PHYSICS 121 FALL Homework #3 - Solutions. Problems from Chapter 5: 3E, 7P, 11E, 15E, 34P, 45P
|
|
- Shavonne Ellis
- 5 years ago
- Views:
Transcription
1 PHYSICS 121 FALL Homework #3 - Solutions Problems from Chapter 5: 3E, 7P, 11E, 15E, 34P, 45P
2 3 We are only concerned with horizontal forces in this problem (gravity plays no direct role) We take East as the +x direction and North as +y This calculation is efficiently implemented on a vector capable calculator, using magnitude-angle notation (with SI units understood) F a = m = (90 0 )+( ) =(29 53 ) 30 Therefore, the acceleration has a magnitude of 29 m/s 2
3 7 We denote the two forces F 1 and F 2 According to Newton s second law, F 1 + F 2 = m a, so F 2 = m a F 1 (a) In unit vector notation F 1 =(200N)îand a = (12 sin 30 m/s 2 )î (12 cos 30 m/s 2 )ĵ= (60m/s 2 )î (104m/s 2 )ĵ Therefore, ( F 2 = (20kg) 60m/s 2) ( î+(20kg) 104m/s 2) ĵ (200N)î (b) The magnitude of F 2 is = ( 32 N)î (21 N)ĵ F 2 = F 2 2x + F 2 2y = ( 32) 2 +( 21) 2 =38N (c) The angle that F 2 makes with the positive x axis is found from tan θ = F 2y /F 2x =21/32 = 0656 Consequently, the angle is either 33 or = 213 Since both the x and y components are negative, the correct result is 213
4 11 We apply Eq 5-12 (a) The mass is m = W/g =(22N)/(98m/s 2 )=22kg At a place where g =49m/s 2, the mass is still 22 kg but the gravitational force is F g = mg =(22kg)(49m/s 2 )=11N (b) As noted, m =22 kg (c) At a place where g = 0 the gravitational force is zero (d) The mass is still 22kg
5 15 We note that the free-body diagram is shown in Fig 5-18 of the text (a) Since the acceleration of the block is zero,the components of the Newton s second law equation yield T mg sin θ =0andN mg cos θ = 0 Solving the first equation for the tension in the string, we find T = mg sin θ =(85 kg)(98m/s 2 ) sin 30 =42N (b) We solve the second equation in part (a) for the normal force N: N = mg cos θ =(85 kg)(98m/s 2 ) cos 30 =72N (c) When the string is cut,it no longer exerts a force on the block and the block accelerates The x component of the second law becomes mg sin θ = ma,so the acceleration becomes a = g sin θ = 98 sin 30 = 49 in SI units The negative sign indicates the acceleration is down the plane The magnitude of the acceleration is 49 m/s 2
6 34 First, we consider all the penguins (1 through 4, counting left to right) as one system, to which we apply Newton s second law: F net = (m 1 + m 2 + m 3 + m 4 )a 222 N = (20 kg +15 kg +m kg)a Second, we consider penguins 3 and 4 as one system, for which we have F net = (m 3 + m 4 )a 111 N = (m kg)a We solve these two equations for m 3 to obtain m 3 = 23 kg The solution step can be made a little easier, though, by noting that the net force on penguins 1 and 2 is also 111 N and applying Newton s law to them as a single system to solve first for a
7 45 The free-body diagram is shown below N is the normal force of the plane on the block and m g is the force of gravity on the block We N take the +x direction to be down the incline, in the direction of the acceleration, and the +y direction to be in the direction of the normal force exerted by the incline on the block The x com- (+x) θ ponent of Newton s second law is then mg sin θ = ma; thus,theacceleration is a = g sin m g θ (a) Placing the origin at the bottom of the plane, the kinematic equations (Table 2-1) for motion along the x axis which we will use are v 2 = v ax and v = v 0 + at The block momentarily stops at its highest point, where v = 0; according to the second equation, this occurs at time t = v 0 /a The position where it stops is v 2 0 x = 1 2 ( a ) = 1 ( 350 m/s) 2 2 (98m/s 2 ) sin 320 = 118 m (b) The time is t = v 0 a = v 0 g sin θ = 350 m/s (98m/s 2 ) sin 320 =0674 s (c) That the return-speed is identical to the initial speed is to be expected since there are no dissipative forces in this problem In order to prove this, one approach is to set x = 0 and solve x = v 0 t at2 for the total time (up and back down) t The result is The velocity when it returns is therefore t = 2v 0 a = 2v 0 g sin θ = 2( 350 m/s) (98m/s 2 ) sin 320 =135 s v = v 0 + at = v 0 + gt sin θ = (98)(135) sin 32 =350 m/s
8 PHYSICS 121 FALL Homework #3 - Solutions Problems from Chapter 6: 1E, 9P, 16P, 22P, 39E, 45P
9 1 We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion problem (with the person s push F in the +x direction) Applying Newton s second law to the x and y axes, we obtain F f s,max = ma N mg = 0 respectively The second equation yields the normal force N = mg, whereupon the maximum static friction is found to be (from Eq 6-1) f s,max = µ s mg Thus, the first equation becomes F µ s mg = ma =0 wherewehaveseta =0 to be consistent with the fact that the static friction is still (just barely) able to prevent the bureau from moving (a) With µ s =045 and m =45 kg, the equation above leads to F =198 N To bring the bureau into a state of motion, the person should push with any force greater than this value Rounding to two significant figures, we can therefore say the minimum required push is F = N (b) Replacing m =45 kg with m =28 kg, the reasoning above leads to roughly F = N
