Physics 201 Homework 5

Save this PDF as:

Size: px
Start display at page:

Transcription

1 Physics 201 Homework 5 Feb 6, The (non-conservative) force propelling a 1500-kilogram car up a mountain joules road does joules of work on the car. The car starts from rest at sea level and has a speed of 27.0 m/s at an altitude of 200 meters above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are non-conservative forces. The propelling force has the effect of increasing the energy level of the car. This energy goes three places: kinetic energy (or speed), potential energy (or altitude), and some is lost to the nonconservative friction. We know the energy going in and we can calculate the amount that goes into kinetic and potential energies. What is left must have been lost to the non-conservative forces. In symbols: We know the kinetic energy gained: And the potential energy gained: W added W lost = E = KE + P E KE = 1 2 mv2 = 1 2 (1500)(27.0)2 = P E = mgh = (1500)(9.80)(200) = Plugging these and the energy added we have: ( ) W lost = = W lost = It s a technicality, but the question asks for the work done by the resistive forces. Since we have calculated the work lost, we need to flip the sign to emphasize that the energy level is decreasing as a consequence of the resistance. 2. A person pushes a 16.0-kilogram shopping cart at a constant velocity for a (a) 54.9 newtons distance of 22.0 meters. She pushes in a direction 29.0 below the horizontal. (b) 1060 joules (c) joules A 48.0-newton frictional force opposes the motion of the cart. (a) What is the (d) none magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force. (a) Since the cart is moving at constant velocity, the system is in equilibrium, so the net force is zero. In this problem there are four forces involved. The pushing force, the weight of the cart, the normal support force, and the kinetic friction. The motion is horizontal, so we should align the coordinate that way. The components of the net force are: F net,x = P x F F net,y = P y W + N Both of these components sum to zero. Since we are given the friction force (48.0 newtons), we know that the x-component of the pushing force is also 48.0 newtons. Since we know that the canonical angle of the push is 331, we have: P x = P cos 331 = 48.0 = P =

2 (b) The work done by any force is the product of the displacement and the component of the force parallel to the displacement. In this case, x = 22.0 meters and P x = 48.0 newtons. Thus, W = xp x = (22.0)(48.0) = 1056 (c) Similarly for the friction, but this force opposes the motion so its component is negative: W = xf x = (22.0)( 48.0) = 1056 (d) The gravitational force has no component in the x direction. Thus, W = xw x = (22.0)(0) = 0 3. A kilogram arrow is fired horizontally. The bowstring exerts an average 39 m/s force of 65 newtons on the arrow over a distance of 0.90 meters. With what speed does the arrow leave the bow? The work done by the bowstring is W = xf x = (0.90)(65) = 58.5 This work has the effect of increasing the energy of the arrow. In this case it increases the kinetic energy of the arrow. The initial kinetic energy is zero, so KE f = 58.5 Using the formula for kinetic energy, KE = 1 2 mv2, we can derive the final resulting speed: 1 2 (0.075)(v)2 = 58.5 = v = An extreme skier, starting from rest, coasts down a mountain slope that makes 10.9 m/s an angle of 25.0 with the horizontal. The coefficient of kinetic friction between her skis and the snow is She coasts down a distance of 10.4 meters before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.50 meters below the edge. How fast is she going just before she lands? There are two phases to the motion in this problem. The first coasting phase, is basically an inclined plane problem from Chapter 4. At the end of this phase the skier will have a certain velocity angled down 25 from the horizontal. The second phase is standard projectile problem which we could solve using the techniques from Chapter 3. However, this since we are only asked about the speed, we could use kinetic energy to answer the question. Either approach will work. But since this is a Chapter 6 problem, it seems appropriate to use energy. In the first phase, there are three forces at work: weight, support, and friction. Using coordinates aligned with the motion (assume the slope is to the right), the canonical angles associated with these forces are 295, 90, and 180, respectively. The magnitudes of these forces are W = mg, N, and F k = (0.200)(N), respectively. We don t know m, but we will assume it cancels out. We don t know N, but it must counter-balance the y-component of the weight. Thus, N + W y = 0. Therefore, N = W y = W sin θ = (9.80)(m) sin 295 = (8.8818)(m) Now there is motion in the x-direction. According to Newton s second law we have F net = ma. The net force is in the x-direction, but friction points in the negative direction, so F net,x = W x F k = (m)(9.80) cos 295 (0.200)(8.8818)(m) = (2.3653)(m) 2

