# Chapter 7: Kinetic Energy and Work

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Chapter 7: Kinetic Energy and Work In this chapter, we will study the important concepts of kinetic energy and the closely related concept of work and power. A- Kinetic Energy Kinetic energy is a physical quantity, which is associated with the moving objects and defined as: K = ½ mv 2 If the body is stationary (v=0), its kinetic energy is zero. The SI unit of kinetic energy is kg.m 2 /s 2 or Joule (J), where 1 J=1 kg.m 2 /s 2. Kinetic energy is a scalar quantity. B- Work The work is defined as the ability to perform a force along a certain displacement. There are different types of work as follows: 1- Work done by a constant force The work done by the constant force F is given by the scalar product of the force F and the displacement d. W F = F.d = Fd cosθ where θ is the angle between the force and displacement. The above equation means that the work is the product of the displacement magnitude by component of the force parallel to the displacement. Therefore, work is a scalar quantity (only magnitude, no direction) and can be positive, negative, or zero. The SI unit of work is (N.m) or joule (J) where 1 N.m = 1 J. Special cases and remarks: If the angle between the force and displacement is zero (parallel), the work is W F = F d (maximum work)

2 If the angle between the force and displacement is 90, the work is W F = 0 If the angle between the force and displacement is 180 (anti-parallel), the work is W F = -F d (minimum work) Positive work: the work done by a system is positive. Negative work: the work done on a system is negative. 2- Work done by the gravity force The work performed by the gravity force (F=mg) is given by W g = F. d = mg. d=mgd cosθ Special cases and remarks: If the object is moving down (the angle between the force and displacement is zero), the work is W g = mg d (maximum work) If the object is moving up (the angle between the force and displacement is 180), the work is W g = -mg d (minimum work) The position where W g = 0 is arbitrary. W g is a function of position only. (It depends only on the relative positions of the earth and the block.) The work W g depends only on the initial and final heights, NOT on the path. 3- Work done by the normal force The work performed by the normal force (reaction force) is always zero, because the angle between the normal force and displacement is 90. W N = 0 Rule: Forces perpendicular to the motion do no work!

3 4- Work done by the friction force The work performed by the frictional force is always negative (at least in this stage), because the angle between the frictional force and displacement is 180. W f = -f k d 5- Work done by the spring force The work performed by a spring force is given by the following expression: W s = ½ kx 2 6- Net Work (the total work) The total work is the summation of all types of work done on an object. W T = W F + W g + W f The total work done on an object determines its motion. If the total work is zero, the object moves at constant speed or is at rest. The object accelerates if the total work has a value other than zero. C- Kinetic Energy and Work The work can be defined as the change in the kinetic energy. If a body changes its velocity from v 1 to v 2, the work done is defined as: 2 W =ΔK =K 2 K 1 = ½ m (v 2 v 2 1 ) This is known as the work-energy theorem: The change of kinetic energy of a body is equal to the work of all the forces acting upon it. D- Mechanical Power Power is defined as the rate of doing work or rate of using energy. It is defined as P = W/t

4 It is also defined as P = F.v The SI unit of power is J/s, N.m/s, or Watt (W), where 1 W=1 J/s=1 N.m/s. Occasionally it is expressed by the mechanical horsepower (hp) where 1 hp = 746 W. However power is a scalar quantity.

5 Examples: 1. A 5 kg object moving at a speed of 6 m/s. Calculate its kinetic energy. The kinetic energy is defined as K = ½ mv 2 = ½ = 90 J 2. A car of mass 1000 kg accelerates at 2 m/s 2 for 10 s from an initial speed of 5 m/s. Determine the work done by the car. The work is defined as the change in the kinetic energy. Therefore W =ΔK =K 2 K 1 = ½ m (v 2 2 v 2 1 ) We know the initial speed (v 1 = 5 m/s) but we have to find out the final speed (v 2 ) using the equation of motion. We know that v 2 = v 1 + at Therefore v 2 = = 25 m/s Hence the work done by the car is W = ½ 1000 ( ) = J 3. A force, F = 2i -3j +k (N) is applied on a box. If the displacement of the box due to the force is d = i +3k (m), find the work done. The work done by the force is the scalar product of the force and its

