# HW#4b Page 1 of 6. I ll use m = 100 kg, for parts b-c: accelerates upwards, downwards at 5 m/s 2 A) Scale reading is the same as person s weight (mg).

Save this PDF as:

Size: px
Start display at page:

Download "HW#4b Page 1 of 6. I ll use m = 100 kg, for parts b-c: accelerates upwards, downwards at 5 m/s 2 A) Scale reading is the same as person s weight (mg)."

## Transcription

1 HW#4b Page 1 of 6 Problem 1. A 100 kg person stands on a scale. a.) What would be the scale readout? b.) If the person stands on the scale in an elevator accelerating upwards at 5 m/s, what is the scale readout? c.) If the person stands on the scale in an elevator accelerating downwards at 5 m/s, what is the scale readout? d.) If the person stands on the scale in an elevator accelerating downwards at 9.8 m/s, what is the scale readout? e.) Explain why your result from part d makes sense. Is the actual weight of the person any different? What does the person perceive his weight to be? f.) If the person stands on the scale in an elevator moving up at a constant speed of 100 m/s, what is the scale readout? Explain why during most of the time during an elevator ride, a person will perceive nothing unusual. I ll use m = 100 kg, for parts b-c: accelerates upwards, downwards at 5 m/s A) Scale reading is the same as person s weight (mg). B) Scale reading (1480 N) is larger than person s actual weight (980 N). Person will feel heavier (feel more support) than they really are (because the floor is pushing up on them with a force larger than their weight). C) Scale reading (480 N) is less than person s actual weight (980 N). Person will feel lighter than they really are (because floor is pushing up on them with a force smaller than their weight).

2 HW#4b Page of 6 D) Scale reading (0 N) is less than person s actual weight (980 N). E) The scale reads zero. Accordingly, the person will feel weightless. It s not that the actual weight (mg = 980 N) of the person is any different from normal. Gravity still exerts the same force on them. But now, since a y = -9.8 m/s, the person and elevator are in freefall. They will fall together, and there is no danger of the person smashing the surface. (Normal force is only as large as it needs to be, so it is zero here.) The floor falls out from underneath the person as fast as the person can fall towards it, so they maintain the same distance apart without actually pressing against each other. Then N can be 0. F) Scale reading is same as person s weight. This looks identitcal to part A, since both at rest and constant velocity are equivalent states according to Newton s 1 st Law. For both, F net = 0. Person will feel their weight just as if standing at rest (not feel heavier or lighter). For most of an elevator ride the elevator is going at a constant velocity, so the person will not feel anything unusual. It is only during the short periods when the elevator is accelerating (getting up to speed from rest, and slowing to a stop) that the person feels lighter or heavier than normal. Problem 1. A 35 kg person stands on a scale. a.) What would be the scale readout? b.) If the person stands on the scale in an elevator accelerating upwards at 7 m/s, what is the scale readout? c.) If the person stands on the scale in an elevator accelerating downwards at 7 m/s, what is the scale readout? d.) If the person stands on the scale in an elevator accelerating downwards at 9.8 m/s, what is the scale readout? e.) Explain why your result from part d makes sense. Is the actual weight of the person any different? What does the person perceive his weight to be? f.) If the person stands on the scale in an elevator moving up at a constant speed of 100 m/s, what is the scale readout? Explain why during most of the time during an elevator ride, a person will perceive nothing unusual.

