Problem Set #8 Solutions


 Juliana Fields
 2 years ago
 Views:
Transcription
1 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 of Buildings 8 (3rd floor) and 16 (4th floor). Please put your name and recitation number clearly on the top page of your problem set. SOURCES: University Physics, Volume 1, 13th Edition, by Hugh D. Young and Roger A. Freedman (AddisonWesley, 2012). This is the required textbook for the course. Essentials of Introductory Classical Mechanics, 6th Edition, by Wit Busza, Susan Cartwright, and Alan H. Guth, available to the MIT community at L/studyguide/index.shtml. Also known as the BCG Study Guide. READING: Young and Freedman Chapter 7, Sections 4 5 (POTENTIAL ENERGY AND ENERGY CONSERVATION: Force and Potential Energy; Energy Diagrams) and Chapter 8, Sections 1 2 (MOMENTUM, IMPULSE, AND COLLISIONS: Momentum and Impulse; Conservation of Momentum). OPTIONAL ADDITIONAL READING: Busza, Cartwright, and Guth (BCG), Chapter 5, pp ( Optional means that you should read this only if you find it useful.) NOTE: Your written solutions must include a brief commentary in addition to any equations or graphs used to arrive at your answer. For example, this commentary should explain your basic strategy for solving the problem and also highlight the concept before applying an equation. We also want you to first solve a problem algebraically before you put in the particular numbers relevant to the problem. We ll say that again: solve the answer in terms of variables before you start sticking in numbers! This makes it easier for you when you check your work and also for the grader to follow your logic. 81: Brake Failure Young and Freedman (13th edition) problem 7.66, page 238. (same as Young and Freedman (12th edition) problem 7.66, page 244.) Fig. P7.66: A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle α (Fig. P7.66). Initially the truck is moving downhill at
2 8.01L Problem Set #8 Solutions, p. 2 speed v 0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle β. The truck ramp has a soft sand surface for which the coefficient of rolling friction is µ r. What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods. Let the distance on the truck ramp be d. Also, let s define the point A as the starting point at which the truck is a distance L from the truck ramp and its velocity is v 0, and B the point at which the truck stops. We use energy transformation law on the truck+earth system: Now: E = W non cons (1) E B E A = f k r (2) E B E A = µ k Nd where N = mg cos β (3) E A = K A + U A = 1 2 mv2 0 + mgl sin α (4) E B = K B + U B = mgd sin β (5) since the velocity at point B is zero. Using equation 3 So, from equations 4 and 5: E B E A = µ k mgd cos β (6) mgd sin β mgl sin α 1 2 mv2 0 = µ k mgd cos β (7) gd(sin β + µ cos β) = gl sin α v2 0 (8) Note: as expected, d = L, if d = L sin α v2 0/g sin β + µ cos β α = β, µ = 0 v 0 = 0. (9) (10)
3 8.01L Problem Set #8 Solutions, p : Loop A small block with mass m slides in a vertical circle of radius R on the inside of a circular vertical track. There is no friction between the track and the block. At the bottom of the block s path, the normal force the track exerts on the block has magnitude N b. What is the magnitude of the normal force (call it N t ) that the track exerts on the block when it is at the top of its path? A block is on the inside of a circular track, moving at speed v. Let the speed at the bottom be v b and that of the top be v t We want expressions for the normal force at top & bottom. Use Newton s 2nd law. Top: Bottom: Fy = mg N t = mv2 t R N t = mv2 t R (11) mg (12) Fy = N b mg = mv2 b R N b = mg + mv2 b (14) R Now we need to solve for the velocities. Since we are given N b and m, we can use equation 14 to solve for v b : (13) v 2 b = R m (N b mg) (15) Now to find the solutions to the normal force at the top, we need to find v t. Using conservation of energy between the top and the bottom: KE b + U b = KE t + U t (16) 1 2 mv2 b + 0 = 1 2 mv2 t + 2mgR (17) vt 2 = vb 2 4gR (18)
