 c) 45 d) 60 Solution: Using an inertial frame on the ground, the bob has a centripetal acceleration of v 2 /r. The force for this comes from the horizontal component of the tension in the rod as shown in the force diagram. The force responsible for the centripetal acceleration is called the centripetal force. θ x T θ y mg T cos mg 0 (2nd law for y components) 2 mv T sin r (2nd law for x components) 2 2 v 10 tan So the answer is c) rg SCIMS Academy 16

17 Example 5 (Banked curves) When a vehicle (say a car) moves around a circular path, some force must act on the car towards the centre, to produce centripetal acceleration. This force is static friction, since there is no motion in the radial direction; and the car moves tangential to the circle. Sometimes the road can be banked, that is the road slopes down from the outer radius to the inner radius, so that the horizontal component of the normal force also contributes to the centripetal force. What is the maximum speed the car can have, to maintain circular motion (and not skid)? Solution: R is the radius of the path. θ is the banking angle and v is the speed of the car. Let us represent the frictional force F s in the direction shown by μn, where μ s μ μ s, where μ s is the coefficient of static friction. This may seem strange, but it only means that the frictional force can be downwards as shown, or act upwards (μ < 0), and its maximum magnitude is μ s N. v 2 /R θ x F s N θ θ y mg 1) N cos N sin mg 0 (2nd law for y components) 2 2) Nsin N cos mv / R (2nd law for x components) From 1) we have N mg (cos sin ). Substituting for N in 2) v Rg(sin cos ) (3) v cos sin 2 2 max When 0 (no banking) v max Rg( s tan ) (maximum value when s ) 1 tan Rg s Setting 0 in 3) gives the speed v Rg tan at which the normal force alone 0 provides the required centripetal force. If v v 0, then 0 in 3), and the frictional force acts up the slope. s SCIMS Academy 17

18 Example 6 (Apparent weight) A person of mass m stands on a bathroom (weighing) scale in an accelerating elevator. What is the scale reading? Solution: The person is acted on by two forces his weight (mg) and the normal force N from the scale. By Newton s 3 rd law, he applies an equal force N on the scale, and this is what the scale reads. From Newton s 2 nd law N mg = ma y which implies N = m(g + a y ) If the lift (and hence the person) is accelerating up (a y positive); N (and hence the scale reading) is more than mg. If the lift is accelerating down (a y negative), the scale shows a lower weight than mg. When a y = -g (free fall), N = 0 (apparently weightless). The scale reading is called the apparent weight. N y mg SCIMS Academy 18

19 Newton s law in a non-inertial frame We have seen that the relative velocity of a point P as per two observers A and B is related by v PA = v PB + v BA. Differentiating this with respect to time, we have a PA = a PB + a BA. Let A be an inertial frame, and a BA is not zero (that is B is accelerating relative to A, and hence is a non-inertial frame). So a PB = a PA a BA. Multiply by mass M of point P, and note that Ma PA is equal to F, the true force acting on P, since A is an inertial frame. We get Ma PB = F Ma BA. This means that Ma in B s frame is equal to F (the actual force) minus Ma BA where a BA is the acceleration of B relative to an inertial frame. Thus while applying Newton s 2nd law in a non-inertial frame B, we need to add Ma BA to the force side of the equation. This quantity Ma BA is called a pseudo or fictitious force, because it is not a true force arising from interaction between bodies. In a BA, the acceleration of B can be relative to any inertial frame, because if A and A are two inertial frames, we have seen that they have a constant velocity relative to each other. Differentiating v BA = v BA + v A A with respect to t, we get a BA = a BA since v A A is constant (implies acceleration of B is the same in all inertial frames). SCIMS Academy 19

20 Example 7 [IIT 2006]: A circular disc with a groove along its diameter is placed on a horizontal surface. A block of mass 1 kg is placed in the groove. The coefficient of friction between the block and the groove surfaces is μ = 2/5. The disc moves with an acceleration of 25 m/s 2 as shown. What is the acceleration of the block with respect to the disc? Take g = 10 m/s 2 and sinθ = 3/5 and cosθ = 4/5. Solution: The disc is a non-inertial frame with acceleration A = 25 m/s 2, hence we introduce a pseudo force -ma in the force diagram for the block, where m is the mass of the block. In the direction perpendicular to the page, the weight mg is balanced by the normal force N. Additionally there is a normal force N 1 from the sides of the groove. The xy axes is parallel and perpendicular to the groove as shown. x A = 25 m/s 2 F s = μ(n+n 1 ) ma θ Block θ ( mg N ) macos ma (2nd law for x components) N 1 1 masin 0 (2nd law for y components) ax ( g Asin ) Acos (10 25 ) ms (negative sign implies its directed along -x axis). x N 1 y SCIMS Academy 20