3 HW#4b Page 3 of 6 Solutions. (a) The person's weight is mass times g: W = (35 kg)(9.8 m/s ) = 343 N. This is exactly the person's weight, though we usually measure our bulk in mass, in kg, rather than in force. (I am not sure where you find a scale that reads out in newtons.) (b) The scale is actually measuring the normal force on the person, and we now have to add the acceleration of the elevator to g: N = m(g+a)=(35 kg)(9.8 m/s + 7 m/s ) = 588 N. (c) Now the acceleration is negative: N = m(g+a)=(35 kg)(9.8 m/s - 7 m/s ) = 98 N. (d) Here the elevator is in free fall; N = 0 N. (e) Well - in free fall, you certainly would not expect to press downwards on the scale. (f) If the elevator moves at any constant speed, the acceleration is equal to zero, and the weight is the same as if the elevator were at rest. N = W = 343 N. (g) Most of the time in an elevator ride the elevator is moving at constant velocity, so there is no acceleration, and ones weight "feels" normal. Problem. Static and Kinetic Friction: A 10 kg crate is at rest on a horizontal table ( s =0.9, k =0.). Then various amounts of horizontal pushing force are applied to it. a.) The largest pushing force that can be applied before the crate starts moving is of size: b.)-d.) If the pushing force were instead the two amounts shown below, please fill in the rest of the values on the chart below: horizontal pushing force: friction force: (size only) (indicate static or kinetic): net force on acceleration of 11 N 11 static 0 N 0 m/s 115 N 19.6 kinetic 95.4 N 9.54 m/s Solutions. (a) this is just equal to the maximum static frictional force, fstatic - max = µ sn = µ smg = ( 0.9)( 10 kg)( 9.8 m/s ) = 88. N Static friction will oppose the pushing force, keeping the object stationary, up until some maximum amount. (If the pushing force is greater than this maximum amount, the object will move.)

4 HW#4b Page 4 of 6 (b,). With a horizontal pushing force of only 11 less than the maximum static frictional force, the crate will not move, and all forces balance out, in horizontal and vertical directions. So, the frictional force just equals the pushing force (in magnitude); the net force is zero; and the acceleration is equal to zero. The object will stay stationary static friction. Force of static friction can take on any value from 0 to 88. N. In this case, it only needs to be 11 N to prevent motion. (c,d). With a horizontal pushing force of 115 greater than the maximum static frictional force, the crate will break loose and accelerate away. The object will move use kinetic friction. The frictional force is kinetic friction now. There is no static friction now. Friction will be equal to fkinetic = µ kn = µ kmg = ( 0.)( 10 kg)( 9.8 m/s ) = 19.6 N The net force is the pushing force, minus this amount, 115 N N = 95.4 causing an acceleration of a = F/m = (95.4 N)/(10 kg) = 9.54 m/s. Problem 3. Friction Patch: A 10 kg crate is sliding at a constant speed of 13 m/s along a frictionless horizontal table, when it encounters a friction patch ( s =0.9, k =0.3). Once the crate enters the friction patch, a.) What are the sizes of the net force and acceleration of the crate? size of net force: N size of acceleration: m/s b.) How fast will the crate be going, when it has traveled 5 m across the friction patch? m/s Solution. (a) The normal force throughout is equal to the weight of the crate, mg = (10 kg)(9.8 m/s ) = 98 N. And since the crate is sliding, the friction is kinetic, not static. So, the frictional force is f = µ kmg = ( 0.3)( 10 kg)( 9.8 m/s ) = 9.4 N This is the net force, since the weight and the normal force cancel out in vertical direction and in horizontal direction there is only kinetic friction and no other forces. The resulting acceleration is a = f / m =.94 m/s (b) We can use equation (4) of the constant -acceleration equation set, (this equation doesn t need to know time, and it relates v, v0, a and x. v = v + a x 0 Here v 0 = 13 m/s, x = 5 m, and a (opposite to the direction of motion) = -.94 m/s. So, ( ) ( )( ) v = 13 m/s.94 m/s 5 m = 11.8 m/s =11.8 m/s