4 8.01L Problem Set #8 Solutions, p. 4 Now finally we find N t, using equation 14 one more time: N t = mv2 b R 5 mg = N b 6mg. (19) 83: Pendulum Young and Freedman (13th edition) problem 7.82, page 239. (same as Young and Freedman (12th edition) problem 7.14, page 240.) Note: Young and Freedman stated this problem with numerical values m = 0.12 kg, l = 0.80 m, and θ = 45. You, however, are being asked to solve the problem symbolically. A small rock with mass m is fastened to a massless string with length l to form a pendulum. The pendulum is swinging so as to make a maximum angle of θ with the vertical. Air resistance is negligible. (a) What is the speed of the rock when the string passes through the vertical position? Choose y = 0 at the lowest point of the swing. We want to use energy conservation. The tension T is perpendicular to the direction of motion, thus does no work; air resistance is negligible; only gravity does work. E 1 = E 2, K 1 + U 1 = K 2 + U 2 (20) Since velocity is 0 at the top of the swing, and gravitational potential energy is so U 1 = mgl(1 cos θ), (21) mgl(1 cos θ) = 1 2 mv2 2 (22) v 2 = 2gl(1 cos θ) (23)
5 8.01L Problem Set #8 Solutions, p. 5 (b) What is the tension in the string when it makes an angle of θ with the vertical (i.e., when it is at its maximum angle)? At θ from the vertical, the rock is at the top of the swing, thus has 0 speed. Apply F = m a to the radial direction T mg cos θ = 0. The tension in the spring is T = mg cos θ (24) (c) What is the tension in the string as it passes through the vertical? We now consider when it passes through the vertical (point 2). Again, let s apply Newton s 2nd law. Fy = ma y (25) Thus The tension increases from point 1 to 2. T mg = m v2 2 l. (26) T = mg + m v2 2 l (27) T = mg(3 2 cos θ) (28) 84: Potential Energy Young and Freedman (13th edition) problem 7.86, page 239. (same as Young and Freedman (12th edition) problem 7.86, page 246.) A particle moves along the xaxis while acted on by a single conservative force parallel to the xaxis. The force corresponds to the potentialenergy function graphed in Fig. P7.86. The particle is released from rest at point A. (a) What is the direction of the force on the particle when it is at point A? F (x A ) = du x=xa = positive, i.e. rightward. (b) At point B? dx Figure P7.86
6 8.01L Problem Set #8 Solutions, p. 6 F (x B ) = du dx x=xb = negative, i.e. leftward. (c) At what value of x is the kinetic energy of the particle a maximum? E = K + U is fixed, so K is maximum when U is minimized x 0.75 m. (d) What is the force on the particle when it is at point C? F (x C ) = du x=xc = 0. dx (e) What is the largest value of x reached by the particle during its motion? x max is the value of x when all energy is potential again, i.e. when a horizontal line through the starting point A intersects blue curve. That is x max 2.1 m. (f) What value or values of x correspond to points of stable equilibrium? The minima of U are stable equilibrium: (g) Of unstable equilibrium? x 0.75 m, and (29) x 1.8 m. (30) Maxima & saddle points of U are unstable equilibria x 1.4 m (31) 85: Force of a Baseball Swing Adapted from Young and Freedman (13th edition) exercise 8.8, page 269. (a) A baseball has mass m b. If the velocity of a pitched ball has a magnitude of v i and the batted ball s velocity is v f in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. The initial and final momentum (xcomponent) of the ball are p 1,x = m b v i (32) p 2,x = m b v f (33)
7 8.01L Problem Set #8 Solutions, p. 7 Apply impulsemomentum theorem, J = p 2 p 1, the xcomponent of impulse equals the change in the xmomentum J x = p 2x p 1x = m(v f + v i ) (34) Therefore, the magnitude of the change in momentum of the ball is J x = m(v f + v i ) which is equal to the magnitude of the impulse applied to it by the bat. (b) If the ball has a mass of kg, an initial speed of 45.0 m/s, a final speed of 55.0 m/s, and the ball remains in contact with the bat for 2.00 ms (milliseconds), find the magnitude of the average force applied by the bat. (F av ) x = J x t The collision time is t = 2.00 ms = s. From J x = (F av ) x t, = (0.145 kg)(45.0 m/s m/s) s = 14.5 N s s = 7250 N (35) Thus the bat applies an average force of 7250 N to the ball in the xdirection. 86: Throwing a Rock Adapted from Young and Freedman (13th edition) exercise 8.28, page 270. (a) You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity v r relative to the earth at an angle θ above the horizontal. If your mass is M and the rock s mass is m r, what is your speed after you throw the rock? Consider the rock and yourself together as one system. Then since there is no horizontal external force acting on this system, the xcomponent of the momentum of the system is conserved, p x1 = p 2x. Since initially you and the rock are at rest, p 1x = 0. After you throw away the rock, the total momentum (xcomponent) is the sum of the momentum of you and the rock.
