2a y. ) f 2 =(v y. (v y. ) i


 Milo Horn
 2 years ago
 Views:
Transcription
1 Jumping problems: Two phases: Jump with forces from gravity and leg muscles, acceleration upward. Flight, animal has left ground, gravity only force, acceleration downward. Treat these separately, each with it s own acceleration. Final velocity from jump is the initial velocity for the flight. P.53. When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.50 mm. a) What is the flea s acceleration during the jump phase? b) How long does the acceleration phase last? c) If the flea jumps straight up, how high will it go? Ignore air resistance here, which in reality would be very large and the flea will not reach this height. Prepare: The motion has two parts: The jump, with an acceleration upward, and the flight after the flea leaves the ground, with a downward acceleration. We treat these parts separately and use a constant acceleration during the jump (which would be the average acceleration in reality). Fleas are amazing jumpers; they can jump several times their body height something we cannot do. We assume constant acceleration so we can use the equations in Table.4. The last of the three relates the three variables we are concerned with in part (a): speed, distance (which we know), and acceleration (which we want). = a y y In part (b) we use the first equation in Table.4 because it relates the initial and final velocities and the acceleration (which we know) with the time interval (which we want). = a y t Part (cs about the phase of the jump after the flea reaches takeoff speed and leaves the ground. So now it is, that is 1.0 m/s instead of. And the acceleration is not the same as in part (a) it is now g (with the positive direction up) since we are ignoring air resistance. We do not know the time it takes the flea to reach maximum height, so we employ the last equation in Table.4 again because we know everything in that equation except y. Solve: (a) Use = 0.0 m/s and rearrange the last equation in Table.4. a y y (1.0 m/s) (0.50 mm) 1000 mm 1 m 1000 m/s (b) Having learned the acceleration from part (a) we can now rearrange the first equation in Table.4 to find the time it takes to reach takeoff speed. Again use =0.0 m/s. t a y 1.0 m/s 1000 m/s.0010 s (c) This time =0.0m/s as the flea reaches the top of its trajectory. Rearrange the last equation in Table.4 to get y v y i a y (1.0 m/s) ( 9.8 m/s ) m 5.1 cm Assess: Just over 5 cm is pretty good considering the size of a flea. It is about 10 0 times the size of a typical flea. Check carefully to see that each answer ends up in the appropriate units. The height of the flea at the top will round to 5. cm above the ground if you include the cm during the initial acceleration phase before the feet actually leave the ground.
2 dimensional problems Leafy spurge disperses its seeds with an explosive ejection from the seed capsule and can expel its seeds to distances up to 5 meters. The seeds have a mass of 3.04 mg each and are accelerated through a distance of mm. The seed pods are at the top of the plants which are up to 1 meter tall. Assuming the optimal 45º angle for the maximum distance, find the velocity with which the seed is ejected to reach the 5 meter range. Ignore air resistance (which is likely a poor assumption). Knowns: y o = 1m x o = 0 θ = 45º y f = 0 x f = 0 a = g in flight Because of the change in height, we cannot use the range formula from lab and the maximum height is no longer reached at half the time for the entire flight; the symmetry of the trajectory is destroyed. Instead, we will split the motion into vertical and horizontal components and solve them independently. Vertical: a = g, v yo = v o sin(45), y o = 1m, y f = 0 y = y o + v yo t + ½ (g) t = 1m + v o sin(45) t g/ t Horizontal: a=0, v xo = v o cos(45), x o = 0, x f = 5m x = x o + v xo t = v o cos(45) t v x = v o cos(45) v y = v yo + a y t = v o sin(45)  gt When the seed hits the ground at t f, y = 0 = 1m + v o sin(45) t f g/ t f x = 5m = v o cos(45) t f solve for t f (simpler than quadratic) drop units to save space, equations consistent. Use t f from x equation, drop units: 0 = 1 +( v o / )(5 /v o ) (g/)(5 /v o ) = g/v o t f = 5m/ v o cos(45) = 5 / v o sin(45)=cos(45)=1/ 5g/v o = 6 v o = 5g/6 = 5(9.8)/6 v o = 6.4 m/s t = 5 / v o = (5 [m]) / (6.4 [m/s]) = 1.1 s We can now determine the acceleration of the seed as it is ejected from the pod.