5 HW#4b Page 5 of 6 Problem 4. Static and Kinetic Friction on an incline: A 10 kg crate is placed at rest on an incline plane ( s =0.7, k =0.5), whose angle can be adjusted. a.) The maximum angle of the incline for which the crate will stay at rest is: b.)-e.) If the angle of the incline were instead the two amounts shown below, please fill in the rest of the values on the chart below: angle of incline: component of weight acting along incline: (size only) friction force: (size only) (indicate static or kinetic): net force on acceleration of 15 o N N m/s 65 o N N m/s f.) For the 65 o angle, find how long it would take the crate to slide down 5 m along the incline. g.) Explain why it is NOT possible for an object which is placed at rest on an inclined plane to then start sliding down with constant speed. Solution. First we need a diagram and a free-body diagram. We choose x direction to be along the incline and y direction to be perpendicular to the incline. (a) When it is not moving, the friction is static friction. Total forces in all directions add to zero. From the free-body diagram we can see that, since the x and the y forces must cancel (body at rest, no acceleration!), N = mgcosθ f = mgsinθ and, for the maximum angle without the mass sliding, f = µ sn = µ smgcosθ This much will always be true if the object is at rest. (N = mg cos Θ is generally always true on an inclined plane). mg sin Θ tries to pull object down incline, but if the object is able to stay at rest, that means force of static friction must be the same size to cancel it off (hence the statement of: mg sin Θ= F F,s ). At the max angle, just before the object would slide f,s is as large as it possible can be, which is µ s N. Setting the two expressions for f=f s,max (at maximum angle) equal gives mg sinθ = µ smg cosθ sinθ µ s = = tanθ cosθ

6 HW#4b Page 6 of 6 For a static coefficient of friction of 0.7, this gives that the maximum angle will be: 1 1 θ = tan µ = tan 0.7 = = 35 degree s (b-e) Now fill in the table. If the angle is equal to 15 degrees, the crate will not slide. We still only have static friction. The component of the weight along the incline is (always, in fact) equal to mg sin θ, which in the fifteen-degree case is equal to mg*sin θ= 5.4 N. The frictional force (static friction) just cancels it out. And all the forces cancel out in pairs, so the net force is zero and the acceleration is zero. Do not use f = s N to calculate static friction. Because s N is the maximum of f s, not f s You should note that, in this case, f s is far less than its maximum amount. Remember f,s µ s N. So f s can be anywhere from 0 to µ s N. IN this case the force trying to get it moving (mg sin Θ) is only 5.4 so f s only needs to be that much to stay at rest. F net = 0, a = 0 Total force in the direction along the incline is mg*sin15 f s=0, so f k = mg*sin15=5.4 N In the 65-degree case, the maximum static force of friction has been exceeded. It moves. And there is only kinetic friction now. f k = k N, always. Normal force is still N = mg cos 65. Because the acceleration in y direction is still zero. The component of the weight along the incline is mg sin(65 )=88.8 N. The frictional force is equal to The net force on the crate is equal to the component of the weight along the incline, minus the frictional force. (The forces perpendicular to the incline cancel out.) And this net force causes the acceleration. F net = mg sinθ - f k = mg sinθ k mg cosθ Fnet = 88.8 N 0.71 N=68.11 N Fnet ( N) Round off to or 3 significant figures. a = 6.61 m/ s a= = = m/s m 10 kg ( ) (f) Now in the second case, with an acceleration (from rest) of 6.61 m/s, the crate goes a distance of 5 m. How long does it take?, v0=0. x0=0, 1 x = at x 5 ( m) =1.3 s t = = = 1.30 s a m/s ( ) (g) If an object is placed at rest on the incline, it will remain at rest unless the angle is great enough for the component of the weight down the incline to overcome static friction. From then on, the retarding force is due to kinetic friction. The force needed to get object moving from rest is larger than the force needed to keep object at constant velocity. I mean the maximum static friction force is larger than the kinetic friction. The force needed to get object moving must overcome maximum static friction, whereas the force needed to keep an object at constant velocity must be equal to the kinetic friction. This is always true since µ s > µ k for all materials. So, if the force trying to get object moving is large enough to get the object moving from rest, it is definitely larger than the forced needed to keep it moving at constant velocity. So there will be a net for, and acceleration, down incline. In other words, if mg sin Θ is enough to exceed F F,s maximum (=µ s N), so there will be net force, hence acceleration, down incline constant velocity from rest impossible. While the driving force is the component of gravity along the incline s direction mg sinθ. Once the angle is larger enough and mgsinθ is already more than the maximum static friction, mgsinθ is definitely larger than the kinetic friction force, so the object has to accelerate.