10 9 (a) The free-body diagram for the block is shown below F is the applied force, N is the normal force of the wall on the block, f is the force of friction, and m g is the force of gravity To determine if the block falls, we find the magnitude f of the force of friction required to hold it without accelerating and also find the normal force of the wall on the block Wecomparef and µ s N If f < µ s N, the block does f not slide on the wall but if f>µ s N, the block does slide The horizontal component of N F Newton s second law is F N =0,soN = F =12Nand µ s N =(060)(12 N) = 72N The vertical component is f mg =0,sof = mg =50N Since f<µ s N the block does m g not slide (b) Since the block does not move f =50N and N = 12 N The force of the wall on the block is F w = N î+f ĵ= (12 N) î+(50n)ĵ where the axes are as shown on Fig 6-21 of the text
11 16 We choose +x horizontally rightwards and +y upwards and observe that the 15 N force has components F x = F cos θ and F y = F sin θ (a)we apply Newton s second law to the y axis: N F sin θ mg =0 = N = (15)sin 40 +(35)(98)= 44 in SI units With µ k =025, Eq 6-2 leads to f k =11N (b)we apply Newton s second law to the x axis: F cos θ f k = ma = a = (15)cos =014 in SI units (m/s 2 ) Since the result is positive-valued, then the block is accelerating in the +x (rightward)direction
12 22 The free-body diagrams are shown below T is the magnitude of the tension force of the string, f is the magnitude of the force of friction on block A, N is the magnitude of the normal force T T of the plane on block A, m A g N is the force of gravity on body A (where m A = 10 kg), and A m B g is the force of gravity on f B block B θ = 30 is the angle of incline For A we take θ the +x to be uphill and +y to m be in the direction of the normal force; the positive direction A g m B g is chosen downward for block B Since A is moving down the incline, the force of friction is uphill with magnitude f k = µ k N (where µ k =020) Newton s second law leads to T f k + m A g sin θ = m A a =0 N m A g cos θ = 0 m B g T = m B a =0 for the two bodies (where a = 0 is a consequence of the velocity being constant) We solve these for the mass of block B m B = m A (sin θ µ k cos θ) =33 kg
13 39 The magnitude of the acceleration of the cyclist as it rounds the curve is given by v 2 /R, wherev is the speed of the cyclist and R is the radius of the curve Since the road is horizontal, only the frictional force of the road on the tires makes this acceleration possible The horizontal component of Newton s second law is f = mv 2 /R IfN is the normal force of the road on the bicycle and m is the mass of the bicycle and rider, the vertical component of Newton s second law leads to N = mg Thus, using Eq 6-1, the maximum value of static friction is f s,max = µ s N = µ s mg If the bicycle does not slip, f µ s mg This means v 2 R µ sg = R v2 µ s g Consequently, the minimum radius with which a cyclist moving at 29 km/h =81 m/s can round the curve without slipping is R min = v2 µ s g = 81 2 (032)(98) =21 m
14 45 The free-body diagram (for the airplane of mass m) is shown below We note that F l is the force of aerodynamic lift and a points rightwards in the figure We also note that a = v 2 /R where v = 480km/h = 133 m/s Applying Newton s law to the axes of the problem (+x rightward and +y upward) we obtain F l sin θ = m v2 R F l cos θ = mg where θ = 40 Eliminating mass from these equations leads to θ F l tan θ = v2 gr which yields R = v 2 /g tan θ = m m g
15 PHYSICS 121 FALL Homework #3 - Solutions Additional Problems: Drag =? m = 2000 kg Lift =? Force of Gravity Thrust 5000N I The airplane shown above flies with constant velocity in the horizontal direction There are four forces acting on the airplane The thrust force and drag force act in the horizontal direction The force of gravity and the lift force act in the vertical direction a If the thrust force has a magnitude of 5000N, what is the magnitude of drag force? Since the airplane flies at constant velocity the net force on the airplane must be zero Consequently both the horizontal component and the vertical component of the net force must be zero F netx = 0 = Thrust - Drag Consequently, Drag = Thrust = 5000N b If the mass of the airplane is 2000kg, what is the magnitude of the lift force? F nety = 0 = Lift - mg So, Lift = mg = 2000kg x 98 m/s 2 = 19600N or 20000N
16 50 kg n = 40N II A normal force with a magnitude of 40N pushes on a 50 kg mass vertically upward as shown above The only other force acting on the mass is the force of gravity a What is the magnitude of the acceleration of the 50 kg mass? F nety = n - mg = 40N - 50kg(98m/s2) = -9N F netx = 0 F net = 9N a = F net /m = 9N/50kg = 196 m/s 2 = 2 m/s 2 b What is the direction of the acceleration of the 50 kg mass? The direction of the acceleration is the same as the direction of the net force The net force is directed vertically down or - y direction
Easy. P5.3 For equilibrium: f = F and n = F g. Also, f = n, i.e., f n F F g. (a) 75.0 N N N N (b) ma y.