### Chapter 4 Newton s Laws: Explaining Motion

Chapter 4 Newton s s Laws: Explaining Motion Newton s Laws of Motion The concepts of force, mass, and weight play critical roles. A Brief History! Where do our ideas and theories about motion come from?!

### Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION

1 P a g e Inertia Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION The property of an object by virtue of which it cannot change its state of rest or of uniform motion along a straight line its own, is

### Newton s Laws of Motion

Physics Newton s Laws of Motion Newton s Laws of Motion 4.1 Objectives Explain Newton s first law of motion. Explain Newton s second law of motion. Explain Newton s third law of motion. Solve problems

### Chapter 5 Newton s Laws of Motion

Chapter 5 Newton s Laws of Motion Force and Mass Units of Chapter 5 Newton s First Law of Motion Newton s Second Law of Motion Newton s Third Law of Motion The Vector Nature of Forces: Forces in Two Dimensions

### Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion. Physics is about forces and how the world around us reacts to these forces.

Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion Physics is about forces and how the world around us reacts to these forces. Whats a force? Contact and non-contact forces. Whats a

### Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting

### v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

### Newton s Laws of Motion

Section 3.2 Newton s Laws of Motion Objectives Analyze relationships between forces and motion Calculate the effects of forces on objects Identify force pairs between objects New Vocabulary Newton s first

### Chapter 4 Dynamics: Newton s Laws of Motion. Copyright 2009 Pearson Education, Inc.

Chapter 4 Dynamics: Newton s Laws of Motion Force Units of Chapter 4 Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal

### Newton s Laws. Physics 1425 lecture 6. Michael Fowler, UVa.

Newton s Laws Physics 1425 lecture 6 Michael Fowler, UVa. Newton Extended Galileo s Picture of Galileo said: Motion to Include Forces Natural horizontal motion is at constant velocity unless a force acts:

### Physics 11 Assignment KEY Dynamics Chapters 4 & 5

Physics Assignment KEY Dynamics Chapters 4 & 5 ote: for all dynamics problem-solving questions, draw appropriate free body diagrams and use the aforementioned problem-solving method.. Define the following

### AP Physics - Chapter 8 Practice Test

AP Physics - Chapter 8 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A single conservative force F x = (6.0x 12) N (x is in m) acts on

### University Physics 226N/231N Old Dominion University. Newton s Laws and Forces Examples

University Physics 226N/231N Old Dominion University Newton s Laws and Forces Examples Dr. Todd Satogata (ODU/Jefferson Lab) satogata@jlab.org http://www.toddsatogata.net/2012-odu Wednesday, September

### Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces Units of Chapter 5 Applications of Newton s Laws Involving Friction Uniform Circular Motion Kinematics Dynamics of Uniform Circular

### Lecture Outline Chapter 5. Physics, 4 th Edition James S. Walker. Copyright 2010 Pearson Education, Inc.

Lecture Outline Chapter 5 Physics, 4 th Edition James S. Walker Chapter 5 Newton s Laws of Motion Dynamics Force and Mass Units of Chapter 5 Newton s 1 st, 2 nd and 3 rd Laws of Motion The Vector Nature

### Chapter 4 Dynamics: Newton s Laws of Motion

Chapter 4 Dynamics: Newton s Laws of Motion Units of Chapter 4 Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal

### There are three different properties associated with the mass of an object:

Mechanics Notes II Forces, Inertia and Motion The mathematics of calculus, which enables us to work with instantaneous rates of change, provides a language to describe motion. Our perception of force is

### VELOCITY, ACCELERATION, FORCE

VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how

### Forces: Equilibrium Examples

Physics 101: Lecture 02 Forces: Equilibrium Examples oday s lecture will cover extbook Sections 2.1-2.7 Phys 101 URL: http://courses.physics.illinois.edu/phys101/ Read the course web page! Physics 101:

Week 8 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

### Announcements. Dry Friction

Announcements Dry Friction Today s Objectives Understand the characteristics of dry friction Draw a FBD including friction Solve problems involving friction Class Activities Applications Characteristics

### 2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration.

2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration. Dynamics looks at the cause of acceleration: an unbalanced force. Isaac Newton was

### QUESTIONS : CHAPTER-5: LAWS OF MOTION

QUESTIONS : CHAPTER-5: LAWS OF MOTION 1. What is Aristotle s fallacy? 2. State Aristotlean law of motion 3. Why uniformly moving body comes to rest? 4. What is uniform motion? 5. Who discovered Aristotlean

### circular motion & gravitation physics 111N

circular motion & gravitation physics 111N uniform circular motion an object moving around a circle at a constant rate must have an acceleration always perpendicular to the velocity (else the speed would

### 5. Forces and Motion-I. Force is an interaction that causes the acceleration of a body. A vector quantity.

5. Forces and Motion-I 1 Force is an interaction that causes the acceleration of a body. A vector quantity. Newton's First Law: Consider a body on which no net force acts. If the body is at rest, it will

### 1. Newton s Laws of Motion and their Applications Tutorial 1

1. Newton s Laws of Motion and their Applications Tutorial 1 1.1 On a planet far, far away, an astronaut picks up a rock. The rock has a mass of 5.00 kg, and on this particular planet its weight is 40.0

### Friction and Newton s 3rd law

Lecture 4 Friction and Newton s 3rd law Pre-reading: KJF 4.8 Frictional Forces Friction is a force exerted by a surface. The frictional force is always parallel to the surface Due to roughness of both

### Newton s Third Law. object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1

Newton s Third Law! If two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1!! Note on notation: is

### PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?

1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always

### Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4 Forces and Newton s Laws of Motion continued Clicker Question 4.3 A mass at rest on a ramp. How does the friction between the mass and the table know how much force will EXACTLY balance the gravity

### Lecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.8-4.12, second half of section 4.7

Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.8-4.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal

### 6: Applications of Newton's Laws

6: Applications of Newton's Laws Friction opposes motion due to surfaces sticking together Kinetic Friction: surfaces are moving relative to each other a.k.a. Sliding Friction Static Friction: surfaces

### Chapter 13, example problems: x (cm) 10.0

Chapter 13, example problems: (13.04) Reading Fig. 13-30 (reproduced on the right): (a) Frequency f = 1/ T = 1/ (16s) = 0.0625 Hz. (since the figure shows that T/2 is 8 s.) (b) The amplitude is 10 cm.

### PHY121 #8 Midterm I 3.06.2013

PHY11 #8 Midterm I 3.06.013 AP Physics- Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension

### Chapter 24 Physical Pendulum

Chapter 4 Physical Pendulum 4.1 Introduction... 1 4.1.1 Simple Pendulum: Torque Approach... 1 4. Physical Pendulum... 4.3 Worked Examples... 4 Example 4.1 Oscillating Rod... 4 Example 4.3 Torsional Oscillator...

### Ch 6 Forces. Question: 9 Problems: 3, 5, 13, 23, 29, 31, 37, 41, 45, 47, 55, 79

Ch 6 Forces Question: 9 Problems: 3, 5, 13, 23, 29, 31, 37, 41, 45, 47, 55, 79 Friction When is friction present in ordinary life? - car brakes - driving around a turn - walking - rubbing your hands together

### B) 40.8 m C) 19.6 m D) None of the other choices is correct. Answer: B

Practice Test 1 1) Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.60 s and reaches the level of the top of the pole after a total elapsed time

### C B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N

Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a

### PHY231 Section 1, Form B March 22, 2012

1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate

### Homework 4. problems: 5.61, 5.67, 6.63, 13.21

Homework 4 problems: 5.6, 5.67, 6.6,. Problem 5.6 An object of mass M is held in place by an applied force F. and a pulley system as shown in the figure. he pulleys are massless and frictionless. Find

### Serway_ISM_V1 1 Chapter 4

Serway_ISM_V1 1 Chapter 4 ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. Newton s second law gives the net force acting on the crate as This gives the kinetic friction force as, so choice (a) is correct. 2. As

### Chapter 18 Static Equilibrium

Chapter 8 Static Equilibrium 8. Introduction Static Equilibrium... 8. Lever Law... Example 8. Lever Law... 4 8.3 Generalized Lever Law... 5 8.4 Worked Examples... 7 Example 8. Suspended Rod... 7 Example

### AP1 Gravity. at an altitude equal to twice the radius (R) of the planet. What is the satellite s speed assuming a perfectly circular orbit?