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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.

### 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

### 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

### B) 286 m C) 325 m D) 367 m Answer: B

Practice Midterm 1 1) When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal velocity. This means that A) the acceleration is equal to g. B) the force of

### 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

### B Answer: neither of these. Mass A is accelerating, so the net force on A must be non-zero Likewise for mass B.

CTA-1. An Atwood's machine is a pulley with two masses connected by a string as shown. The mass of object A, m A, is twice the mass of object B, m B. The tension T in the string on the left, above mass

### 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

### A) N > W B) N = W C) N < W. speed v. Answer: N = W

CTN-12. Consider a person standing in an elevator that is moving upward at constant speed. The magnitude of the upward normal force, N, exerted by the elevator floor on the person's feet is (larger than/same

### 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

### 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

### 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

### 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

### D) A block pulled with a constant force will have a constant acceleration in the same direction as the force.

Phsics 100A Homework 4 Chapter 5 Newton s First Law A)If a car is moving to the left with constant velocit then the net force applied to the car is zero. B) An object cannot remain at rest unless the net

### 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

### Recitation Week 4 Chapter 5

Recitation Week 4 Chapter 5 Problem 5.5. A bag of cement whose weight is hangs in equilibrium from three wires shown in igure P5.4. wo of the wires make angles θ = 60.0 and θ = 40.0 with the horizontal.

### Lecture 7 Force and Motion. Practice with Free-body Diagrams and Newton s Laws

Lecture 7 Force and Motion Practice with Free-body Diagrams and Newton s Laws oday we ll just work through as many examples as we can utilizing Newton s Laws and free-body diagrams. Example 1: An eleator

### University Physics 226N/231N Old Dominion University. Getting Loopy and Friction

University Physics 226N/231N Old Dominion University Getting Loopy and Friction Dr. Todd Satogata (ODU/Jefferson Lab) satogata@jlab.org http://www.toddsatogata.net/2012-odu Friday, September 28 2012 Happy

### 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

### 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

### DISPLACEMENT AND FORCE IN TWO DIMENSIONS

DISPLACEMENT AND FORCE IN TWO DIMENSIONS Vocabulary Review Write the term that correctly completes the statement. Use each term once. coefficient of kinetic friction equilibrant static friction coefficient

### 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

### 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

### 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

### 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

### Physics 100 Friction Lab

Åsa Bradley SFCC Physics Name: AsaB@spokanefalls.edu 509 533 3837 Lab Partners: Physics 100 Friction Lab Two major types of friction are static friction and kinetic (also called sliding) friction. Static

### LAB 6 - GRAVITATIONAL AND PASSIVE FORCES

L06-1 Name Date Partners LAB 6 - GRAVITATIONAL AND PASSIVE FORCES OBJECTIVES And thus Nature will be very conformable to herself and very simple, performing all the great Motions of the heavenly Bodies

### 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?!

### 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

### 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

### 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

### 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

### LAB 6: GRAVITATIONAL AND PASSIVE FORCES

55 Name Date Partners LAB 6: GRAVITATIONAL AND PASSIVE FORCES And thus Nature will be very conformable to herself and very simple, performing all the great Motions of the heavenly Bodies by the attraction

### Problems Worked Out.notebook. February 26, soh cah toa θ. Working with the triangles. Statics Answers Worked out in 3rd block

Working with the triangles slope weight = mass x gravity soh cah toa θ hypotenuse normal force adjacent FN = weight cos θ opposite force downhill F downhill = weight sin θ Statics 12-17 Answers Worked

