Problem Set #8 Solutions


 Juliana Fields
 10 months 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 https://web.mit.edu/ 8.01L/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)
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
More informationChapter 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 information1.7 Work Done, Potential and Kinetic Energy
Students should be able to: 1.7.1 Define work done, potential energy, kinetic energy, efficiency and power 1.7.2 Recognise that when work is done energy is transferred from one form to another 1.7.3 Calculate
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 informationAP 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
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 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 informationWork, 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.
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 informationChapter 9. is gradually increased, does the center of mass shift toward or away from that particle or does it remain stationary.
Chapter 9 9.2 Figure 937 shows a three particle system with masses m 1 3.0 kg, m 2 4.0 kg, and m 3 8.0 kg. The scales are set by x s 2.0 m and y s 2.0 m. What are (a) the x coordinate and (b) the y coordinate
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 informationChapter 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
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 informationChapter 7 WORK, ENERGY, AND Power Work Done by a Constant Force Kinetic Energy and the WorkEnergy Theorem Work Done by a Variable Force Power
Chapter 7 WORK, ENERGY, AND Power Work Done by a Constant Force Kinetic Energy and the WorkEnergy Theorem Work Done by a Variable Force Power Examples of work. (a) The work done by the force F on this
More informationChapter 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
More informationChapter 5 Newton s Laws of Motion
Chapter 5 Newton s Laws of Motion Sir Isaac Newton (1642 1727) Developed a picture of the universe as a subtle, elaborate clockwork slowly unwinding according to welldefined rules. The book Philosophiae
More informationLab 8: Ballistic Pendulum
Lab 8: Ballistic Pendulum Equipment: Ballistic pendulum apparatus, 2 meter ruler, 30 cm ruler, blank paper, carbon paper, masking tape, scale. Caution In this experiment a steel ball is projected horizontally
More informationPhysics 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
More informationReview D: Potential Energy and the Conservation of Mechanical Energy
MSSCHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 Fall 2005 Review D: Potential Energy and the Conservation of Mechanical Energy D.1 Conservative and Nonconservative Force... 2 D.1.1 Introduction...
More informationChapter 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 nonzero speed carries energy
More informationPhysics 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
More informationProblem Set #13 Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.0L: Physics I January 3, 06 Prof. Alan Guth Problem Set #3 Solutions Due by :00 am on Friday, January in the bins at the intersection of Buildings
More informationEDUH 1017  SPORTS MECHANICS
4277(a) Semester 2, 2011 Page 1 of 9 THE UNIVERSITY OF SYDNEY EDUH 1017  SPORTS MECHANICS NOVEMBER 2011 Time allowed: TWO Hours Total marks: 90 MARKS INSTRUCTIONS All questions are to be answered. Use
More informationChapter 3.8 & 6 Solutions
Chapter 3.8 & 6 Solutions P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequency are inverses according to Equation 3.26. To find speed we need to know the distance traveled
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 informationWork 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
More informationMechanics 1: Conservation of Energy and Momentum
Mechanics : Conservation of Energy and Momentum If a certain quantity associated with a system does not change in time. We say that it is conserved, and the system possesses a conservation law. Conservation
More informationPractice Exam Three Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01T Fall Term 2004 Practice Exam Three Solutions Problem 1a) (5 points) Collisions and Center of Mass Reference Frame In the lab frame,
More information10.1 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
More informationSerway_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
More informationLecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is
Lecture 17 Rotational Dynamics Rotational Kinetic Energy Stress and Strain and Springs Cutnell+Johnson: 9.49.6, 10.110.2 Rotational Dynamics (some more) Last time we saw that the rotational analog of
More informationQUESTIONS : CHAPTER5: LAWS OF MOTION
QUESTIONS : CHAPTER5: LAWS OF MOTION 1. What is Aristotle s fallacy? 2. State Aristotlean law of motion 3. Why uniformly moving body comes to rest? 4. What is uniform motion? 5. Who discovered Aristotlean
More information5. Forces and MotionI. Force is an interaction that causes the acceleration of a body. A vector quantity.
5. Forces and MotionI 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
More informationUniversity Physics 226N/231N Old Dominion University. Getting Loopy and Friction
University Physics 226N/231N Old Dominion University Getting Loopy and Friction Dr. Todd Satogata (ODU/Jefferson Lab) satogata@jlab.org http://www.toddsatogata.net/2012odu Friday, September 28 2012 Happy
More informationAP Physics 1 Midterm Exam Review
AP Physics 1 Midterm Exam Review 1. The graph above shows the velocity v as a function of time t for an object moving in a straight line. Which of the following graphs shows the corresponding displacement
More informationAP1 Dynamics. Answer: (D) foot applies 200 newton force to nose; nose applies an equal force to the foot. Basic application of Newton s 3rd Law.
1. A mixed martial artist kicks his opponent in the nose with a force of 200 newtons. Identify the actionreaction force pairs in this interchange. (A) foot applies 200 newton force to nose; nose applies
More informationAP1 Oscillations. 1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false?