Chapter 5 Homework Solutions Easy P5.3 For equilibrium: f = F and n = F g. Also, f = n, i.e., (a) f n F F g s k 75.0 N 25.09.80 N 0.306 60.0 N 25.09.80 N 0.245 ANS. FIG. P5.3 P5.4 F y ma y : n mg 0 f s
More informationPHYSICS 221, FALL 2010 EXAM #1 Solutions WEDNESDAY, SEPTEMBER 29, 2010
PHYSICS 1, FALL 010 EXAM 1 Solutions WEDNESDAY, SEPTEMBER 9, 010 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively. In
More informationPHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009
PHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively.
More informationPhys101 First Major-111 Zero Version Monday, October 17, 2011 Page: 1
Monday, October 17, 011 Page: 1 Q1. 1 b The speed-time relation of a moving particle is given by: v = at +, where v is the speed, t t + c is the time and a, b, c are constants. The dimensional formulae
More informationWS-CH-4 Motion and Force Show all your work and equations used. Isaac Newton ( )
AP PHYSICS 1 WS-CH-4 Motion and Force Show all your work and equations used. Isaac Newton (1643-1727) Isaac Newton was the greatest English mathematician of his generation. He laid the foundation for differential
More information1. An excellent discussion and equation development related to this problem is given in Sample Problem 6-3. We merely quote (and apply) their main
1. An excellent discussion and equation development related to this problem is given in Sample Problem 6-3. We merely quote (and apply) their main result (Eq. 6-13) 1 θ = tan µ = tan 1 004.. s . The free-body
More informationAP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force).
AP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force). 1981M1. A block of mass m, acted on by a force of magnitude F directed horizontally to the
More informationChapter 4 Dynamics: Newton s Laws of Motion
Chapter 4 Dynamics: Newton s Laws of Motion 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 Force Applications
More informationPhys 1401: General Physics I
1. (0 Points) What course is this? a. PHYS 1401 b. PHYS 1402 c. PHYS 2425 d. PHYS 2426 2. (0 Points) Which exam is this? a. Exam 1 b. Exam 2 c. Final Exam 3. (0 Points) What version of the exam is this?
More informationPhys101 Second Major-131 Zero Version Coordinator: Dr. A. A. Naqvi Sunday, November 03, 2013 Page: 1
Coordinator: Dr. A. A. Naqvi Sunday, November 03, 2013 Page: 1 Q1. Two forces are acting on a 2.00 kg box. In the overhead view of Figure 1 only one force F 1 and the acceleration of the box are shown.
More informationPH 2213 : Chapter 05 Homework Solutions
PH 2213 : Chapter 05 Homework Solutions Problem 5.4 : The coefficient of static friction between hard rubber and normal street pavement is about 0.90. On how steep a hill (maximum angle) can you leave
More informationAP Physics 1 Dynamics Free Response Problems ANS KEY
AP Physics 1 Dynamics ree Response Problems ANS KEY 1. A block of mass m, acted on by a force directed horizontally, slides up an inclined plane that makes an angle θ with the horizontal. The coefficient
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) You are standing in a moving bus, facing forward, and you suddenly fall forward as the
More informationd. Determine the power output of the boy required to sustain this velocity.
AP Physics C Dynamics Free Response Problems 1. A 45 kg boy stands on 30 kg platform suspended by a rope passing over a stationary pulley that is free to rotate. The other end of the rope is held by the
More informationPhysics 2210 Homework 18 Spring 2015
Physics 2210 Homework 18 Spring 2015 Charles Jui April 12, 2015 IE Sphere Incline Wording A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle
More informationPhysics 2211 M Quiz #2 Solutions Summer 2017
Physics 2211 M Quiz #2 Solutions Summer 2017 I. (16 points) A block with mass m = 10.0 kg is on a plane inclined θ = 30.0 to the horizontal, as shown. A balloon is attached to the block to exert a constant
More information( ) ( ) 2. We apply Newton s second law (specifically, Eq. 5-2). (a) We find the x component of the force is. (b) The y component of the force is
1. We are only concerned with horizontal forces in this problem (gravity plays no direct role). We take East as the +x direction and North as +y. This calculation is efficiently implemented on a vector-capable
More informationFigure 5.1a, b IDENTIFY: Apply to the car. EXECUTE: gives.. EVALUATE: The force required is less than the weight of the car by the factor.