1. A satellite of mass m S orbits a planet of mass m P at an altitude equal to twice the radius (R) of the planet. What is the satellite s speed assuming a perfectly circular orbit? (A) v = Gm P R (C)

### Newton s Law of Motion

chapter 5 Newton s Law of Motion Static system 1. Hanging two identical masses Context in the textbook: Section 5.3, combination of forces, Example 4. Vertical motion without friction 2. Elevator: Decelerating

### More of Newton s Laws

More of Newton s Laws Announcements: Tutorial Assignments due tomorrow. Pages 19-21, 23, 24 (not 22,25) Note Long Answer HW due this week. CAPA due on Friday. Have added together the clicker scores so

### Physics I Honors: Chapter 4 Practice Exam

Physics I Honors: Chapter 4 Practice Exam Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Which of the following statements does not describe

### Chapter 5 Newton s Laws of Motion

Chapter 5 Newton s Laws of Motion Sir Isaac Newton (1642 1727) Developed a picture of the universe as a subtle, elaborate clockwork slowly unwinding according to well-defined rules. The book Philosophiae

### SOLUTIONS TO PROBLEM SET 4

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01X Fall Term 2002 SOLUTIONS TO PROBLEM SET 4 1 Young & Friedman 5 26 A box of bananas weighing 40.0 N rests on a horizontal surface.

### Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion

Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion Conceptual Questions 1) Which of Newton's laws best explains why motorists should buckle-up? A) the first law

### Chapter 6 Work and Energy

Chapter 6 WORK AND ENERGY PREVIEW Work is the scalar product of the force acting on an object and the displacement through which it acts. When work is done on or by a system, the energy of that system

### AP1 Dynamics. Answer: (D) foot applies 200 newton force to nose; nose applies an equal force to the foot. Basic application of Newton s 3rd Law.

1. A mixed martial artist kicks his opponent in the nose with a force of 200 newtons. Identify the action-reaction force pairs in this interchange. (A) foot applies 200 newton force to nose; nose applies

### Force. A force is a push or a pull. Pushing on a stalled car is an example. The force of friction between your feet and the ground is yet another.

Force A force is a push or a pull. Pushing on a stalled car is an example. The force of friction between your feet and the ground is yet another. Force Weight is the force of the earth's gravity exerted

### Copyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass

Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of

### Work, Energy and Power Practice Test 1

Name: ate: 1. How much work is required to lift a 2-kilogram mass to a height of 10 meters?. 5 joules. 20 joules. 100 joules. 200 joules 5. ar and car of equal mass travel up a hill. ar moves up the hill

### NEWTON S LAWS OF MOTION

NEWTON S LAWS OF MOTION Background: Aristotle believed that the natural state of motion for objects on the earth was one of rest. In other words, objects needed a force to be kept in motion. Galileo studied

### UNIT 2D. Laws of Motion

Name: Regents Physics Date: Mr. Morgante UNIT 2D Laws of Motion Laws of Motion Science of Describing Motion is Kinematics. Dynamics- the study of forces that act on bodies in motion. First Law of Motion

### www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x

Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity

### Mass, energy, power and time are scalar quantities which do not have direction.

Dynamics Worksheet Answers (a) Answers: A vector quantity has direction while a scalar quantity does not have direction. Answers: (D) Velocity, weight and friction are vector quantities. Note: weight and

### VIII. Magnetic Fields - Worked Examples

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.0 Spring 003 VIII. Magnetic Fields - Worked Examples Example : Rolling rod A rod with a mass m and a radius R is mounted on two parallel rails

### Chapter 3.8 & 6 Solutions

Chapter 3.8 & 6 Solutions P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequency are inverses according to Equation 3.26. To find speed we need to know the distance traveled

### Uniform Circular Motion III. Homework: Assignment (1-35) Read 5.4, Do CONCEPT QUEST #(8), Do PROBS (20, 21) Ch. 5 + AP 1997 #2 (handout)

Double Date: Objective: Uniform Circular Motion II Uniform Circular Motion III Homework: Assignment (1-35) Read 5.4, Do CONCEPT QUEST #(8), Do PROBS (20, 21) Ch. 5 + AP 1997 #2 (handout) AP Physics B

### Solution Derivations for Capa #11

Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform

### Curso2012-2013 Física Básica Experimental I Cuestiones Tema IV. Trabajo y energía.

1. A body of mass m slides a distance d along a horizontal surface. How much work is done by gravity? A) mgd B) zero C) mgd D) One cannot tell from the given information. E) None of these is correct. 2.