8 8.01L Problem Set #8 Solutions, p. 8 p 2x = m r v rx + m p v px where m r, m p are the masses of the rock and you, respectively, and v rx, v px are the xcomponents of the velocities. v px = p 2x m r v rx m p = m r m p v rx = m r m p v r cos θ. (36) (b) Calculate for the following values: v r = 12.0 m/s, θ = 35.0, M = 70.0 kg, m r = 15.0 kg. v rx = v r cos 35 = 12.0 m/s cos 35 = 9.83 m/s (37) v px = 15.0 kg 9.83 m/s = 2.11 m/s 70.0 kg (38) (a negative sign means you are moving in the opposite direction of the rock). Since you are confined to move on the horizontal surface v p = v px î = (2.11 m/s) î (39) Your speed after throwing the rock is 2.11 m/s. 87: Speed of a Bullet Adapted from Young and Freedman (13th edition) exercise 8.42, page 271. (a) A bullet with mass m b is fired horizontally into a wooden target block of mass M t resting on a horizontal surface. The coefficient of kinetic friction between block and surface is µ. The bullet remains embedded in the block, which is observed to slide a distance d along the surface before stopping. What was the initial speed of the bullet? Set the initial speed of the bullet to be v b. Consider the bullet and the block together as the system. Assume during the first stage, the bullet embeds itself in the block so quickly that the impulse due to the friction between the surface and block (external force) is negligible. Then we can apply momentum conservation to find the velocity of the system after the bullet embeds in the block and they
9 8.01L Problem Set #8 Solutions, p. 9 move together with the same speed v 1. By momentum conservation, p 1 = p 0. p 1 = (m b + m)v 1, p 0 = m b v b (40) v 1 = m bv b (41) m b + m For the second stage, we use the workenergy theorem. (since the potential energy doesn t change). Final speed is 0 K = 1 2 mv2 K = W f (42) W f = fd, f = µn = µmg (43) 1 2 mv2 1 = µmgd (44) ( mb v b m b + m v1 2 ) 2 = 2µdg (45) = 2µgd (46) v b = 2µgd m b + m m b (47) (b) Calculate using m b = 5 g, M t = 1.20 kg, µ = 0.20, d = m. v b = 2µgd m b + m (48) m b = m/s m kg kg (49) kg = 229 m/s (50) Therefore the initial speed of the bullet is 229 m/s. 88: The Roller Skater and the Football BCG problem 5B.4, page 175. This problem has a hint on page 194, an answer to the hint on page 196, and an answer to the problem on page 197. You are rollerskating peacefully down a street at a constant speed of 3 m/s when someone suddenly throws a football at you from directly ahead. If your mass is 65 kg and the football s is 0.40 kg, what is your speed afterwards if (a) you catch the ball, which was thrown at you with a horizontal velocity of 15 m/s; (b) you miss it and it bounces off you with a velocity of 10 m/s (relative to the street) in the opposite direction? In each case, what is the total kinetic energy of the system comprising you and the ball before and after your interaction? (Remember: solve the answer in terms of variables before you start sticking in numbers!)
10 8.01L Problem Set #8 Solutions, p. 10 In this problem we will use conservation of momentum! Let s name our variables first. Let the mass of the person be M, the mass of the football be m. The initial velocity of the person is v and that of the ball v b. The final speed of the person is v f and if the ball is not caught by the person, the speed will be v b,2 but of course if the person catches the ball, the latter will have the same speed v f as the skater. A good habit is to write down the momenta before and after the collision. For the first case, quantity Before collision After collision p x Mv mv b (M + m)v f 1 KE 2 Mv mv2 b = J 1 2 m)v2 f By conservation of momentum, in the first case, the final velocity is: v f = Mv mv b M + m = 65 kg 3 m/s 0.4 kg 15 m/s 65 kg kg = 2.89 m/s (51) For the second case: quantity Before collision After collision p x Mv mv b Mv f + mv 2 1 KE 2 Mv mv2 b = J 1 2 Mv2 f mv2 2 v f = Mv m(v b + v 2 ) M = 65 kg 3 m/s 0.4 kg(15 m/s + 10 m/s) 65 kg = 2.85 m/s (52) Knowing these values for v f, we can calculate the final kinetic energy for each of these cases: Case 1: K f = 1 2 (M + m)v2 f = J (53) Case 2: K f = 1 2 Mv2 f mv2 2 = J. (54)
Chapter 9. particle is increased.