51 IDENTIFY: for each object Apply to each weight and to the pulley SET UP: Take upward The pulley has negligible mass Let be the tension in the rope and let be the tension in the chain EXECUTE: (a) The
More informationExam 2 Phys Fall 2002 Version A. Name ID Section
Closed book exam - Calculators are allowed. Only the official formula sheet downloaded from the course web page can be used. You are allowed to write notes on the back of the formula sheet. Use the scantron
More informationb) What does each letter (or symbol) stand for in this equation? c) What are the corresponding SI units? (Write: symbol $ unit).
First Name: Last Name: 1. a) What is Newton s Second Law in formula form? b) What does each letter (or symbol) stand for in this equation? c) What are the corresponding SI units? (Write: symbol $ unit).
More information1. A sphere with a radius of 1.7 cm has a volume of: A) m 3 B) m 3 C) m 3 D) 0.11 m 3 E) 21 m 3
1. A sphere with a radius of 1.7 cm has a volume of: A) 2.1 10 5 m 3 B) 9.1 10 4 m 3 C) 3.6 10 3 m 3 D) 0.11 m 3 E) 21 m 3 2. A 25-N crate slides down a frictionless incline that is 25 above the horizontal.
More informationChapter 5 Circular Motion; Gravitation
Chapter 5 Circular Motion; Gravitation Kinematics of Uniform Circular Motion Dynamics of Uniform Circular Motion Highway Curves, Banked and Unbanked Non-uniform Circular Motion Centrifugation Will be covered
More informationPHYSICS FORMULAS. A. B = A x B x + A y B y + A z B z = A B cos (A,B)
PHYSICS FORMULAS A = A x i + A y j Φ = tan 1 A y A x A + B = (A x +B x )i + (A y +B y )j A. B = A x B x + A y B y + A z B z = A B cos (A,B) linear motion v = v 0 + at x - x 0 = v 0 t + ½ at 2 2a(x - x
More informationPhys 1401: General Physics I
1. (0 Points) What course is this? a. PHYS 1401 b. PHYS 1402 c. PHYS 2425 d. PHYS 2426 2. (0 Points) Which exam is this? a. Exam 1 b. Exam 2 c. Final Exam 3. (0 Points) What version of the exam is this?
More informationChecking Understanding
Newton s First Law If there is no net force, the velocity of a mass remains constant (neither the magnitude nor the direction of the velocity changes). Objects at rest feel no net force. Objects in motion
More informationTopic: Force PHYSICS 231
Topic: Force PHYSICS 231 Current Assignments Homework Set 2 due this Thursday, Jan 27, 11 pm Reading for next week: Chapters 10.1-6,10.10,8.3 2/1/11 Physics 231 Spring 2011 2 Key Concepts: Force Free body
More informationPhysics 18 Spring 2011 Homework 4 Wednesday February 9, 2011
Physics 18 Spring 2011 Homework 4 Wednesday February 9, 2011 Make sure your name is on your homework, and please box your final answer. Because we will be giving partial credit, be sure to attempt all
More informationYear 11 Physics Tutorial 84C2 Newton s Laws of Motion
Year 11 Physics Tutorial 84C2 Newton s Laws of Motion Module Topic 8.4 Moving About 8.4.C Forces Name Date Set 1 Calculating net force 1 A trolley was moved to the right by a force applied to a cord attached
More informationy(t) = y 0 t! 1 2 gt 2. With y(t final ) = 0, we can solve this for v 0 : v 0 A ĵ. With A! ĵ =!2 and A! = (2) 2 + (!
1. The angle between the vector! A = 3î! 2 ĵ! 5 ˆk and the positive y axis, in degrees, is closest to: A) 19 B) 71 C) 90 D) 109 E) 161 The dot product between the vector! A = 3î! 2 ĵ! 5 ˆk and the unit
More informationDynamic equilibrium: object moves with constant velocity in a straight line. = 0, a x = i
Dynamic equilibrium: object moves with constant velocity in a straight line. We note that F net a s are both vector quantities, so in terms of their components, (F net ) x = i (F i ) x = 0, a x = i (a
More informationFriction is always opposite to the direction of motion.