3 So, the acceleration of the skier down the slope is m/s 2. Now we have a constant acceleration problem. The quantities are a = v 0 = 0 v =? x = 10.4 The equation to use is v 2 = v ax. Thus, (v) 2 = (0) 2 + (2)(2.3653)(10.4) = v = This is the endpoint of the first phase of motion, but it is also the beginning of the second phase. We decided to use energy considerations to work this phase. We know that she is 3.50 meters above her landing point. This means she has a certain amout of potential energy: And her initial kinetic energy is P E = mgh = (m)(9.80)(3.50) = (34.300)(m) KE = 1 2 mv2 = 1 2 (m)(7.0141)2 = (24.599)(m) Since no energy is lost, the potential energy is converted into additional kinetic energy at the end of the drop. So the final kinetic energy is KE = (34.300)(m) + (24.599)(m) = (58.899)(m) Using the formula for kinetic energy, KE = 1 2 mv2, we can determine her final speed: (58.899)(m) = 1 2 (m)(v)2 = v = Rocket man has a propulsion unit strapped to his back. He starts from rest 23 kj on the ground, fires the unit, and is propelled straight upward. At a height of 16 meters, his speed is 5.0 m/s. His mass, including the propulsion unit, has the approximately constant value of 136 kilograms. Find the work done by the force generated by the propulsion unit. The gravitational force is conservative, so we can say that W = E = KE + P E Now, initially Rocket man has no kinetic or potential energy because he is at rest and on the ground. But at the end point, he has both kinetic and potential energy. Thus, KE = 1 2 mv2 = 1 2 (136)(5.0)2 = 1700 and P E = mgh = (136)(9.80)(16) = Now, his initial mechanical energy is zero. But the final mechanical energy is E = KE + P E = = The work done by the propulsion unit is where this energy came from. 6. In 2.0 minutes, a ski lift raises four skiers at constant speed to a height of 3000 watts 140 meters. The average mass of each skier is 65 kilograms. What is the average power provided by the tension in the cable pulling the lift? 3

4 Since the work done by the lift has the effect of increasing the potential energy of the skiers, we know by the work-energy theorem that W = P E = mgh = (4 65)(9.80)(140) = Since this quantity of work is done over a 2.0 minute time period (or 120 seconds), by definition, the power involved is P = W ork t = = A small lead ball, attached to a 1.5 meter rope, is being whirled in a circle 42 meters that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 0.75 meters above the ground. The fall travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise? At the top of its trajectory, the kinetic energy is momentarily zero all the energy is in the gravitational potential energy mgh. From this we can calculate the height. Since we know the initial height, all we need to determine is the initial speed. The ball moves in uniform circular motion with a radius of 1.5 meters. Each revolution is 2πr = meters. The speed is three times this value per second: So, the energy of the ball is v = (3)(2π)(1.5) = KE = 1 2 mv2 = 1 2 (m)(28.274)2 = PE = mgh = (m)(9.8)(0.75) = 7.35 For a total of (407.07)(m) joules of energy. Therefore the maximum height must be PE = mgh = (m)(9.8)(h) = (407.07)(m) = h = A pendulum consists of a small object hanging from the ceiling at the end 43 of a string of negligible mass. The string has a length of 0.75 meters. With the string hanging vertically, the object is given an initial velocity of 2.0 m/s parallel to the ground and swings upward on a circular arc. Eventually, the object comes to a momentary halt at a point where the string makes an angle θ with its initial vertical orientation and then swings back downward. Find the angle θ. Initially the height of the pendulum is zero, so its potential energy is zero. The kinetic energy is KE = 1 2 mv2 = 1 2 (m)(2.0)2 = (2.0)(m) We can use this to determine the height the pendulum rises, since the energy will all be converted to potential: PE = mgh = (m)(9.8)(h) = (2.0)(m) = h = From this we should be able to calculate the angle in question using some trig. I will help to draw a little picture here: 4