Chapter 9 9. Figure 936 shows a three particle system. What are (a) the x coordinate and (b) the y coordinate of the center of mass of the three particle system. (c) What happens to the center of mass
More informationAP 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
More informationMidterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m
Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of
More informationPhysics 201 Homework 5
Physics 201 Homework 5 Feb 6, 2013 1. The (nonconservative) force propelling a 1500kilogram car up a mountain 1.21 10 6 joules road does 4.70 10 6 joules of work on the car. The car starts from rest
More informationChapter 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
More informationChapter 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
More informationA +TB v A + v B = 0. Av A B 2 = m>s = 1.27 m>sc Av B B 2 = 1.27 m>st
15 5. If cylinder is given an initial downward speed of 2 m>s, determine the speed of each cylinder when t = 3 s. Neglect the mass of the pulleys. Freeody Diagram: The freebody diagram of blocks and
More informationExam 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
More informationAP 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
More informationTHE NATURE OF FORCES Forces can be divided into two categories: contact forces and noncontact forces.
SESSION 2: NEWTON S LAWS Key Concepts In this session we Examine different types of forces Review and apply Newton's Laws of motion Use Newton's Law of Universal Gravitation to solve problems Xplanation
More informationExercises 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
More informationPhysics 1000 Final Examination. December A) 87 m B) 46 m C) 94 m D) 50 m
Answer all questions. The multiple choice questions are worth 4 marks and problems 10 marks each. 1. You walk 55 m to the north, then turn 60 to your right and walk another 45 m. How far are you from where
More informationF 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 250N force is directed horizontally as shown to push a 29kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force,
More informationPHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013
PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be
More informationCHAPTER 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
More informationMULTIPLE 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
More informationChapter 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
More informationPhysics 125 Practice Exam #3 Chapters 67 Professor Siegel
Physics 125 Practice Exam #3 Chapters 67 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
More informationChapter 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
More informationLecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.84.12, second half of section 4.7
Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.84.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal
More informationPhysics Midterm Review. MultipleChoice Questions
Physics Midterm Review MultipleChoice Questions 1. A train moves at a constant velocity of 90 km/h. How far will it move in 0.25 h? A. 10 km B. 22.5 km C. 25 km D. 45 km E. 50 km 2. A bicyclist moves
More informationPHY231 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
More informationPhysics 100 prac exam2
Physics 100 prac exam2 Student: 1. Earth's gravity attracts a person with a force of 120 lbs. The force with which the Earth is attracted towards the person is B. small but not zero. C. billions and billions
More informationPhysics: 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 buckleup? A) the first law
More informationAP1 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
More informationLecture 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,
More informationB) 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
More information1 of 6 10/17/2009 2:32 PM
1 of 6 10/17/2009 2:32 PM Chapter 9 Homework Due: 9:00am on Monday, October 19, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]
More informationPhysics 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
More informationP211 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
More informationCh 8 Potential energy and Conservation of Energy. Question: 2, 3, 8, 9 Problems: 3, 9, 15, 21, 24, 25, 31, 32, 35, 41, 43, 47, 49, 53, 55, 63
Ch 8 Potential energ and Conservation of Energ Question: 2, 3, 8, 9 Problems: 3, 9, 15, 21, 24, 25, 31, 32, 35, 41, 43, 47, 49, 53, 55, 63 Potential energ Kinetic energ energ due to motion Potential energ
More informationv 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
More informationVELOCITY, 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
More informationCollege Physics 140 Chapter 4: Force and Newton s Laws of Motion
College Physics 140 Chapter 4: Force and Newton s Laws of Motion We will be investigating what makes you move (forces) and how that accelerates objects. Chapter 4: Forces and Newton s Laws of Motion Forces
More informationPhysics 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
More informationB) 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
More informationMechanics 1. Revision Notes
Mechanics 1 Revision Notes July 2012 MECHANICS 1... 2 1. Mathematical Models in Mechanics... 2 Assumptions and approximations often used to simplify the mathematics involved:... 2 2. Vectors in Mechanics....