6 displacement, such as W F = F. d = (2x1) +(-3x0) + (1x3) = 5 J 4. A box is pulled across the floor by a horizontal force of magnitude 80 N. How much work does the force exert to pull the object 5 m? The work done by the force is W F = F d cosθ But the force and displacement are parallel; the angle between them is zero. Therefore W F = F d = 80 x 5= 400 J 5. A box is dragged across the floor by a force of magnitude 50 N directed 60 o above the horizontal. Calculate the work done by the force to pull the object 6 m? The work done by the force is W F = F d cosθ = 50 x 6 x cos(60)=150 J 6. A horizontal force F is applied to move a 5 kg carton across the floor. If the acceleration of the carton is measured to be 2 m/s 2, how much work does F do in moving the carton 7 m? While the force and displacement are parallel, the work done is

7 W F = F d The force can be determined from Newton s second law as F =ma = 5x2 = 10 N Therefore the work is W F = F d = 10x 7 = 70 J 7. A horizontal force F is applied to move a 6 kg carton across the floor. If the carton starts from rest and its speed after 3 sec is 6 m/s, how much work does F do in moving the carton 4 m? While the force and displacement are parallel, the work done is W F = F d However the force, F, can be determined from Newton s second law as F =ma The acceleration cab be found using the motion equation v=v 0 +at Since the carton starts its motion from rest (v 0 =0), the acceleration is a=v/t =6/3 =2 m/s 2 Therefore the force is F =ma = 6x2 = 12 N Hence the work done by the force F is W F = F d = 12 x 4 = 48 J

8 8. A 5.0-kg box is raised a distance of 2.5 m from rest by a vertical applied force of 90 N. Find (a) the work done on the box by the applied force, and (b) the work done on the box by gravity. (a) Using the definition of work in which the force and displacement are in the same direction (parallel), we get W F = F d =90 x 2.5 = 225 J (b) Using the definition of work done by gravity for moving up object (the force and displacement are anti-parallel), we get W g = -mg d = - 5 x 9.8 x 2.5= J 9. A box rests on a horizontal, frictionless surface. Ali pushes on the box with a force of 18 N to the right and Dima pushes on the box with a force of 12 N to the left. The box moves 4.0 m to the right. Find the work done by (a) Ali, (b) Dima, and (c) the net force. (a) While the force applied by Ali is in the same direction of displacement (their angle is zero), the work done by Ali is W Ali = F Ali d =18 x 4 = 72 J (b) While the force applied by Dima is in the opposite direction of displacement (their angle is 180), the work done by Dima is W Dima = - F Dima d =- 12 x 4 = - 48 J (c) The net force is F net = F Ali + F Dima = = 6 N

9 We can also calculate the net force using the work concept where the net work is W net = F net d But the total (net) work is W net = W Ali + W Dima = = 24 J Therefore we get F net = W net / d =24/4 = 6 N 10. A 40 kg box is pulled 30 m on a horizontal floor by applying a force (F p ) of magnitude 100 N directed by an angle of 60º above the horizontal. If the floor exerts a friction force (f r ) of magnitude 20 N, calculate the work done by each one of these forces. Calculate the work done by the weight and the normal force. Calculate also the total work done on the box. (a) The work done by the force is W F = F p d cosθ=100 x 30 x cos60= 1500 J (b) The work done by the friction force is W f = f r d cosθ=20 x 30 x cos180= -600 J Note that the angle between f r and displacement is 180º because they point

10 in opposite directions. (c) The work done by the gravity force is W g = 0 J Note that the box is moving along x-axis and there is no displacement along the vertical direction, therefore its work is zero. (c) The work done by the normal force is W N = 0 J The normal force does no work because it is normal (perpendicular) to the displacement. (d) The total work done on the box is W T = W F + W f + W g + W N = =900 J 11. A stationary block of 10 kg is pulled up along a smooth incline of length 10 m and height 5 m by applying an external force of 50 N parallel to the incline. At the end of incline, find (i) work done by the external force, (ii) work done by gravity, and (iii) speed of the block. (i) The work done by the force is W F = F d cosθ= cos0= 500 J