5 0.75 θ Figure 1: Problem 6.46 From the picture we can see the triangle defined by the angle in which we are interested. We know the hypotenuse is 0.75 meters and the adjacent side is ( ) = These sides are connected via the cosine function. Thus: cos θ = = θ = A semitrailer is coasting downhill along a mountain highway when its brakes 27.0 m/s fail. The driver pulls onto a runaway-truck ramp that is inclined at an angle of 14.0 above the horizontal. The semitrailer coasts to a stop after traveling 154 meters along the ramp. What was the truck s initial speed? Neglect air resistance and friction We are asked about the initial state, so let s start with the end state instead. We know the semi comes to a stop, so there is no kinetic energy. It is at a certain height, which is potential energy. But how high? The height is actually the opposite side of the triangle formed by the ramp. The hypotenuse is 154 meters and the angle is Thus: sin 14.0 = h 154 = h= So, the potential energy is PE = mgh = (m)(9.8)(37.256) = (365.11)(m) This is how much kinetic energy it has when it hits the ramp. Thus, (365.11)(m) = 1 2 mv2 = v = A 1300-kilogram car is to accelerate from rest to a speed of 30.0 m/s in a 132 horsepower time of 12.0 seconds as it climbs a 15 hill. Assuming uniform acceleration, what minimum horsepower is needed to accelerate the car in this way? We could use the techniques from Chapter 4 to solve this, but using energy is much easier. The initial energy of the car is zero: it starts at rest (no kinetic) and is at the bottom of the hill (no potential). As the engine works on the car, it gains energy. At the top of the hill it has kinetic energy: KE = 1 2 mv2 = 1 2 (1300)(30.0)2 = Its potential energy is governed by it height. In order to figure this out we need to determine how far it travels. We can use x = 1 2 (v + v 0)(t) = 1 2 ( )(12.0) = 180 5

6 Since the hill is at a 15 incline, the total elevation change is So the car s final potential energy is h = x sin θ = (180) sin 15 = P E = mgh = (1300)(9.8)(46.587) = The work done by the engine is the combination of the kinetic and potential energies: W = Finally, power is defined as work done per second, so: P = W t = = Since there are 746 watts in one horsepower, we have P = 98210/746 = hp 6

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

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

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

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

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

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

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

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

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

Chapter 6. Work and Energy

Chapter 6 Work and Energy ENERGY IS THE ABILITY TO DO WORK = TO APPLY A FORCE OVER A DISTANCE= Example: push over a distance, pull over a distance. Mechanical energy comes into 2 forms: Kinetic energy

= 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

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

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

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

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

Work-Energy Bar Charts

Name: Work-Energy Bar Charts Read from Lesson 2 of the Work, Energy and Power chapter at The Physics Classroom: http://www.physicsclassroom.com/class/energy/u5l2c.html MOP Connection: Work and Energy:

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

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

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

Conservative vs. Non-conservative forces Gravitational Potential Energy. Work done by non-conservative forces and changes in mechanical energy

Next topic Conservative vs. Non-conservative forces Gravitational Potential Energy Mechanical Energy Conservation of Mechanical energy Work done by non-conservative forces and changes in mechanical energy

KE =? v o. Page 1 of 12

Page 1 of 12 CTEnergy-1. A mass m is at the end of light (massless) rod of length R, the other end of which has a frictionless pivot so the rod can swing in a vertical plane. The rod is initially horizontal

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

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...

10.1 Quantitative. Answer: A Var: 50+

Chapter 10 Energy and Work 10.1 Quantitative 1) A child does 350 J of work while pulling a box from the ground up to his tree house with a rope. The tree house is 4.8 m above the ground. What is the mass

F13--HPhys--Q5 Practice

Name: Class: Date: ID: A F13--HPhys--Q5 Practice Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A vector is a quantity that has a. time and direction.

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

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

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

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

Conservation of Energy Workshop. Academic Resource Center

Conservation of Energy Workshop Academic Resource Center Presentation Outline Understanding Concepts Kinetic Energy Gravitational Potential Energy Elastic Potential Energy Example Conceptual Situations

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

Problem Set #8 Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.01L: Physics I November 7, 2015 Prof. Alan Guth Problem Set #8 Solutions Due by 11:00 am on Friday, November 6 in the bins at the intersection

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

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

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

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.

Worksheet #1 Free Body or Force diagrams

Worksheet #1 Free Body or Force diagrams Drawing Free-Body Diagrams Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.

Name Period WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS. 1. Stored energy or energy due to position is known as energy.