6. Forces and Motion-II Friction: The resistance between two surfaces when attempting to slide one object across the other. Friction is due to interactions at molecular level where rough edges bond together:
More informationPhysics 8 Wednesday, October 11, 2017
Physics 8 Wednesday, October 11, 2017 HW5 due Friday. It s really Friday this week! Homework study/help sessions (optional): Bill will be in DRL 2C6 Wednesdays from 4 6pm (today). Grace will be in DRL
More informationCHAPTER 4 NEWTON S LAWS OF MOTION
62 CHAPTER 4 NEWTON S LAWS O MOTION CHAPTER 4 NEWTON S LAWS O MOTION 63 Up to now we have described the motion of particles using quantities like displacement, velocity and acceleration. These quantities
More informationPhysics 125: Classical Physics A. 1 Practice Problems for Midterm Exam 1
Physics 125: Classical Physics A 1 Practice Problems for Midterm Exam 1 Problem 1 The Figure 1 depicts velocity as a function of time for a short run. Find: a) The acceleration at t = 5 seconds. b) The
More informationw = mg Use: g = 10 m/s 2 1 hour = 60 min = 3600 sec
The exam is closed book and closed notes. Part I: There are 1 multiple choice questions, 1 point each. The answers for the multiple choice questions are to be placed on the SCANTRON form provided. Make
More information2. We apply Newton s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the chopping block is F F F
Chapter 5 1. We are only concerned with horizontal forces in this problem (gravity plays no direct role). We take East as the +x direction and North as +y. This calculation is efficiently implemented on
More informationExam 1 Solutions. PHY 2048 Spring 2014 Acosta, Rinzler. Note that there are several variations of some problems, indicated by choices in parentheses.
Exam 1 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 Let vector a! = 4î + 3 ĵ and vector b! = î + 2 ĵ (or b! = î + 4 ĵ ). What is the
More informationForces and Motion in One Dimension
Nicholas J. Giordano www.cengage.com/physics/giordano Forces and Motion in One Dimension Applications of Newton s Laws We will learn how Newton s Laws apply in various situations We will begin with motion
More informationPhysics. TOPIC : Friction. 1. To avoid slipping while walking on ice, one should take smaller steps because of the
TOPIC : Friction Date : Marks : 0 mks Time : ½ hr. To avoid slipping while walking on ice, one should take smaller steps because of the Friction of ice is large (b Larger normal reaction (c Friction of
More informationLecture PowerPoints. Chapter 5 Physics for Scientists & Engineers, with Modern Physics, 4 th edition. Giancoli
Lecture PowerPoints Chapter 5 Physics for Scientists & Engineers, with Modern Physics, 4 th edition 2009 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely
More informationChapter 8: Newton s Laws Applied to Circular Motion
Chapter 8: Newton s Laws Applied to Circular Motion Centrifugal Force is Fictitious? F actual = Centripetal Force F fictitious = Centrifugal Force Center FLEEing Centrifugal Force is Fictitious? Center
More informationThe Laws of Motion. Newton s first law Force Mass Newton s second law Gravitational Force Newton s third law Examples
The Laws of Motion Newton s first law Force Mass Newton s second law Gravitational Force Newton s third law Examples Gravitational Force Gravitational force is a vector Expressed by Newton s Law of Universal
More information24/06/13 Forces ( F.Robilliard) 1
R Fr F W 24/06/13 Forces ( F.Robilliard) 1 Mass: So far, in our studies of mechanics, we have considered the motion of idealised particles moving geometrically through space. Why a particular particle
More informationChapter 5 Gravitation Chapter 6 Work and Energy
Chapter 5 Gravitation Chapter 6 Work and Energy Chapter 5 (5.6) Newton s Law of Universal Gravitation (5.7) Gravity Near the Earth s Surface Chapter 6 (today) Work Done by a Constant Force Kinetic Energy,
More information(a) What is the normal (compression) component of the force by the slide on the girl?
Question (52) Normal force on a girl sliding down a waterslide Wet n Wild has a waterslide called Daredevil Drop. At the steepest part of the slide, the angle is about 65. A girl goes down the slide. Her
More information= constant of gravitation is G = N m 2 kg 2. Your goal is to find the radius of the orbit of a geostationary satellite.
Problem 1 Earth and a Geostationary Satellite (10 points) The earth is spinning about its axis with a period of 3 hours 56 minutes and 4 seconds. The equatorial radius of the earth is 6.38 10 6 m. The
More informationHSC PHYSICS ONLINE B F BA. repulsion between two negatively charged objects. attraction between a negative charge and a positive charge
HSC PHYSICS ONLINE DYNAMICS TYPES O ORCES Electrostatic force (force mediated by a field - long range: action at a distance) the attractive or repulsion between two stationary charged objects. AB A B BA
More informationFriction forces. Lecture 8. Chapter 6. Physics I. Course website:
Lecture 8 Physics I Chapter 6 Friction forces Course website: http://faculty.uml.edu/andriy_danylov/teaching/physicsi Today we are going to discuss: Chapter 6: Some leftover (Ch.5) Kinetic/Static Friction:
More informationGet Solution of These Packages & Learn by Video Tutorials on FRICTION
1. FRICTION : When two bodies are kept in contact, electromagnetic forces act between the charged particles (molecules) at the surfaces of the bodies. Thus, each body exerts a contact force of the other.