11 Note that both the force and displacement are in the same direction. (ii) The work done by the gravity force is W g = F g d cosθ= F g d cosθ= cos180 = - 98 J (iii) The total work done is W T = W F + W g = =402 J Therefore the change in the kinetic energy is 2 W =ΔK =K 2 K 1 = ½ m v 2 Where the initial speed is zero. Hence the final speed is v 22 = 2ΔK/m = 2 402/10 = 80.4 (m/s) 2 v 2 = 9 m/s 12. Maha slides a 1-kg toy for 3 m along the floor to her sister Reem. The toy is moving at 4 m/s when Maha lets go, and at 2 m/s when Reem catches it. (a) Find the work done by friction and (b) calculate the force of friction. (a) We know that the total work done on a body equals the change in its kinetic energy. Since we have only a friction force, then the total work is due to it. Hence, we have W f =ΔK =K 2 K 1 = ½ m (v 22 v 12 ) = ½ 1 ( ) = -6 J (b) It is well known that the work done by the friction force is given by W f = -f k d Therefore the friction force is f k = -W f /d = -(-6)/3 =2 N

12 13. An 8.0 kg block initially at rest is pulled to the right for 3.0 m with a force of 12 N over a surface. Determine its final speed if: (a) the surface has no friction and (b) the surface has a coefficient of kinetic friction of 0.15 (a) In the case of frictionless surface (no friction force), the total work of the system is only the work done by the external force. We have W T = W F =F d =12 3=36 J Therefore the change in kinetic energy is equivalent to the total work. Hence the final speed of the block is W T =ΔK =K 2 K 1 = ½ m (v 22 v 12 ) But the block starts from rest; therefore its initial speed is zero. The final speed is then 2 W T = ½ m v 2 Or v 22 = 2 W T /m = 2 36/8.0 =9 (m/s) 2 Or v 2 = 3 m/s (b) In the case of frictional surface (there is friction force), the total work of the system is the work done by the external force and friction force. We have W T = W F + W f =F d f k d The friction force is defined (from the previous chapter) as: f k =µ k N =µ k mg = = N

13 The total work, then, is W T =F d f k d That gives W T = =0.72 J The final speed can be determined using the work-energy theorem as 2 W T = ½ m v 2 Or v 2 2 = 2 W T /m = /8.0 =0.18 (m/s) 2 Or v 2 = 0.42 m/s 14. A block of mass 1.6 kg resting on a frictionless surface is attached to a horizontal spring with a spring constant k= N/m. The spring is compressed to 2.0 cm and released from rest. Find the velocity of the block when released. We know that the work done by a spring is given by W s = ½ k x 2 = ½ (0.02) 2 = 0.2 J The total work is (there is only one kind of work due to the spring) W T = W s = 0.2 J Therefore the final speed is W T = ½ m v 2 That will give us the following value v= 0.5 m/s

14 15. A 5 kg ball is freely dropped from a height of 20 m above the ground. Calculate its speed when it hits the ground. We know that the work done by the gravity is given by W g = mgd= = 980 J The total work is (there is only gravity work) W T = W g = 980 J Using the work-energy theorem, we get W T = ½ m v 2 Or v= (2W T /m) ½ = (2 980/5) ½ = 19.8 m/s 16. Zain does 240 J of work in pushing a box. If he does the work in 4 s, what is his power output? The output power is defined as P = W/t = 240/4 = 60 W 17. Ahmad pushes a box along a smooth floor using a force of 210 N and a power output of 350 W. How long does it take Ahmad to push the box 20 m? The output power is defined as P = W/t

15 Therefore the time taken for this operation is t = W/P But the work is W = F d = = 4200 J Hence the time is t = W/P = 4200 / 350 = 12 s 17. A box is lifted at a constant speed of 7.0 m/s by a machine. If the power of the machine is W, find the force applied. The output power is defined as P = F.v But the force and velocity are parallel (both upwards), the force is F = P/ v = 21000/7.0 = 3000 N

### 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

### WORK DONE BY A CONSTANT FORCE

WORK DONE BY A CONSTANT FORCE The definition of work, W, when a constant force (F) is in the direction of displacement (d) is W = Fd SI unit is the Newton-meter (Nm) = Joule, J If you exert a force of

### Springs. Spring can be used to apply forces. Springs can store energy. These can be done by either compression, stretching, or torsion.