3 During ejection, the velocity changes from 0 to 6.4 m/s over a distance of mm. If we initially ignore gravity, we can use v f = v o +ad a = (v f  v o )/d = 6.4 /[(.00)] = 1.0 x 10 4 m/s Comparing this acceleration to g, it appears that ignoring g was justified, since it is much smaller than the value calculated. Using Newton s second law, we can calculate the force applied to the seed to eject it. F = ma = (3.04 x 106 kg)(1.0 x 10 4 m/s ) = 3.10 x 10  kg m/s = 3.10 x 10  N Ramp problems: Consider a block on a ramp making an angle of 30º with the horizontal. Under normal conditions, the block will move along the ramp surface. The coordinates which make the calculations the simplest are parallel to the ramp and perpendicular to it. This follows the general strategy of Choose one coordinate axis along the direction of the acceleration or motion and the other perpendicular to that. This means that the gravitational force needs to be split into components. I have given you some geometry tools on the second exam data sheet, anticipating that the geometry you learned in high school probably did not all stay in your ready recall memory. Applying these tools to the gravitational force (mg) gives the components in the diagram to the right. The gravitational force can be replaced by tits components, redrawing the forces with the ramp direction horizontal gives the diagram on the next page.
4 Unless the block is sinking into the ramp or jumping up off it, the acceleration perpendicular to the ramp is zero, as is the velocity. The acceleration along the ramp is downward or to the right in the diagram. Force balance: Perpendicular F N  mg cos(30) = 0 or F N = mg cos(30) Along ramp: F net = mg sin(30) f =ma (a) If the ramp is frictionless, then Fnet = mg sin(30) and the acceleration of the block is given by a = g sin(30) = g/. This gives v = v o + a t = v o + (g/)t and d = d o + v o t + ½ a t = d o + v o t + ½ (g/) t For displacement, take d o = 0 = v o t + ½ (g/) t (b) If there is friction, the friction force can be represented by f = μ F N where μ is the coefficient of friction. Let s see what we can determine from this and our force balance equations above. f = μ F N and, from the perpendicular force balance, F N = mg cos(30) giving f = μ mg cos(30) Then F net = mg sin(30) f = mg sin(30) μ mg cos(30) = mg [sin(30) μ cos(30)] = ma or a = g [sin(30) μ cos(30)]. Does this suggest a way to measure the coefficient of static friction (μ s ) when the block is standing still on the ramp?
