Homework 6. problems: 8.-, 8.38, 8.63
|
|
- Clifford Dixon
- 6 years ago
- Views:
Transcription
1 Homework 6 problems: 8.-, 8.38, 8.63
2 Problem A circus trapeze consists of a bar suspended by two parallel ropes, each of length l. allowing performers to swing in a vertical circular arc. Suppose a performer with mass m holds the bar and steps off an elevated platform, starting from rest with the ropes at an angle θ i with respect to the vertical. Assume the size of the performer s body is small compared to the length l, she does not pump the trapeze to swing higher, and air resistance is negligible. (a) Show that when the ropes make an angle θ with the vertical, the performer must exert a force mg ( 3 cosθ - cosθi ) so as to hang on. (b) Determine the angle θ i for which the force needed to hang on at the bottom of the swing is twice as large as the gravitational force exerted on the performer. T θ l θ W v h a) According to Newton s third law of motion, the performer exert a force on the rope opposite to the force that the rope exerts on the performer. Motion of the performer is affected by two forces the gravitational force exerted by the earth and the tension force exerted by the ropes. The tension force adjusts itself to such a value that the radial component of the net force results in the centripetal acceleration of the performer. The radial component of the tension force is equal to tension while radial component of the gravitational force is equal to the projection of the weigh on the radial direction
3 () mac Fc T mgcosθ In circular motion, the centripetal acceleration depends on the speed of the performer and the radius of the path v () a c l It can be determined from the work-energy theorem. With the reference for potential energy at the lowest point, the her potential energy at an arbitrary point of the path is (3) U g mgh mg( l - l cosθ) Since the work due to tension is zero (at any point the tension force is perpendicular to the differential displacement), mechanical energy on the performed is conserved (4) mg( - l cosθ ) + 0 mg( l - l cosθ) i + l. (Note. One can also relate the change in kinetic energy to the net work (4 ) 0 ΔW net ΔW g mg( l - l cosθ ) mg( l - l cosθ) ) The rest is algebra. Solving (4) for square of speed v gl (cosθ cosθi) Substituting square of speed in equation (), the centripetal acceleration at the considered location of the performer is a c g(cosθ cosθ Finally solving equation () for the unknown tension T ma c i ) + mgcosθ 3mgcosθ mgcosθ b) At the lowest point θ 0. For tension to be twice as large as the weight of the performer, the initial angle satisfies the following equation mg 3mg mgcosθi from which θi aeccos 60 i i
4 Problem 8.38 A 659-kg elevator starts from rest. It moves upward for 3.0 s with constant acceleration until it reaches its cruising speed of.75 m/s. (a) What is the average power of the elevator motor during this time? (b) How does this power compare with the motor power when the elevator moves at its cruising speed? a) Solution. Using the definition of the average value of a function, the average power delivered by the motor over a certain time interval is related to the work done in this time interval t t t dw P dt M M dt dwm t dt t t ΔW P M av.m t t t t t t Δt There are two forces exerted on the elevator: the earth exerts the gravitational force and the cords (motor) exert the tensile force. According to the work-energy theorem, the resultant net work is equal to the change in the kinetic energy of the elevator. The gravitational work can be found from the displacement and weight of the elevator a( Δt) Δv Δt Δ Wg mgδh mg mg Hence Δv Δt 0 Pav,M Δt mg and P m.75 v v s m Δ g 659kg + t Δ 3s.75 m s av,m m s 5987W T W
5 Solution From the differential form of the work-energy theorem, we can determine the instantaneous value of the power delivered by the motor at an instant when the elevator has speed v and accelerates at rate a dwm + dwg dk from which dw dw dk d d P M g M + ( mgh) a + mgv dt dt dt dt dt From the definition of the average value of a function t t t dv ( a + mgv) dt m v dt + mg vdt t t dt t Pav,M t t t t Δt mgδh + Δt Δt ( Δt) mga + Δt Δt mgv + Δt The motor delivers power to balance the negative gravitational work and to increase the kinetic energy of the elevator. b) When the elevator reaches its cruising speed the motor is working at a constant power represented by an earlier function m m PM mgv 659kg W s s Despite at constant speed, of the elevator does not increase the kinetic energy of the elevator, at the higher speed the elevator requires higher power to increase its gravitational potential energy at higher rate.