Work-Energy Part 2 Springs Spring can be used to apply forces Springs can store energy These can be done by either compression, stretching, or torsion. Springs Ideal, or linear springs follow a rule called:

### Chapter 8: Conservation of Energy

Chapter 8: Conservation of Energy This chapter actually completes the argument established in the previous chapter and outlines the standing concepts of energy and conservative rules of total energy. I

### PHYS101 The Laws of Motion Spring 2014

The Laws of Motion 1. An object of mass m 1 = 55.00 kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging object of mass m 2

### 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

### 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

### A Review of Vector Addition

Motion and Forces in Two Dimensions Sec. 7.1 Forces in Two Dimensions 1. A Review of Vector Addition. Forces on an Inclined Plane 3. How to find an Equilibrant Vector 4. Projectile Motion Objectives Determine

### 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

### 103 PHYS - CH7 - Part2

Work due to friction If friction is involved in moving objects, work has to be done against the kinetic frictional force. This work is: W = f d = f d cos180 = f d o f k k k Physics 1 Example f n r θ F

### 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

### 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

### Newton's laws of motion

Newton's laws of motion Forces Forces as vectors Resolving vectors Explaining motion - Aristotle vs Newton Newton s first law Newton s second law Weight Calculating acceleration Newton s third law Moving

### Chapter 7 WORK, ENERGY, AND Power Work Done by a Constant Force Kinetic Energy and the Work-Energy Theorem Work Done by a Variable Force Power

Chapter 7 WORK, ENERGY, AND Power Work Done by a Constant Force Kinetic Energy and the Work-Energy Theorem Work Done by a Variable Force Power Examples of work. (a) The work done by the force F on this

### Example (1): Motion of a block on a frictionless incline plane

Firm knowledge of vector analysis and kinematics is essential to describe the dynamics of physical systems chosen for analysis through ewton s second law. Following problem solving strategy will allow

### 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

### Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel

Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel Name: Lab Day: 1. A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the

### 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

### Work, Power, Energy Multiple Choice. PSI Physics. Multiple Choice Questions

Work, Power, Energy Multiple Choice PSI Physics Name Multiple Choice Questions 1. A block of mass m is pulled over a distance d by an applied force F which is directed in parallel to the displacement.

### 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

### Lecture 9. Friction in a viscous medium Drag Force Quantified

Lecture 9 Goals Describe Friction in Air (Ch. 6) Differentiate between Newton s 1 st, 2 nd and 3 rd Laws Use Newton s 3 rd Law in problem solving Assignment: HW4, (Chap. 6 & 7, due 10/5) 1 st Exam Thurs.,

### 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

### AP1 WEP. Answer: E. The final velocities of the balls are given by v = 2gh.

1. Bowling Ball A is dropped from a point halfway up a cliff. A second identical bowling ball, B, is dropped simultaneously from the top of the cliff. Comparing the bowling balls at the instant they reach

### Work Energy & Power. September 2000 Number 05. 1. Work If a force acts on a body and causes it to move, then the force is doing work.

PhysicsFactsheet September 2000 Number 05 Work Energy & Power 1. Work If a force acts on a body and causes it to move, then the force is doing work. W = Fs W = work done (J) F = force applied (N) s = distance

### 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

### 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

### 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

### Work, Kinetic Energy and Potential Energy

Chapter 6 Work, Kinetic Energy and Potential Energy 6.1 The Important Stuff 6.1.1 Kinetic Energy For an object with mass m and speed v, the kinetic energy is defined as K = 1 2 mv2 (6.1) Kinetic energy

### Ground Rules. PC1221 Fundamentals of Physics I. Force. Zero Net Force. Lectures 9 and 10 The Laws of Motion. Dr Tay Seng Chuan

PC1221 Fundamentals of Physics I Lectures 9 and 10 he Laws of Motion Dr ay Seng Chuan 1 Ground Rules Switch off your handphone and pager Switch off your laptop computer and keep it No talking while lecture