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 informationexample Δy gravity Δy can
Physic 3 Lecture 5 Main points of today s lecture: Newton s st law: If there is no net force, the velocity of a mass remains constant (neither the magnitude nor the direction of the velocity changes).
More informationWeek 4 Homework/Recitation: 9/21/2017 Chapter4: Problems 3, 5, 11, 16, 24, 38, 52, 77, 78, 98. is shown in the drawing. F 2
Week 4 Homework/Recitation: 9/1/017 Chapter4: Problems 3, 5, 11, 16, 4, 38, 5, 77, 78, 98. 3. Two horizontal forces, F 1 and F, are acting on a box, but only F 1 is shown in the drawing. F can point either
More informationConcept of Force and Newton s Laws of Motion
Concept of Force and Newton s Laws of Motion 8.01 W02D2 Chapter 7 Newton s Laws of Motion, Sections 7.1-7.4 Chapter 8 Applications of Newton s Second Law, Sections 8.1-8.4.1 Announcements W02D3 Reading
More informationForce - a push or a pull A force described by its strength and by the direction in which it acts The SI unit for force is the newton (N)
Forces Force - a push or a pull A force described by its strength and by the direction in which it acts The SI unit for force is the newton (N) The direction and strength of forces can be represented by
More informationEQUATIONS OF MOTION: RECTANGULAR COORDINATES
EQUATIONS OF MOTION: RECTANGULAR COORDINATES Today s Objectives: Students will be able to: 1. Apply Newton s second law to determine forces and accelerations for particles in rectilinear motion. In-Class
More information= 40 N. Q = 60 O m s,k
Sample Exam #2 Technical Physics Multiple Choice ( 6 Points Each ): F app = 40 N 20 kg Q = 60 O = 0 1. A 20 kg box is pulled along a frictionless floor with an applied force of 40 N. The applied force
More informationVersion PREVIEW Semester 1 Review Slade (22222) 1
Version PREVIEW Semester 1 Review Slade () 1 This print-out should have 48 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Holt SF 0Rev 10A
More informationWiley Plus. Final Assignment (5) Is Due Today: Before 11 pm!
Wiley Plus Final Assignment (5) Is Due Today: Before 11 pm! Final Exam Review December 9, 009 3 What about vector subtraction? Suppose you are given the vector relation A B C RULE: The resultant vector
More informationGeneral Physics I Spring Applying Newton s Laws
General Physics I Spring 2011 Applying Newton s Laws 1 Equilibrium An object is in equilibrium if the net force acting on it is zero. According to Newton s first law, such an object will remain at rest
More information= v 0 x. / t = 1.75m / s 2.25s = 0.778m / s 2 nd law taking left as positive. net. F x ! F
Multiple choice Problem 1 A 5.-N bos sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the bos sliding to the right at 1.75 m/s
More informationPhys101 First Major-061 Zero Version Coordinator: Abdelmonem Monday, October 30, 2006 Page: 1
Coordinator: Abdelmonem Monday, October 30, 006 Page: 1 Q1. An aluminum cylinder of density.70 g/cm 3, a radius of.30 cm, and a height of 1.40 m has the mass of: A) 6.8 kg B) 45.1 kg C) 13.8 kg D) 8.50
More information= M. L 2. T 3. = = cm 3
Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 1 Q1. Work is defined as the scalar product of force and displacement. Power is defined as the rate of change of work with time. The dimension
More informationPS113 Chapter 4 Forces and Newton s laws of motion
PS113 Chapter 4 Forces and Newton s laws of motion 1 The concepts of force and mass A force is described as the push or pull between two objects There are two kinds of forces 1. Contact forces where two
More information4.1 Forces. Chapter 4 The Laws of Motion
4.1 Forces Chapter 4 he Laws of Motion 4.2 Newton s First Law it s not the nature of an object to stop, once set in motion, but rather to continue in its original state of motion. An object moves with
More informationExample. F and W. Normal. F = 60cos 60 N = 30N. Block accelerates to the right. θ 1 F 1 F 2
Physic 3 Lecture 7 Newton s 3 d Law: When a body exerts a force on another, the second body exerts an equal oppositely directed force on the first body. Frictional forces: kinetic friction: fk = μk N static
More informationCourse Name : Physics I Course # PHY 107. Lecture-5 : Newton s laws - Part Two
Course Name : Physics I Course # PHY 107 Lecture-5 : Newton s laws - Part Two Abu Mohammad Khan Department of Mathematics and Physics North South University https://abukhan.weebly.com Copyright: It is
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Diagram 1 A) B - A. B) A - B. C) A + B. D) A B.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) In the diagram shown, the unknown vector is 1) Diagram 1 A) B - A. B) A - B. C) A + B.