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

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

AP Physics C Fall Final Web Review

Name: Class: _ Date: _ AP Physics C Fall Final Web Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. On a position versus time graph, the slope of

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

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

Work and Energy continued

Chapter 6 Work and Energy continued Requested Seat reassignments (Sec. 1) Gram J14 Weber C22 Hardecki B5 Pilallis B18 Murray B19 White B20 Ogden C1 Phan C2 Vites C3 Mccrate C4 Demonstrations Swinging mass,

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

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

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

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,

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

P211 Midterm 2 Spring 2004 Form D

1. An archer pulls his bow string back 0.4 m by exerting a force that increases uniformly from zero to 230 N. The equivalent spring constant of the bow is: A. 115 N/m B. 575 N/m C. 1150 N/m D. 287.5 N/m

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

Vectors; 2-D Motion. Part I. Multiple Choice. 1. v

This test covers vectors using both polar coordinates and i-j notation, radial and tangential acceleration, and two-dimensional motion including projectiles. Part I. Multiple Choice 1. v h x In a lab experiment,

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

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

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

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

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Vector A has length 4 units and directed to the north. Vector B has length 9 units and is directed

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.

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

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

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

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

Tennessee State University

Tennessee State University Dept. of Physics & Mathematics PHYS 2010 CF SU 2009 Name 30% Time is 2 hours. Cheating will give you an F-grade. Other instructions will be given in the Hall. MULTIPLE CHOICE.

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

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

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

Lesson 3 - Understanding Energy (with a Pendulum)

Lesson 3 - Understanding Energy (with a Pendulum) Introduction This lesson is meant to introduce energy and conservation of energy and is a continuation of the fundamentals of roller coaster engineering.

Physics 1A Lecture 10C

Physics 1A Lecture 10C "If you neglect to recharge a battery, it dies. And if you run full speed ahead without stopping for water, you lose momentum to finish the race. --Oprah Winfrey Static Equilibrium

Review Assessment: Lec 02 Quiz

COURSES > PHYSICS GUEST SITE > CONTROL PANEL > 1ST SEM. QUIZZES > REVIEW ASSESSMENT: LEC 02 QUIZ Review Assessment: Lec 02 Quiz Name: Status : Score: Instructions: Lec 02 Quiz Completed 20 out of 100 points

Gravitational Potential Energy

Gravitational Potential Energy Consider a ball falling from a height of y 0 =h to the floor at height y=0. A net force of gravity has been acting on the ball as it drops. So the total work done on the

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 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

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

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

physics 111N work & energy

physics 111N work & energy conservation of energy entirely gravitational potential energy kinetic energy turning into gravitational potential energy gravitational potential energy turning into kinetic

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

The Ballistic Pendulum

1 The Ballistic Pendulum Introduction: By this time, you have probably become familiar with the concepts of work, energy, and potential energy, in the lecture part of the course. In this lab, we will be

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

Energy - Key Vocabulary

Energy - Key Vocabulary Term Potential Energy Kinetic Energy Joules Gravity Definition The energy an object possesses due to its position. PE = mgh The energy an object possesses when it is in motion.

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

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

BHS Freshman Physics Review. Chapter 2 Linear Motion Physics is the oldest science (astronomy) and the foundation for every other science.

BHS Freshman Physics Review Chapter 2 Linear Motion Physics is the oldest science (astronomy) and the foundation for every other science. Galileo (1564-1642): 1 st true scientist and 1 st person to use

Exercises on Work, Energy, and Momentum. A B = 20(10)cos98 A B 28

Exercises on Work, Energy, and Momentum Exercise 1.1 Consider the following two vectors: A : magnitude 20, direction 37 North of East B : magnitude 10, direction 45 North of West Find the scalar product

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 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?!

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

Instructor Now pick your pencils up and get this important equation in your notes.

Physics 605 Mechanical Energy (Read objectives on screen.) No, I haven t been playing with this toy the whole time you ve been gone, but it is kind of hypnotizing, isn t it? So where were we? Oh yes, we

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

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

Problem Set 5 Work and Kinetic Energy Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department o Physics Physics 8.1 Fall 1 Problem Set 5 Work and Kinetic Energy Solutions Problem 1: Work Done by Forces a) Two people push in opposite directions on

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

AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s

AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s Answer the multiple choice questions (2 Points Each) on this sheet with capital

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,

1. Mass, Force and Gravity

STE Physics Intro Name 1. Mass, Force and Gravity Before attempting to understand force, we need to look at mass and acceleration. a) What does mass measure? The quantity of matter(atoms) b) What is the

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

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

Work and Conservation of Energy

Work and Conservation of Energy Topics Covered: 1. The definition of work in physics. 2. The concept of potential energy 3. The concept of kinetic energy 4. Conservation of Energy General Remarks: Two

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