1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false? (A) The displacement is directly related to the acceleration. (B) The
More informationAcceleration due to Gravity
Acceleration due to Gravity 1 Object To determine the acceleration due to gravity by different methods. 2 Apparatus Balance, ball bearing, clamps, electric timers, meter stick, paper strips, precision
More informationProblem 6.40 and 6.41 Kleppner and Kolenkow Notes by: Rishikesh Vaidya, Physics Group, BITSPilani
Problem 6.40 and 6.4 Kleppner and Kolenkow Notes by: Rishikesh Vaidya, Physics Group, BITSPilani 6.40 A wheel with fine teeth is attached to the end of a spring with constant k and unstretched length
More informationSupplemental Questions
Supplemental Questions The fastest of all fishes is the sailfish. If a sailfish accelerates at a rate of 14 (km/hr)/sec [fwd] for 4.7 s from its initial velocity of 42 km/h [fwd], what is its final velocity?
More informationPhysics 11 Assignment KEY Dynamics Chapters 4 & 5
Physics Assignment KEY Dynamics Chapters 4 & 5 ote: for all dynamics problemsolving questions, draw appropriate free body diagrams and use the aforementioned problemsolving method.. Define the following
More informationTHE NOT SO SIMPLE PENDULUM
INTRODUCTION: THE NOT SO SIMPLE PENDULUM This laboratory experiment is used to study a wide range of topics in mechanics like velocity, acceleration, forces and their components, the gravitational force,
More informationKE = ½mv 2 PE = mgh W = Fdcosθ THINK ENERGY! (KE F + PE F ) = (KE 0 + PE 0 ) + W NC. Tues Oct 6 Assign 7 Fri Preclass Thursday
Tues Oct 6 Assign 7 Fri Preclass Thursday Conservation of Energy Work, KE, PE, Mech Energy Power To conserve total energy means that the total energy is constant or stays the same. With Work, we now have
More informationSample Questions for the AP Physics 1 Exam
Sample Questions for the AP Physics 1 Exam Sample Questions for the AP Physics 1 Exam Multiplechoice Questions Note: To simplify calculations, you may use g 5 10 m/s 2 in all problems. Directions: Each
More informationG U I D E T O A P P L I E D O R B I T A L M E C H A N I C S F O R K E R B A L S P A C E P R O G R A M
G U I D E T O A P P L I E D O R B I T A L M E C H A N I C S F O R K E R B A L S P A C E P R O G R A M CONTENTS Foreword... 2 Forces... 3 Circular Orbits... 8 Energy... 10 Angular Momentum... 13 FOREWORD
More informationUnit 3 Work and Energy Suggested Time: 25 Hours
Unit 3 Work and Energy Suggested Time: 25 Hours PHYSICS 2204 CURRICULUM GUIDE 55 DYNAMICS Work and Energy Introduction When two or more objects are considered at once, a system is involved. To make sense
More informationNewton 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
More informationChapter 7 Momentum and Impulse
Chapter 7 Momentum and Impulse Collisions! How can we describe the change in velocities of colliding football players, or balls colliding with bats?! How does a strong force applied for a very short time
More informationExam Three Momentum Concept Questions
Exam Three Momentum Concept Questions Isolated Systems 4. A car accelerates from rest. In doing so the absolute value of the car's momentum changes by a certain amount and that of the Earth changes by:
More informationPhysics 41 HW Set 1 Chapter 15
Physics 4 HW Set Chapter 5 Serway 8 th OC:, 4, 7 CQ: 4, 8 P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59, 67, 74 OC CQ P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59,
More informationTennessee 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 Fgrade. Other instructions will be given in the Hall. MULTIPLE CHOICE.
More informationConservative vs. Nonconservative forces Gravitational Potential Energy. Work done by nonconservative forces and changes in mechanical energy
Next topic Conservative vs. Nonconservative forces Gravitational Potential Energy Mechanical Energy Conservation of Mechanical energy Work done by nonconservative forces and changes in mechanical energy
More informationChapter #7 Giancoli 6th edition Problem Solutions
Chapter #7 Giancoli 6th edition Problem Solutions ü Problem #8 QUESTION: A 9300 kg boxcar traveling at 5.0 m/s strikes a second boxcar at rest. The two stick together and move off with a speed of 6.0 m/s.
More informationF13HPhysQ5 Practice
Name: Class: Date: ID: A F13HPhysQ5 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.
More informationWork 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,
More informationSimple Harmonic Motion
Simple Harmonic Motion 1 Object To determine the period of motion of objects that are executing simple harmonic motion and to check the theoretical prediction of such periods. 2 Apparatus Assorted weights
More informationForces. Definition Friction Falling Objects Projectiles Newton s Laws of Motion Momentum Universal Forces Fluid Pressure Hydraulics Buoyancy
Forces Definition Friction Falling Objects Projectiles Newton s Laws of Motion Momentum Universal Forces Fluid Pressure Hydraulics Buoyancy Definition of Force Force = a push or pull that causes a change
More information