More informationChapter 4: Newton s Second Law F = m a. F = m a (4.2)
Lecture 7: Newton s Laws and Their Applications 1 Chapter 4: Newton s Second Law F = m a First Law: The Law of Inertia An object at rest will remain at rest unless, until acted upon by an external force.
More informationPHYSICS 231 INTRODUCTORY PHYSICS I
PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 6 Last Lecture: Gravity Normal forces Strings, ropes and Pulleys Today: Friction Work and Kinetic Energy Potential Energy Conservation of Energy Frictional Forces
More informationRutgers University Department of Physics & Astronomy. 01:750:271 Honors Physics I Fall Lecture 8. Home Page. Title Page. Page 1 of 35.
Rutgers University Department of Physics & Astronomy 01:750:271 Honors Physics I Fall 2015 Lecture 8 Page 1 of 35 Midterm 1: Monday October 5th 2014 Motion in one, two and three dimensions Forces and Motion
More informationPhysics 53 Summer Exam I. Solutions
Exam I Solutions In questions or problems not requiring numerical answers, express the answers in terms of the symbols for the quantities given, and standard constants such as g. In numerical questions
More informationMechanics II. Which of the following relations among the forces W, k, N, and F must be true?
Mechanics II 1. By applying a force F on a block, a person pulls a block along a rough surface at constant velocity v (see Figure below; directions, but not necessarily magnitudes, are indicated). Which
More informationPhysics 2211 A & B Quiz #2 Solutions Fall P sin θ = µ k P cos θ + mg
Physics 2211 A & B Quiz #2 Solutions Fall 2016 I. (16 points) A block of mass m is sliding up a vertical wall at constant non-zero velocity v 0, due to an applied force P pushing against it at an angle
More informationForces and Newton s Laws Reading Notes. Give an example of a force you have experienced continuously all your life.
Forces and Newton s Laws Reading Notes Name: Section 4-1: Force What is force? Give an example of a force you have experienced continuously all your life. Give an example of a situation where an object
More informationAP Physics Free Response Practice Dynamics
AP Physics Free Response Practice Dynamics 14) In the system shown above, the block of mass M 1 is on a rough horizontal table. The string that attaches it to the block of mass M 2 passes over a frictionless
More informationPhysics 2211 A & B Quiz #4 Solutions Fall 2016
Physics 22 A & B Quiz #4 Solutions Fall 206 I. (6 points) A pendulum bob of mass M is hanging at rest from an ideal string of length L. A bullet of mass m traveling horizontally at speed v 0 strikes it
More informationEnergy present in a variety of forms. Energy can be transformed form one form to another Energy is conserved (isolated system) ENERGY
ENERGY Energy present in a variety of forms Mechanical energy Chemical energy Nuclear energy Electromagnetic energy Energy can be transformed form one form to another Energy is conserved (isolated system)
More informationDynamics; Newton s Laws of Motion
Dynamics; Newton s Laws of Motion Force A force is any kind of push or pull on an object. An object at rest needs a force to get it moving; a moving object needs a force to change its velocity. The magnitude
More informationPHYSICS 231 Laws of motion PHY 231
PHYSICS 231 Laws of motion 1 Newton s Laws First Law: If the net force exerted on an object is zero the object continues in its original state of motion; if it was at rest, it remains at rest. If it was
More informationPhysics 101 Lecture 5 Newton`s Laws
Physics 101 Lecture 5 Newton`s Laws Dr. Ali ÖVGÜN EMU Physics Department The Laws of Motion q Newton s first law q Force q Mass q Newton s second law q Newton s third law qfrictional forces q Examples
More informationPYP 001 FIRST MAJOR EXAM CODE: TERM: 151 SATURDAY, OCTOBER 17, 2015 PAGE: 1
TERM: 151 SATURDAY, OCTOBER 17, 2015 PAGE: 1 *Read the following (20) questions and choose the right answer: 1 The figure below represents the speed-time graph for the motion of a vehicle during a 7.0-minute
More informationTopic 2: Mechanics 2.2 Forces
Representing forces as vectors A force is a push or a pull measured in Newtons. One force we are very familiar with is the force of gravity, AKA the weight. The very concepts of push and pull imply direction.