5 1. A student brings in her physics homework, dragging it across campus in a wooden box. She is pulling with a force of 450 N on a rope, which inclined at 38 degrees to the horizontal. The ground exerts a horizontal force of 15N that opposes the motion. Calculate the acceleration of the crate if (at mass is 310 kg and (bts weight is 310 N. First, the crate will move (be acceleratedn the horizontal direction. Use coordinate axes which are horizontal and vertical. Resolve the 450N pull into horizontal and vertical components. The simplified force diagram is now: Vertical: F N + 450sin(38) = mg Horizontal: 450cos(38) 15 = ma a) m = 350 kg: a (450cos(38) 15) N 350kg 0.66 m / s b) wt = 350N = mg m = 350/g = 35.7 kg a (450cos(38) 15) N 35.7kg 6.43 m / s You can also calculate the Normal force from the ground: a) F N + 450sin(38) = mg = (350 kg)(9.8m/s ) = 630 N F N = sin(38) = 353 N b) F N + 450sin(38) = mg = 350 N F N = sin(38) = 73 N
6 . m1 is 3.70 kg on a frictionless inclined plane. Pulley is massless, and frictionless. m is.30 kg. Show force diagrams for each mass. Key ideas for solving text problems: (a) What is magnitude of acceleration of each block? (b) What is direction of acceleration of hanging block, m? (c) What is value of the tension in the cord? (1) The tension in a single piece of rope is the same anywhere in the rope and at both ends, as long as it is massless or nearly so. () The only effect of a massless, frictionless pulley is to change the direction the force, it does not affect the size of the force. of (3) Objects connected by string will have the same acceleration. (What would happen if this were not true?) From the diagram of the block on the ramp, the simplest axes are going to be along the ramp and perpendicular to it (you only need to break up one vector that way). This gives the following diagram with the components of the weight (force of gravityncluded: For this block, and F N = m 1 g cos(θrom the vertical balance T m 1 g sin(θ) = m 1 a along the ramp. For the hanging block, m g T = m a For connected objects like this, it is frequently fruitful to look at the entire set of connected objects as one total mass and set up the forces for that. Here, that gives m g m 1 g sin(θ) = (m 1 + m )a Which will allow us to solve for the acceleration a. Let s look at the forces along the ramp and for the hanging mass. T m 1 g sin(θ) = m 1 a m g T = m a We want to solve for tension T and acceleration a, so let me rearrange these to give T  m 1 a = m 1 g sin(θ) T + m a = m g Subtracting the first equation from the second leaves the equation above for the two masses together: m a + m 1 a = m g m 1 g sin(θ) (The T s subtract out). Once we solve this for a, we can use that value in either equation above to find the tension T.
7 If it is helpful, there is a not on the web site describing various ways to work with two equations in two unknowns. Some miscelaneous force diagrams:
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
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 informationTwoBody System: Two Hanging Masses
Specific Outcome: i. I can apply Newton s laws of motion to solve, algebraically, linear motion problems in horizontal, vertical and inclined planes near the surface of Earth, ignoring air resistance.
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 informationNewton s Third Law. object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1
Newton s Third Law! If two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1!! Note on notation: is
More informationSOLUTIONS TO PROBLEM SET 4
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01X Fall Term 2002 SOLUTIONS TO PROBLEM SET 4 1 Young & Friedman 5 26 A box of bananas weighing 40.0 N rests on a horizontal surface.
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 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 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 informationTEACHER ANSWER KEY November 12, 2003. Phys  Vectors 11132003
Phys  Vectors 11132003 TEACHER ANSWER KEY November 12, 2003 5 1. A 1.5kilogram lab cart is accelerated uniformly from rest to a speed of 2.0 meters per second in 0.50 second. What is the magnitude
More informationForces: Equilibrium Examples
Physics 101: Lecture 02 Forces: Equilibrium Examples oday s lecture will cover extbook Sections 2.12.7 Phys 101 URL: http://courses.physics.illinois.edu/phys101/ Read the course web page! Physics 101:
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 informationFRICTION, 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
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 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 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
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
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 informationConceptual Questions: Forces and Newton s Laws
Conceptual Questions: Forces and Newton s Laws 1. An object can have motion only if a net force acts on it. his statement is a. true b. false 2. And the reason for this (refer to previous question) is
More informationHomework 4. problems: 5.61, 5.67, 6.63, 13.21
Homework 4 problems: 5.6, 5.67, 6.6,. Problem 5.6 An object of mass M is held in place by an applied force F. and a pulley system as shown in the figure. he pulleys are massless and frictionless. Find
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 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 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 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 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 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 informationIMPORTANT NOTE ABOUT WEBASSIGN:
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
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 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 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: 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 informationAP Physics Applying Forces
AP Physics Applying Forces This section of your text will be very tedious, very tedious indeed. (The Physics Kahuna is just as sorry as he can be.) It s mostly just a bunch of complicated problems and
More informationLAB 6: GRAVITATIONAL AND PASSIVE FORCES
55 Name Date Partners LAB 6: GRAVITATIONAL AND PASSIVE FORCES And thus Nature will be very conformable to herself and very simple, performing all the great Motions of the heavenly Bodies by the attraction
More informationLAB 6  GRAVITATIONAL AND PASSIVE FORCES
L061 Name Date Partners LAB 6  GRAVITATIONAL AND PASSIVE FORCES OBJECTIVES And thus Nature will be very conformable to herself and very simple, performing all the great Motions of the heavenly Bodies
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 informationThis week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension. PHYS 2: Chap.