### This week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension. PHYS 2: Chap.

This week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension PHYS 2: Chap. 19, Pg 2 1 New Topic Phys 1021 Ch 7, p 3 A 2.0 kg wood box slides down a vertical

### Chapter 11 Equilibrium

11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium of

### 8. Newton's Law of Gravitation

2 8. Newton's Law Gravitation Rev.nb 8. Newton's Law of Gravitation Introduction and Summary There is one other major law due to Newton that will be used in this course and this is his famous Law of Universal

### Solution: (a) For a positively charged particle, the direction of the force is that predicted by the right hand rule. These are:

Problem 1. (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields as shown in the figure. (b) Repeat part (a), assuming the moving particle is

### PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be

### 7. Kinetic Energy and Work

Kinetic Energy: 7. Kinetic Energy and Work The kinetic energy of a moving object: k = 1 2 mv 2 Kinetic energy is proportional to the square of the velocity. If the velocity of an object doubles, the kinetic

### AP Physics Newton's Laws Practice Test

AP Physics Newton's Laws Practice Test Answers: A,D,C,D,C,E,D,B,A,B,C,C,A,A 15. (b) both are 2.8 m/s 2 (c) 22.4 N (d) 1 s, 2.8 m/s 16. (a) 12.5 N, 3.54 m/s 2 (b) 5.3 kg 1. Two blocks are pushed along a

### If you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ

Experiment 4 ormal and Frictional Forces Preparation Prepare for this week's quiz by reviewing last week's experiment Read this week's experiment and the section in your textbook dealing with normal forces

### Lab 7: Rotational Motion

Lab 7: Rotational Motion Equipment: DataStudio, rotary motion sensor mounted on 80 cm rod and heavy duty bench clamp (PASCO ME-9472), string with loop at one end and small white bead at the other end (125

### Mechanics 1: Conservation of Energy and Momentum

Mechanics : Conservation of Energy and Momentum If a certain quantity associated with a system does not change in time. We say that it is conserved, and the system possesses a conservation law. Conservation

### Lab 5: Conservation of Energy

Lab 5: Conservation of Energy Equipment SWS, 1-meter stick, 2-meter stick, heavy duty bench clamp, 90-cm rod, 40-cm rod, 2 double clamps, brass spring, 100-g mass, 500-g mass with 5-cm cardboard square

### Mechanics 2. Revision Notes

Mechanics 2 Revision Notes November 2012 Contents 1 Kinematics 3 Constant acceleration in a vertical plane... 3 Variable acceleration... 5 Using vectors... 6 2 Centres of mass 8 Centre of mass of n particles...

### Chapter 4: Newton s Laws: Explaining Motion

Chapter 4: Newton s Laws: Explaining Motion 1. All except one of the following require the application of a net force. Which one is the exception? A. to change an object from a state of rest to a state

### Conceptual Questions: Forces and Newton s Laws

Conceptual Questions: Forces and Newton s Laws 1. An object can have motion only if a net force acts on it. his statement is a. true b. false 2. And the reason for this (refer to previous question) is

### Physics Midterm Review Packet January 2010

Physics Midterm Review Packet January 2010 This Packet is a Study Guide, not a replacement for studying from your notes, tests, quizzes, and textbook. Midterm Date: Thursday, January 28 th 8:15-10:15 Room:

### CHAPTER 6 WORK AND ENERGY

CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS. REASONING AND SOLUTION The work done by F in moving the box through a displacement s is W = ( F cos 0 ) s= Fs. The work done by F is W = ( F cos θ). s From

### Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)

Gravitational Field A gravitational field as a region in which an object experiences a force due to gravitational attraction Gravitational Field Strength The gravitational field strength at a point in