### 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

### PHYSICS 149: Lecture 15

PHYSICS 149: Lecture 15 Chapter 6: Conservation of Energy 6.3 Kinetic Energy 6.4 Gravitational Potential Energy Lecture 15 Purdue University, Physics 149 1 ILQ 1 Mimas orbits Saturn at a distance D. Enceladus

### Physics 101 Prof. Ekey. Chapter 5 Force and motion (Newton, vectors and causing commotion)

Physics 101 Prof. Ekey Chapter 5 Force and motion (Newton, vectors and causing commotion) Goal of chapter 5 is to establish a connection between force and motion This should feel like chapter 1 Questions

### Physics 201 Fall 2009 Exam 2 October 27, 2009

Physics 201 Fall 2009 Exam 2 October 27, 2009 Section #: TA: 1. A mass m is traveling at an initial speed v 0 = 25.0 m/s. It is brought to rest in a distance of 62.5 m by a force of 15.0 N. The mass is

### 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

### 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

### 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

### Summary Notes. to avoid confusion it is better to write this formula in words. time

National 4/5 Physics Dynamics and Space Summary Notes The coloured boxes contain National 5 material. Section 1 Mechanics Average Speed Average speed is the distance travelled per unit time. distance (m)

### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) The following four forces act on a 4.00 kg object: 1) F 1 = 300 N east F 2 = 700 N north

### 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

### 56 Chapter 5: FORCE AND MOTION I

Chapter 5: FORCE AND MOTION I 1 An example of an inertial reference frame is: A any reference frame that is not accelerating B a frame attached to a particle on which there are no forces C any reference

### 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

### What is a force? Identifying forces. What is the connection between force and motion? How are forces related when two objects interact?

Chapter 4: Forces What is a force? Identifying forces. What is the connection between force and motion? How are forces related when two objects interact? Application different forces (field forces, contact

### 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 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

### 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

### 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

### 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

### PHYSICS 111 HOMEWORK#6 SOLUTION. February 22, 2013

PHYSICS 111 HOMEWORK#6 SOLUTION February 22, 2013 0.1 A block of mass m = 3.20 kg is pushed a distance d = 4.60 m along a frictionless, horizontal table by a constant applied force of magnitude F = 16.0

### Physics 2101, First Exam, Fall 2007

Physics 2101, First Exam, Fall 2007 September 4, 2007 Please turn OFF your cell phone and MP3 player! Write down your name and section number in the scantron form. Make sure to mark your answers in 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:

### 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,

### Version 001 Quest 3 Forces tubman (20131) 1

Version 001 Quest 3 Forces tubman (20131) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. l B Conceptual

### Assignment Work (Physics) Class :Xi Chapter :04: Motion In PLANE

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Assignment Work (Physics) Class :Xi Chapter :04: Motion In PLANE State law of parallelogram of vector addition and derive expression for resultant of two vectors

### 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

### f max s = µ s N (5.1)

Chapter 5 Forces and Motion II 5.1 The Important Stuff 5.1.1 Friction Forces Forces which are known collectively as friction forces are all around us in daily life. In elementary physics we discuss the

### 3 Work, Power and Energy

3 Work, Power and Energy At the end of this section you should be able to: a. describe potential energy as energy due to position and derive potential energy as mgh b. describe kinetic energy as energy

### Work and Direction. Work and Direction. Work and Direction. Work and Direction

Calculate the net gravitational force on the shaded ball. Be sure to include the magnitude and direction. Each ball has a mass of 20,000 kg. (0.79N, 22.5 o N of E) Chapter Six Work = Force X distance W

### 8. Potential Energy and Conservation of Energy Potential Energy: When an object has potential to have work done on it, it is said to have potential

8. Potential Energy and Conservation of Energy Potential Energy: When an object has potential to have work done on it, it is said to have potential energy, e.g. a ball in your hand has more potential energy

### 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

### 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,

### 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

### 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

### 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 8: Potential Energy and Conservation of Energy. Work and kinetic energy are energies of motion.