6 Problem 8.63 A 0-kg block is released from point A as shown in Figure The track is frictionless except for the portion between B and C, which is 6 m long. The block travels down the track, hits a spring of force constant k,50 N/m and compresses the spring 0.3 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C. and the block. A h 3m B l 6m C x Let's assume that the block is a particle. During the motion, the block interacts with the earth (weight), the surface (friction), and the spring (tensile-compressive stress). Solution. We could use the first version (for the kinetic energy) of the work-energy theorem. The change in the kinetic energy of the block is zero. According to the work-energy theorem it must be equal to the work done by all forces exerted on the block. From the definition of potential energy, we can find the gravitational work and the work done by the spring. ) W ΔU ( 0 mgh) mgh g g ) Ws ΔUs 0 Directly from the definitions of work, the work due to the frictional force can be expressed in terms of the length of the rough part of the track. r r 3) W f d fl μnl μmgl f track
7 From the first version of the work-energy theorem we obtain 4) mgh μmgl 0 The rest is math. Solving for the coefficient of kinetic friction we find N mgh 50 ( 0.3m) 5) m μ mgl h l mg 6m 3m m 0kg 9.8 s 0.33 Solution. Two of the exerted forces, the gravitational and the normal (elastic), are conservative. The frictional force is not conservative. I will include the conservative interactions into the mechanical energy and use the second version (concerning the mechanical energy of the particle) of the work-energy theorem in the solution. Which position we choose for the reference of potential energy is arbitrary. I will choose the level of the horizontal part of the track as the reference potential energy for the gravitational interaction and a relaxed spring for the elastic interaction. We should have no problem determining the change in the mechanical energy between the two positions when the block's speed is zero. At point A, the kinetic energy and the potential energy due to the spring are zero. Therefore the mechanical energy is equal to the gravitational potential energy at this position 6) E K + U + U mgh A A g,a s, A The kinetic and gravitational potential energy at point D, where the block rests momentarily, are zero. The mechanical energy is equal to the elastic potential energy of the spring. 7) ED KD + Ug,D + Us,D where x is the compression of the spring from the relaxed position. There is only one nonconservative force exerted on the block during its motion, the frictional force. From the definition of work and
8 the nature of frictional force, we can relate the frictional work with the length of the rough part of the track. r r 8) W f d fl μnl μmgl f track I used the fact that the vertical component of acceleration is zero, therefore the magnitude of the normal force is equal to the magnitude of the weight. Now I can relate the expression for the work done by the frictional force with the change in the mechanical energy 9) W W ΔE f nc The rest is math. Combining all three equations, I can find the unknown friction coefficient: mgh Wf ΔE 0) μ h mg mg mg l l l l mg
Work and energy. 15 m. c. Find the work done by the normal force exerted by the incline on the crate.
Work and energy 1. A 10.0-kg crate is pulled 15.0 m up along a frictionless incline as shown in the figure below. The crate starts at rest and has a final speed of 6.00 m/s. motor 15 m 5 a. Draw the free-body
More informationPhys101 Second Major-152 Zero Version Coordinator: Dr. W. Basheer Monday, March 07, 2016 Page: 1
Phys101 Second Major-15 Zero Version Coordinator: Dr. W. Basheer Monday, March 07, 016 Page: 1 Q1. Figure 1 shows two masses; m 1 = 4.0 and m = 6.0 which are connected by a massless rope passing over a
More informationP8.14. m 1 > m 2. m 1 gh = 1 ( 2 m 1 + m 2 )v 2 + m 2 gh. 2( m 1. v = m 1 + m 2. 2 m 2v 2 Δh determined from. m 2 g Δh = 1 2 m 2v 2.