More informationCircular Motion Dynamics Concept Questions
Circular Motion Dynamics Concept Questions Problem 1: A puck of mass m is moving in a circle at constant speed on a frictionless table as shown above. The puck is connected by a string to a suspended bob,
More informationLECTURE 11 FRICTION AND DRAG
LECTURE 11 FRICTION AND DRAG 5.5 Friction Static friction Kinetic friction 5.6 Drag Terminal speed Penguins travel on ice for miles by sliding on ice, made possible by small frictional force between their
More informationNewton s Three Laws. F = ma. Kinematics. Gravitational force Normal force Frictional force Tension More to come. k k N
Newton s Three Laws F = ma Gravitational force Normal force Frictional force Tension More to come Kinematics f s,max = µ sfn 0 < fs µ sfn k k N f = µ F Rules for the Application of Newton s Laws of Motion
More information66 Chapter 6: FORCE AND MOTION II
Chapter 6: FORCE AND MOTION II 1 A brick slides on a horizontal surface Which of the following will increase the magnitude of the frictional force on it? A Putting a second brick on top B Decreasing the
More informationLaws of Motion. A fighter aircraft is looping in a vertical plane. The minimum velocity at the highest point is (Given r = radius of the loop) a) gr b) gr c) gr d) 3gr. In non-inertial frame, the second
More informationPHYSICS 231 INTRODUCTORY PHYSICS I
PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 4 Main points of last lecture Scalars vs. Vectors Vectors A: (A x, A y ) or A & θ Addition/Subtraction Projectile Motion X-direction: a x = 0 (v x = constant)
More informationPHYSICS 221 SPRING EXAM 1: February 20, 2014; 8:15pm 10:15pm
PHYSICS 221 SPRING 2014 EXAM 1: February 20, 2014; 8:15pm 10:15pm Name (printed): Recitation Instructor: Section # INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit questions,
More informationChapter Work, Energy and Power. Q1. The co-efficient of restitution e for a perfectly elastic collision is [1988] (a) 1 (b) 0 (c) (d) 1 Ans: (a)
Chapter Work, Energy and Power Q1. The co-efficient of restitution e for a perfectly elastic collision is [1988] (a) 1 (b) 0 (c) (d) 1 Q2. A bullet of mass 10g leaves a rifle at an initial velocity of
More informationForce 10/01/2010. (Weight) MIDTERM on 10/06/10 7:15 to 9:15 pm Bentley 236. (Tension)
Force 10/01/2010 = = Friction Force (Weight) (Tension), coefficient of static and kinetic friction MIDTERM on 10/06/10 7:15 to 9:15 pm Bentley 236 2008 midterm posted for practice. Help sessions Mo, Tu
More informationChapter 5. Force and Motion-I
Chapter 5 Force and Motion-I 5.3 Newton s First Law Newton s First Law: If no force acts on a body, the body s velocity cannot change The purpose of Newton s First Law is to introduce the special frames
More informationAP Physics First Nine Weeks Review
AP Physics First Nine Weeks Review 1. If F1 is the magnitude of the force exerted by the Earth on a satellite in orbit about the Earth and F2 is the magnitude of the force exerted by the satellite on the
More informationPhysics 2514 Lecture 13
Physics 2514 Lecture 13 P. Gutierrez Department of Physics & Astronomy University of Oklahoma Physics 2514 p. 1/18 Goals We will discuss some examples that involve equilibrium. We then move on to a discussion
More informationPhysics 8 Monday, October 9, 2017
Physics 8 Monday, October 9, 2017 Pick up a HW #5 handout if you didn t already get one on Wednesday. It s due this Friday, 10/13. It contains some Ch9 (work) problems, some Ch10 (motion in a plane) problems,
More informationBell Ringer: What is Newton s 3 rd Law? Which force acts downward? Which force acts upward when two bodies are in contact?
Bell Ringer: What is Newton s 3 rd Law? Which force acts downward? Which force acts upward when two bodies are in contact? Does the moon attract the Earth with the same force that the Earth attracts the
More information(35+70) 35 g (m 1+m 2)a=m1g a = 35 a= =3.27 g 105
Coordinator: Dr. W. L-Basheer Monday, March 16, 2015 Page: 1 Q1. 70 N block and a 35 N block are connected by a massless inextendable string which is wrapped over a frictionless pulley as shown in Figure
More informationExamples Newton's Laws and Friction
Examples Newton's Laws and Friction 1. A 10.0 kg box is sitting on a table. (A) If a 49 N force is required to overcome friction and start the block moving, calculate the coefficient of static friction.
More informationSolution of HW4. and m 2
Solution of HW4 9. REASONING AND SOLUION he magnitude of the gravitational force between any two of the particles is given by Newton's law of universal gravitation: F = Gm 1 m / r where m 1 and m are the
More informationDynamics: Forces and Newton s Laws of Motion
Lecture 7 Chapter 5 Dynamics: Forces and Newton s Laws of Motion Course website: http://faculty.uml.edu/andriy_danylov/teaching/physicsi Today we are going to discuss: Chapter 5: Force, Mass: Section 5.1
More information