Chapter 8: Potential Energy and Conservation of Energy Work and kinetic energy are energies of motion. Consider a vertical spring oscillating with mass m attached to one end. At the extreme ends of travel

### 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

### Motion Lesson 1: Review of Basic Motion

Motion in one and two dimensions: Lesson 1 Semi-notes Motion Lesson 1: Review of Basic Motion Note. For these semi notes we will use the bold italics convention to represent vectors. Complete the following

### Concept Review. Physics 1

Concept Review Physics 1 Speed and Velocity Speed is a measure of how much distance is covered divided by the time it takes. Sometimes it is referred to as the rate of motion. Common units for speed or

### 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

### Forces. -using a consistent system of units, such as the metric system, we can define force as:

Forces Force: -physical property which causes masses to accelerate (change of speed or direction) -a push or pull -vector possessing both a magnitude and a direction and adds according to the Parallelogram

### 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

### 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

### 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,

### Recap. A force is the product of an object s mass and acceleration. Forces are the reason why objects change their velocity. Newton s second law:

Recap A force is the product of an object s mass and acceleration. Forces are the reason why objects change their velocity. Newton s second law: Unit: 1 N = 1 kg m/s 2 Forces are vector quantities, since

### SOLID MECHANICS DYNAMICS TUTORIAL INERTIA FORCES IN MECHANISMS

SOLID MECHANICS DYNAMICS TUTORIAL INERTIA FORCES IN MECHANISMS This work covers elements of the syllabus for the Engineering Council Exam D225 Dynamics of Mechanical Systems C103 Engineering Science. This

### = 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

### Chapter 7: Momentum and Impulse

Chapter 7: Momentum and Impulse 1. When a baseball bat hits the ball, the impulse delivered to the ball is increased by A. follow through on the swing. B. rapidly stopping the bat after impact. C. letting

Section Review Answers Chapter 12 Section 1 1. Answers may vary. Students should say in their own words that an object at rest remains at rest and an object in motion maintains its velocity unless it experiences

### 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

### 1 of 9 10/27/2009 7:46 PM

1 of 9 10/27/2009 7:46 PM Chapter 11 Homework Due: 9:00am on Tuesday, October 27, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy [Return to Standard Assignment View]

### 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

### 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 Exam 1 Review - Chapter 1,2

Physics 1401 - Exam 1 Review - Chapter 1,2 13. Which of the following is NOT one of the fundamental units in the SI system? A) newton B) meter C) kilogram D) second E) All of the above are fundamental

### 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

### PHYSICS MIDTERM REVIEW

1. The acceleration due to gravity on the surface of planet X is 19.6 m/s 2. If an object on the surface of this planet weighs 980. newtons, the mass of the object is 50.0 kg 490. N 100. kg 908 N 2. If

### WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS

WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS 1. Stored energy or energy due to position is known as Potential energy. 2. The formula for calculating potential energy is mgh. 3. The three factors that

### General Physical Science

General Physical Science Chapter 4 Work and Energy Work The work done by a constant force F acting upon an object is the product of the magnitude of the force (or component of the force) and the parallel

### 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

### 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.

### F = ma. F = mg. Forces. Forces. Free Body Diagrams. Find the unknown forces!! Ex. 1 Ex N. Newton s First Law. Newton s Second Law

Forces Free Body Diagrams Push or pull on an object Causes acceleration Measured in Newtons N = Kg m s Shows all forces as vectors acting on an object Vectors always point away from object Used to help

### Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis

* By request, but I m not vouching for these since I didn t write them Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis There are extra office hours today & tomorrow Lots of practice exams

### Explaining Motion:Forces

Explaining Motion:Forces Chapter Overview (Fall 2002) A. Newton s Laws of Motion B. Free Body Diagrams C. Analyzing the Forces and Resulting Motion D. Fundamental Forces E. Macroscopic Forces F. Application

### 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

### 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

### Work, Energy and Power

Work, Energy and Power In this section of the Transport unit, we will look at the energy changes that take place when a force acts upon an object. Energy can t be created or destroyed, it can only be changed

### Exam 2 Review Questions PHY Exam 2

Exam 2 Review Questions PHY 2425 - Exam 2 Section: 4 1 Topic: Newton's First Law: The Law of Inertia Type: Conceptual 1 According to Newton's law of inertia, A) objects moving with an initial speed relative

### SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Exam Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s.

### 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