### Describe the relationship between gravitational force and distance as shown in the diagram.

Name Period Chapter 2 The Laws of Motion Review Describe the relationship between gravitational force and distance as shown in the diagram. Assess the information about gravity, mass, and weight. Read

### 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.

### 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

### Resistance in the Mechanical System. Overview

Overview 1. What is resistance? A force that opposes motion 2. In the mechanical system, what are two common forms of resistance? friction and drag 3. What is friction? resistance that is produced when

### charge is detonated, causing the smaller glider with mass M, to move off to the right at 5 m/s. What is the

This test covers momentum, impulse, conservation of momentum, elastic collisions, inelastic collisions, perfectly inelastic collisions, 2-D collisions, and center-of-mass, with some problems requiring

### 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

### Objective: Equilibrium Applications of Newton s Laws of Motion I

Type: Single Date: Objective: Equilibrium Applications of Newton s Laws of Motion I Homework: Assignment (1-11) Read (4.1-4.5, 4.8, 4.11); Do PROB # s (46, 47, 52, 58) Ch. 4 AP Physics B Mr. Mirro Equilibrium,

### Physics 201 Homework 8

Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kg-m 2. What is the

### 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:

### 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

### 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

### 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

### STAAR Science Tutorial 25 TEK 8.6C: Newton s Laws

Name: Teacher: Pd. Date: STAAR Science Tutorial 25 TEK 8.6C: Newton s Laws TEK 8.6C: Investigate and describe applications of Newton's law of inertia, law of force and acceleration, and law of action-reaction

### Name: Partners: Period: Coaster Option: 1. In the space below, make a sketch of your roller coaster.

1. In the space below, make a sketch of your roller coaster. 2. On your sketch, label different areas of acceleration. Put a next to an area of negative acceleration, a + next to an area of positive acceleration,

### 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 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

### 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

### PHYSICS 149: Lecture 4

PHYSICS 149: Lecture 4 Chapter 2 2.3 Inertia and Equilibrium: Newton s First Law of Motion 2.4 Vector Addition Using Components 2.5 Newton s Third Law 1 Net Force The net force is the vector sum of all

### Steps to Solving Newtons Laws Problems.

Mathematical Analysis With Newtons Laws similar to projectiles (x y) isolation Steps to Solving Newtons Laws Problems. 1) FBD 2) Axis 3) Components 4) Fnet (x) (y) 5) Subs 1 Visual Samples F 4 1) F 3 F

### Two-Body System: Two Hanging Masses

Specific Outcome: i. I can apply Newton s laws of motion to solve, algebraically, linear motion problems in horizontal, vertical and inclined planes near the surface of Earth, ignoring air resistance.

### F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26

Physics 23 Exam 2 Spring 2010 Dr. Alward Page 1 1. A 250-N force is directed horizontally as shown to push a 29-kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force,

### Section 3 Newton s Laws of Motion

Section 3 Newton s Laws of Motion Key Concept Newton s laws of motion describe the relationship between forces and the motion of an object. What You Will Learn Newton s first law of motion states that

### 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

### 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

### 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 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 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7. February 13, 2013

PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7 February 13, 2013 0.1 A 2.00-kg object undergoes an acceleration given by a = (6.00î + 4.00ĵ)m/s 2 a) Find the resultatnt force acting on the object

### Q: Who established the law of universal gravitation? A: Newton. Q: What is a spring scale used for? A: To measure weight

Q: Who established the law of universal gravitation? A: Newton Q: What is a spring scale used for? A: To measure weight Q: What is the Law of Universal Gravitation? A: Everything in the universe has gravity.