. Two objects are connected by a light string passing over a light frictionless pulley as in Figure P8.3. The object of mass m is released from rest at height h. Using the principle of conservation of
More informationPotential Energy & Conservation of Energy
PHYS 101 Previous Exam Problems CHAPTER 8 Potential Energy & Conservation of Energy Potential energy Conservation of energy conservative forces Conservation of energy friction Conservation of energy external
More informationAP Physics. Chapters 7 & 8 Review
AP Physics Chapters 7 & 8 Review 1.A particle moves along the x axis and is acted upon by a single conservative force given by F x = ( 20 4.0x)N where x is in meters. The potential energy associated with
More informationLAST NAME FIRST NAME DATE. Rotational Kinetic Energy. K = ½ I ω 2
LAST NAME FIRST NAME DATE Work, Energy and Power CJ - Assignment 3 6.5 The Conservation of Mechanical Energy Problems 3, 34, 38, 40 page 190 Work Kinetic Energy Rotational Kinetic Energy W = F d cosθ KE
More informationAP PHYSICS 1 UNIT 4 / FINAL 1 PRACTICE TEST
AP PHYSICS 1 UNIT 4 / FINAL 1 PRACTICE TEST NAME FREE RESPONSE PROBLEMS Put all answers on this test. Show your work for partial credit. Circle or box your answers. Include the correct units and the correct
More informationPotential energy functions used in Chapter 7
Potential energy functions used in Chapter 7 CHAPTER 7 CONSERVATION OF ENERGY Conservation of mechanical energy Conservation of total energy of a system Examples Origin of friction Gravitational potential
More informationOld Exams Questions Ch. 8 T072 Q2.: Q5. Q7.
Old Exams Questions Ch. 8 T072 Q2.: A ball slides without friction around a loop-the-loop (see Fig 2). A ball is released, from rest, at a height h from the left side of the loop of radius R. What is the
More informationPhys101 Second Major-162 Zero Version Coordinator: Dr. Kunwar S. Saturday, March 25, 2017 Page: N Ans:
Coordinator: Dr. Kunwar S. Saturday, March 25, 2017 Page: 1 Q1. Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 20 N, acting due east, and the other
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 informationA. B. C. D. E. v x. ΣF x
Q4.3 The graph to the right shows the velocity of an object as a function of time. Which of the graphs below best shows the net force versus time for this object? 0 v x t ΣF x ΣF x ΣF x ΣF x ΣF x 0 t 0
More informationChapter 8 Solutions. The change in potential energy as it moves from A to B is. The change in potential energy in going from A to B is
Chapter 8 Solutions *8. (a) With our choice for the zero level for potential energy at point B, U B = 0. At point A, the potential energy is given by U A = mgy where y is the vertical height above zero
More informationPHYS 101 Previous Exam Problems. Kinetic Energy and
PHYS 101 Previous Exam Problems CHAPTER 7 Kinetic Energy and Work Kinetic energy Work Work-energy theorem Gravitational work Work of spring forces Power 1. A single force acts on a 5.0-kg object in such
More informationPhys101 Second Major-162 Zero Version Coordinator: Dr. Kunwar S. Saturday, March 25, 2017 Page: 1
Coordinator: Dr. Kunwar S. Saturday, March 25, 2017 Page: 1 Q1. Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 20 N, acting due east, and the other
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 informationLesson 5. Luis Anchordoqui. Physics 168. Tuesday, September 26, 17
Lesson 5 Physics 168 1 C. B.-Champagne Luis Anchordoqui 2 2 Work Done by a Constant Force distance moved times component of force in direction of displacement W = Fd cos 3 Work Done by a Constant Force
More information1- A force F = ( 6ˆ i 2ˆ j )N acts on a particle that undergoes a displacement
1- A force F = ( 6ˆ i 2ˆ j )N acts on a particle that undergoes a displacement r = ( 3ˆ i + ˆ j )m. Find (a) the work done by the force on the particle and (b) the angle between F and r. 2- The force acting
More information(a) On the dots below that represent the students, draw and label free-body diagrams showing the forces on Student A and on Student B.
2003 B1. (15 points) A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on
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 informationAP Physics C: Work, Energy, and Power Practice
AP Physics C: Work, Energy, and Power Practice 1981M2. A swing seat of mass M is connected to a fixed point P by a massless cord of length L. A child also of mass M sits on the seat and begins to swing
More informationPOTENTIAL ENERGY AND ENERGY CONSERVATION
7 POTENTIAL ENERGY AND ENERGY CONSERVATION 7.. IDENTIFY: U grav = mgy so ΔU grav = mg( y y ) SET UP: + y is upward. EXECUTE: (a) ΔU = (75 kg)(9.8 m/s )(4 m 5 m) = +6.6 5 J (b) ΔU = (75 kg)(9.8 m/s )(35
More information(A) 10 m (B) 20 m (C) 25 m (D) 30 m (E) 40 m
PSI AP Physics C Work and Energy (Algebra Based) Multiple Choice Questions (use g = 10 m/s 2 ) 1. A student throws a ball upwards from the ground level where gravitational potential energy is zero. At
More informationCHAPTER 8: Conservation of Energy 1.)