This week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension PHYS 2: Chap. 19, Pg 2 1 New Topic Phys 1021 Ch 7, p 3 A 2.0 kg wood box slides down a vertical
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 informationPhysics 590 Homework, Week 6 Week 6, Homework 1
Physics 590 Homework, Week 6 Week 6, Homework 1 Prob. 6.1.1 A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 35 m/s. At the same time it has a horizontal
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 informationPHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 17. February 13, 2013
PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 17 February 13, 2013 0.1 A 2.00kg object undergoes an acceleration given by a = (6.00î + 4.00ĵ)m/s 2 a) Find the resultatnt force acting on the object
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 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 informationA) N > W B) N = W C) N < W. speed v. Answer: N = W
CTN12. Consider a person standing in an elevator that is moving upward at constant speed. The magnitude of the upward normal force, N, exerted by the elevator floor on the person's feet is (larger than/same
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 informationObjective: Equilibrium Applications of Newton s Laws of Motion I
Type: Single Date: Objective: Equilibrium Applications of Newton s Laws of Motion I Homework: Assignment (111) Read (4.14.5, 4.8, 4.11); Do PROB # s (46, 47, 52, 58) Ch. 4 AP Physics B Mr. Mirro Equilibrium,
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 informationForces. When an object is pushed or pulled, we say that a force is exerted on it.
Forces When an object is pushed or pulled, we say that a force is exerted on it. Forces can Cause an object to start moving Change the speed of a moving object Cause a moving object to stop moving Change
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 informationMass, 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
More informationPhysics I Honors: Chapter 4 Practice Exam
Physics I Honors: Chapter 4 Practice Exam Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Which of the following statements does not describe
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 informationRecitation Week 4 Chapter 5
Recitation Week 4 Chapter 5 Problem 5.5. A bag of cement whose weight is hangs in equilibrium from three wires shown in igure P5.4. wo of the wires make angles θ = 60.0 and θ = 40.0 with the horizontal.
More informationChapter 11 Equilibrium
11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium 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 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 informationA Determination of g, the Acceleration Due to Gravity, from Newton's Laws of Motion
A Determination of g, the Acceleration Due to Gravity, from Newton's Laws of Motion Objective In the experiment you will determine the cart acceleration, a, and the friction force, f, experimentally for
More informationIf you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ
Experiment 4 ormal and Frictional Forces Preparation Prepare for this week's quiz by reviewing last week's experiment Read this week's experiment and the section in your textbook dealing with normal forces
More informationPhysics 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 8:1510:15 Room:
More informationMULTIPLE 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
More informationSteps 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
More informationChapter 07 Test A. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.
Class: Date: Chapter 07 Test A Multiple Choice Identify the choice that best completes the statement or answers the question. 1. An example of a vector quantity is: a. temperature. b. length. c. velocity.
More informationNewton s Second Law. ΣF = m a. (1) In this equation, ΣF is the sum of the forces acting on an object, m is the mass of
Newton s Second Law Objective The Newton s Second Law experiment provides the student a hands on demonstration of forces in motion. A formulated analysis of forces acting on a dynamics cart will be developed
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 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 informationNewton s Laws. Physics 1425 lecture 6. Michael Fowler, UVa.