### STATIC AND KINETIC FRICTION

STATIC AND KINETIC FRICTION LAB MECH 3.COMP From Physics with Computers, Vernier Software & Technology, 2000. INTRODUCTION If you try to slide a heavy box resting on the floor, you may find it difficult

### At the skate park on the ramp

At the skate park on the ramp 1 On the ramp When a cart rolls down a ramp, it begins at rest, but starts moving downward upon release covers more distance each second When a cart rolls up a ramp, it rises

### 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

### 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

### 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,

### 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

### Dynamics Pulleys, Ramps, and Friction

Name School Date Purpose To investigate the vector nature of forces. To practice the use free-body diagrams (FBDs). To learn to apply Newton s Second Law to systems of masses connected by pulleys. Equipment

### 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

### 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

### ACTIVITY 1: Gravitational Force and Acceleration

CHAPTER 3 ACTIVITY 1: Gravitational Force and Acceleration LEARNING TARGET: You will determine the relationship between mass, acceleration, and gravitational force. PURPOSE: So far in the course, you ve

### 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

### 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

### HW Set II page 1 of 9 PHYSICS 1401 (1) homework solutions

HW Set II page 1 of 9 4-50 When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 20 km (about the size of the San Francisco

### 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

### Physics 160 Biomechanics. Newton s Laws

Physics 160 Biomechanics Newton s Laws Questions to Think About Why does it take more force to cause an object to start sliding than it does to keep it sliding? Why is a ligament more likely to tear during

### 1) 0.33 m/s 2. 2) 2 m/s 2. 3) 6 m/s 2. 4) 18 m/s 2 1) 120 J 2) 40 J 3) 30 J 4) 12 J. 1) unchanged. 2) halved. 3) doubled.

Base your answers to questions 1 through 5 on the diagram below which represents a 3.0-kilogram mass being moved at a constant speed by a force of 6.0 Newtons. 4. If the surface were frictionless, the

### 356 CHAPTER 12 Bob Daemmrich

Standard 7.3.17: Investigate that an unbalanced force, acting on an object, changes its speed or path of motion or both, and know that if the force always acts toward the same center as the object moves,

### Chapter 6. Work and Energy

Chapter 6 Work and Energy The concept of forces acting on a mass (one object) is intimately related to the concept of ENERGY production or storage. A mass accelerated to a non-zero speed carries energy

### while the force of kinetic friction is fk = µ

19. REASONING AND SOLUION We know that µ s =2.0µ k for a crate in contact with a MAX cement floor. he maximum force of static friction is fs = µ sfn while the force of kinetic friction is fk = µ kfn. As

### Big Science Idea. Forces. Name. When you ride a bike, your foot pushes against the pedal. The push makes the wheels of the bike move.

Forces Worksheet 1 Name Forces When you ride a bike, your foot pushes against the pedal. The push makes the wheels of the bike move. When you drop something, it is pulled to the ground by gravity. A PUSH

### Experiment: Static and Kinetic Friction

PHY 201: General Physics I Lab page 1 of 6 OBJECTIVES Experiment: Static and Kinetic Friction Use a Force Sensor to measure the force of static friction. Determine the relationship between force of static

### 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

### Ch 7 Kinetic Energy and Work. Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43

Ch 7 Kinetic Energy and Work Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43 Technical definition of energy a scalar quantity that is associated with that state of one or more objects The state

### Forces. When an object is pushed or pulled, we say that a force is exerted on it.

Forces When an object is pushed or pulled, we say that a force is exerted on it. Forces can Cause an object to start moving Change the speed of a moving object Cause a moving object to stop moving Change

### Work, Energy & Momentum Homework Packet Worksheet 1: This is a lot of work!

Work, Energy & Momentum Homework Packet Worksheet 1: This is a lot of work! 1. A student holds her 1.5-kg psychology textbook out of a second floor classroom window until her arm is tired; then she releases

### Normal Force Example: Incline

Normal Force Example: Incline α The angle of the frictionless incline is α = 30. Mass slides down the incline, starting from rest. What is the speed of the mass after it slid 10 meters downhill? [use g