### Prelab Exercises: Hooke's Law and the Behavior of Springs

59 Prelab Exercises: Hooke's Law and the Behavior of Springs Study the description of the experiment that follows and answer the following questions.. (3 marks) Explain why a mass suspended vertically

### Chapter 4 Dynamics: Newton s Laws of Motion

Chapter 4 Dynamics: Newton s Laws of Motion Units of Chapter 4 Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the NormalForce

### CHAPTER 15 FORCE, MASS AND ACCELERATION

CHAPTER 5 FORCE, MASS AND ACCELERATION EXERCISE 83, Page 9. A car initially at rest accelerates uniformly to a speed of 55 km/h in 4 s. Determine the accelerating force required if the mass of the car

### Newton s Second Law. ΣF = m a. (1) In this equation, ΣF is the sum of the forces acting on an object, m is the mass of

Newton s Second Law Objective The Newton s Second Law experiment provides the student a hands on demonstration of forces in motion. A formulated analysis of forces acting on a dynamics cart will be developed

### AP Physics C. Oscillations/SHM Review Packet

AP Physics C Oscillations/SHM Review Packet 1. A 0.5 kg mass on a spring has a displacement as a function of time given by the equation x(t) = 0.8Cos(πt). Find the following: a. The time for one complete

### Lecture 07: Work and Kinetic Energy. Physics 2210 Fall Semester 2014

Lecture 07: Work and Kinetic Energy Physics 2210 Fall Semester 2014 Announcements Schedule next few weeks: 9/08 Unit 3 9/10 Unit 4 9/15 Unit 5 (guest lecturer) 9/17 Unit 6 (guest lecturer) 9/22 Unit 7,

### Physics, Chapter 3: The Equilibrium of a Particle

University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1-1958 Physics, Chapter 3: The Equilibrium of a Particle

### = Ps cos 0 = (150 N)(7.0 m) = J F N. s cos 180 = µ k

Week 5 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions o these problems, various details have been changed, so that the answers will come out dierently. The method to ind the solution

### TEACHER ANSWER KEY November 12, 2003. Phys - Vectors 11-13-2003

Phys - Vectors 11-13-2003 TEACHER ANSWER KEY November 12, 2003 5 1. A 1.5-kilogram lab cart is accelerated uniformly from rest to a speed of 2.0 meters per second in 0.50 second. What is the magnitude

### W02D2-2 Table Problem Newton s Laws of Motion: Solution

ASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 W0D- Table Problem Newton s Laws of otion: Solution Consider two blocks that are resting one on top of the other. The lower block

### 9. The kinetic energy of the moving object is (1) 5 J (3) 15 J (2) 10 J (4) 50 J

1. If the kinetic energy of an object is 16 joules when its speed is 4.0 meters per second, then the mass of the objects is (1) 0.5 kg (3) 8.0 kg (2) 2.0 kg (4) 19.6 kg Base your answers to questions 9

### Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

### Name Class Period. F = G m 1 m 2 d 2. G =6.67 x 10-11 Nm 2 /kg 2

Gravitational Forces 13.1 Newton s Law of Universal Gravity Newton discovered that gravity is universal. Everything pulls on everything else in the universe in a way that involves only mass and distance.

### Lecture-IV. Contact forces & Newton s laws of motion

Lecture-IV Contact forces & Newton s laws of motion Contact forces: Force arises from interaction between two bodies. By contact forces we mean the forces which are transmitted between bodies by short-range

### 1) The gure below shows the position of a particle (moving along a straight line) as a function of time. Which of the following statements is true?

Physics 2A, Sec C00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to ll your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

### AP Physics Applying Forces

AP Physics Applying Forces This section of your text will be very tedious, very tedious indeed. (The Physics Kahuna is just as sorry as he can be.) It s mostly just a bunch of complicated problems and

### Units DEMO spring scales masses

Dynamics the study of the causes and changes of motion Force Force Categories ContactField 4 fundamental Force Types 1 Gravity 2 Weak Nuclear Force 3 Electromagnetic 4 Strong Nuclear Force Units DEMO spring

### FRICTION, WORK, AND THE INCLINED PLANE

FRICTION, WORK, AND THE INCLINED PLANE Objective: To measure the coefficient of static and inetic friction between a bloc and an inclined plane and to examine the relationship between the plane s angle