Newton s Laws Physics 1425 lecture 6 Michael Fowler, UVa. Newton Extended Galileo s Picture of Galileo said: Motion to Include Forces Natural horizontal motion is at constant velocity unless a force acts:
More informationHW Set II page 1 of 9 PHYSICS 1401 (1) homework solutions
HW Set II page 1 of 9 450 When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 20 km (about the size of the San Francisco
More informationExam 2 Review Questions PHY Exam 2
Exam 2 Review Questions PHY 2425  Exam 2 Section: 4 1 Topic: Newton's First Law: The Law of Inertia Type: Conceptual 1 According to Newton's law of inertia, A) objects moving with an initial speed relative
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 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 informationFundamental Mechanics: Supplementary Exercises
Phys 131 Fall 2015 Fundamental Mechanics: Supplementary Exercises 1 Motion diagrams: horizontal motion A car moves to the right. For an initial period it slows down and after that it speeds up. Which of
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 informationWeight 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
More informationUNIT 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
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 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 informationParachute Jumping, Falling, and Landing David C. Arney, Barbra S. Melendez, Debra Schnelle 1
Parachute Jumping, Falling, and Landing David C. Arney, Barbra S. Melendez, Debra Schnelle 1 Introduction It is extremely important that leaders of airborne units understand the safety, medical, and operational
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 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 informationcircular motion & gravitation physics 111N
circular motion & gravitation physics 111N uniform circular motion an object moving around a circle at a constant rate must have an acceleration always perpendicular to the velocity (else the speed would
More informationProjectile Motion 1:Horizontally Launched Projectiles
A cannon shoots a clown directly upward with a speed of 20 m/s. What height will the clown reach? How much time will the clown spend in the air? Projectile Motion 1:Horizontally Launched Projectiles Two
More informationChapter 13, example problems: x (cm) 10.0
Chapter 13, example problems: (13.04) Reading Fig. 1330 (reproduced on the right): (a) Frequency f = 1/ T = 1/ (16s) = 0.0625 Hz. (since the figure shows that T/2 is 8 s.) (b) The amplitude is 10 cm.
More informationProblems 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 1217 Answers Worked
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 informationAngular acceleration α
Angular Acceleration Angular acceleration α measures how rapidly the angular velocity is changing: Slide 70 Linear and Circular Motion Compared Slide 7 Linear and Circular Kinematics Compared Slide 7
More information1) The gure below shows the position of a particle (moving along a straight line) as a function of time. Which of the following statements is true?
Physics 2A, Sec C00: Mechanics  Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to ll your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry
More informationSTATIC AND KINETIC FRICTION
STATIC AND KINETIC FRICTION LAB MECH 3.COMP From Physics with Computers, Vernier Software & Technology, 2000. INTRODUCTION If you try to slide a heavy box resting on the floor, you may find it difficult
More informationSolution Derivations for Capa #11
Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform
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 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 informationD) A block pulled with a constant force will have a constant acceleration in the same direction as the force.
Phsics 100A Homework 4 Chapter 5 Newton s First Law A)If a car is moving to the left with constant velocit then the net force applied to the car is zero. B) An object cannot remain at rest unless the net
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 information= 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
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 informationSolution: (a) For a positively charged particle, the direction of the force is that predicted by the right hand rule. These are:
Problem 1. (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields as shown in the figure. (b) Repeat part (a), assuming the moving particle is
More informationRotational Motion: Moment of Inertia
Experiment 8 Rotational Motion: Moment of Inertia 8.1 Objectives Familiarize yourself with the concept of moment of inertia, I, which plays the same role in the description of the rotation of a rigid body
More informationFriction and Newton s 3rd law
Lecture 4 Friction and Newton s 3rd law Prereading: KJF 4.8 Frictional Forces Friction is a force exerted by a surface. The frictional force is always parallel to the surface Due to roughness of both
More information5.1 Vector and Scalar Quantities. A vector quantity includes both magnitude and direction, but a scalar quantity includes only magnitude.
Projectile motion can be described by the horizontal ontal and vertical components of motion. In the previous chapter we studied simple straightline motion linear motion. Now we extend these ideas to
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 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 information