More informationPHY231 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
More informationChapter 9. Center of Mass & Linear Momentum
Chapter 9 Center of Mass & Linear Momentum 9.2 The Center of Mass The center of mass of a system of particles is the point that moves as though: (1) all of the system s mass were concentrated there; (2)
More informationPhysics 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
More informationCh 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
More informationEndofChapter Exercises
EndofChapter Exercises Exercises 1 12 are conceptual questions that are designed to see if you have understood the main concepts of the chapter. 1. Figure 11.20 shows four different cases involving a
More information1) 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 3.0kilogram mass being moved at a constant speed by a force of 6.0 Newtons. 4. If the surface were frictionless, the
More informationGravitational 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
More informationPhys 111 Fall P111 Syllabus
Phys 111 Fall 2012 Course structure Five sections lecture time 150 minutes per week Textbook Physics by James S. Walker fourth edition (Pearson) Clickers recommended Coursework Complete assignments from
More information9. 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
More informationMULTIPLE 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) A container explodes and breaks into three fragments that fly off 120 apart from each
More informationC 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
More informationB Answer: neither of these. Mass A is accelerating, so the net force on A must be nonzero Likewise for mass B.
CTA1. 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
More informationAssignment 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
More informationSHORT 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.
More information8. 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
More informationAt 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
More informationHigher Physics Our Dynamic Universe Notes
Higher Physics Our Dynamic Universe Notes Teachers Booklet Previous knowledge This section builds on the knowledge from the following key areas from Dynamics and Space Booklet 1  Dynamics Velocity and
More informationWork, 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 1.5kg psychology textbook out of a second floor classroom window until her arm is tired; then she releases
More informationAP Physics C. Oscillations/SHM Review Packet
AP Physics C Oscillations/SHM Review Packet 1. A 0.5 kg mass on a spring has a displacement as a function of time given by the equation x(t) = 0.8Cos(πt). Find the following: a. The time for one complete
More informationReview 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
More information2Elastic collisions in
After completing this chapter you should be able to: solve problems about the impact of a smooth sphere with a fixed surface solve problems about the impact of smooth elastic spheres. In this chapter you
More informationAAPT UNITED STATES PHYSICS TEAM AIP CEE 2013
F = ma Exam AAPT UNITED STATES PHYSICS TEAM AIP CEE F = ma Contest 5 QUESTIONS  75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Use g = N/kg throughout this contest. You may
More informationWork. Work = Force distance (the force must be parallel to movement) OR Work = (Force)(cos θ)(distance)
Work Work = Force distance (the force must be parallel to movement) OR Work = (Force)(cos θ)(distance) When you are determining the force parallel to the movement you can do this manually and keep track
More informationForces. Isaac Newton was the first to discover that the laws that govern motions on the Earth also applied to celestial bodies.
Forces Now we will discuss the part of mechanics known as dynamics. We will introduce Newton s three laws of motion which are at the heart of classical mechanics. We must note that Newton s laws describe
More informationProblem 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
More informationPhysics 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
More information1. A tennis ball of mass m moving horizontally with speed u strikes a vertical tennis racket. The ball bounces back with a horizontal speed v.
1. A tennis ball of mass m moving horizontally with speed u strikes a vertical tennis racket. The ball bounces back with a horizontal speed v. The magnitude of the change in momentum of the ball is A.
More informationPHY121 #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
More informationPhysics 101 Exam 1 NAME 2/7
Physics 101 Exam 1 NAME 2/7 1 In the situation below, a person pulls a string attached to block A, which is in turn attached to another, heavier block B via a second string (a) Which block has the larger
More informationWORK 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 Newtonmeter (Nm) = Joule, J If you exert a force of
More information7. 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
More informationKE =? v o. Page 1 of 12
Page 1 of 12 CTEnergy1. 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
More informationVectors; 2D Motion. Part I. Multiple Choice. 1. v
This test covers vectors using both polar coordinates and ij notation, radial and tangential acceleration, and twodimensional motion including projectiles. Part I. Multiple Choice 1. v h x In a lab experiment,
More informationBROCK UNIVERSITY. PHYS 1P21/1P91 Solutions to Midterm test 26 October 2013 Instructor: S. D Agostino
BROCK UNIVERSITY PHYS 1P21/1P91 Solutions to Midterm test 26 October 2013 Instructor: S. D Agostino 1. [10 marks] Clearly indicate whether each statement is TRUE or FALSE. Then provide a clear, brief,
More informationChapter 07: Kinetic Energy and Work
Chapter 07: Kinetic Energy and Work Conservation of Energy is one of Nature s fundamental laws that is not violated. Energy can take on different forms in a given system. This chapter we will discuss work
More informationcharge 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, 2D collisions, and centerofmass, with some problems requiring
More informationElastic and Inelastic Collisions
Elastic and Inelastic Collisions Different kinds of collisions produce different results. Sometimes the objects stick together. Sometimes the objects bounce apart. What is the difference between these
More informationChapter 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
More information8.012 Physics I: Classical Mechanics Fall 2008
MIT OpenCourseWare http://ocw.mit.edu 8.012 Physics I: Classical Mechanics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS INSTITUTE
More informationUnit 1: Vectors. a m/s b. 8.5 m/s c. 7.2 m/s d. 4.7 m/s
Multiple Choice Portion 1. A boat which can travel at a speed of 7.9 m/s in still water heads directly across a stream in the direction shown in the diagram above. The water is flowing at 3.2 m/s. What
More informationGeneral Physics Physics 101 Test #3 Spring 2013 Friday 4/12/13 Prof. Bob Ekey
General Physics Physics 101 Test #3 Spring 2013 Friday 4/12/13 Prof. Bob Ekey Name (print): I hereby declare upon my word of honor that I have neither given nor received unauthorized help on this work.
More informationPhysics 271 FINAL EXAMSOLUTIONS Friday Dec 23, 2005 Prof. Amitabh Lath
Physics 271 FINAL EXAMSOLUTIONS Friday Dec 23, 2005 Prof. Amitabh Lath 1. The exam will last from 8:00 am to 11:00 am. Use a # 2 pencil to make entries on the answer sheet. Enter the following id information
More informationDISPLACEMENT 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
More informationKinetic Energy (A) stays the same stays the same (B) increases increases (C) stays the same increases (D) increases stays the same.
1. A cart full of water travels horizontally on a frictionless track with initial velocity v. As shown in the diagram, in the back wall of the cart there is a small opening near the bottom of the wall
More informationLesson 04: Newton s laws of motion
www.scimsacademy.com Lesson 04: Newton s laws of motion If you are not familiar with the basics of calculus and vectors, please read our freely available lessons on these topics, before reading this lesson.
More informationUniversity 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/2012odu Wednesday, September
More informationGround 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
More informationPHYS 211 FINAL FALL 2004 Form A
1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each
More informationMechanics 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...
More informationName Date Class. The Nature of Force and Motion (pages ) 2. When one object pushes or pulls another object, the first object is
CHAPTER 4 MOTION AND FORCES SECTION 4 1 The Nature of Force and Motion (pages 116121) This section explains how balanced and unbalanced forces are related to the motion of an object. It also explains
More informationNewton s Law of Motion
chapter 5 Newton s Law of Motion Static system 1. Hanging two identical masses Context in the textbook: Section 5.3, combination of forces, Example 4. Vertical motion without friction 2. Elevator: Decelerating
More informationWork, Energy and Power Practice Test 1
Name: ate: 1. How much work is required to lift a 2kilogram 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
More informationPhysics 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 noncontact forces. Whats a
More information2.1 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
More informationNewton s Laws of Motion
Newton s Laws of Motion FIZ101E Kazım Yavuz Ekşi My contact details: Name: Kazım Yavuz Ekşi Email: eksi@itu.edu.tr Notice: Only emails from your ITU account are responded. Office hour: Wednesday 10.0012.00
More informationwww.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
More informationChapter 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?!
More informationPhysics 211 Week 12. Simple Harmonic Motion: Equation of Motion
Physics 11 Week 1 Simple Harmonic Motion: Equation of Motion A mass M rests on a frictionless table and is connected to a spring of spring constant k. The other end of the spring is fixed to a vertical
More informationPHYSICS 111 HOMEWORK SOLUTION #8. March 24, 2013
PHYSICS 111 HOMEWORK SOLUTION #8 March 24, 2013 0.1 A particle of mass m moves with momentum of magnitude p. a) Show that the kinetic energy of the particle is: K = p2 2m (Do this on paper. Your instructor
More informationChapter Test. Teacher Notes and Answers Forces and the Laws of Motion. Assessment
Assessment Chapter Test A Teacher Notes and Answers Forces and the Laws of Motion CHAPTER TEST A (GENERAL) 1. c 2. d 3. d 4. c 5. c 6. c 7. c 8. b 9. d 10. d 11. c 12. a 13. d 14. d 15. b 16. d 17. c 18.
More informationChapter 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
More informationCurso20122013 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.
More information