CHAPTE 8: Conservation of Energy 1.) How We Got Here! We started by noticing that a force component acted along the line of a body s motion will affect the magnitude of the body s velocity. We multiplied
More informationPhysics 1 Second Midterm Exam (AM) 2/25/2010
Physics Second Midterm Eam (AM) /5/00. (This problem is worth 40 points.) A roller coaster car of m travels around a vertical loop of radius R. There is no friction and no air resistance. At the top of
More informationExam 2: Equation Summary
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 Physics Fall Term 2012 Exam 2: Equation Summary Newton s Second Law: Force, Mass, Acceleration: Newton s Third Law: Center of Mass: Velocity
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 informationPhysics 201, Midterm Exam 2, Fall Answer Key
Physics 201, Midterm Exam 2, Fall 2006 Answer Key 1) A constant force is applied to a body that is already moving. The force is directed at an angle of 60 degrees to the direction of the body s velocity.
More informationAP Physics 1: MIDTERM REVIEW OVER UNITS 2-4: KINEMATICS, DYNAMICS, FORCE & MOTION, WORK & POWER
MIDTERM REVIEW AP Physics 1 McNutt Name: Date: Period: AP Physics 1: MIDTERM REVIEW OVER UNITS 2-4: KINEMATICS, DYNAMICS, FORCE & MOTION, WORK & POWER 1.) A car starts from rest and uniformly accelerates
More information13.7 Power Applied by a Constant Force
13.7 Power Applied by a Constant Force Suppose that an applied force F a acts on a body during a time interval Δt, and the displacement of the point of application of the force is in the x -direction by
More informationOld Exam. Question Chapter 7 072
Old Exam. Question Chapter 7 072 Q1.Fig 1 shows a simple pendulum, consisting of a ball of mass M = 0.50 kg, attached to one end of a massless string of length L = 1.5 m. The other end is fixed. If the
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 informationPotential energy and conservation of energy
Chapter 8 Potential energy and conservation of energy Copyright 8.1_2 Potential Energy and Work Potential energy U is energy that can be associated with the configuration (arrangement) of a system of objects
More information(b) The mechanical energy would be 20% of the results of part (a), so (0 20)(920 m) 180 m.
PH Chapter 7 Solutions 7.4. IDENTIFY: The energy from the food goes into the increased gravitational potential energy of the hiker. We must convert food calories to joules. SET P: The change in gravitational
More informationPhys101 Lectures 9 and 10 Conservation of Mechanical Energy
Phys101 Lectures 9 and 10 Conservation of Mechanical Energy Key points: Conservative and Nonconservative Forces Potential Energy Generalized work-energy principle Mechanical Energy and Its Conservation
More informationCHAPTER 6 WORK AND ENERGY
CHAPTER 6 WORK AND ENERGY ANSWERS TO FOCUS ON CONCEPTS QUESTIONS (e) When the force is perpendicular to the displacement, as in C, there is no work When the force points in the same direction as the displacement,
More informationLecture 10. Potential energy and conservation of energy
Lecture 10 Potential energy and conservation of energy Today s Topics: Potential Energy and work done by conservative forces Work done by nonconservative forces Conservation of mechanical energy Potential
More informationl1, l2, l3, ln l1 + l2 + l3 + ln
Work done by a constant force: Consider an object undergoes a displacement S along a straight line while acted on a force F that makes an angle θ with S as shown The work done W by the agent is the product
More informationExam 1 Solutions. Kinematics and Newton s laws of motion
Exam 1 Solutions Kinematics and Newton s laws of motion No. of Students 80 70 60 50 40 30 20 10 0 PHY231 Spring 2012 Midterm Exam 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Raw Score 1. In which
More information- 1 -APPH_MidTerm. Mid - Term Exam. Part 1: Write your answers to all multiple choice questions in this space. A B C D E A B C D E
Name - 1 -APPH_MidTerm AP Physics Date Mid - Term Exam Part 1: Write your answers to all multiple choice questions in this space. 1) 2) 3) 10) 11) 19) 20) 4) 12) 21) 5) 13) 22) 6) 7) 14) 15) 23) 24) 8)
More informationis acting on a body of mass m = 3.0 kg and changes its velocity from an initial
PHYS 101 second major Exam Term 102 (Zero Version) Q1. A 15.0-kg block is pulled over a rough, horizontal surface by a constant force of 70.0 N acting at an angle of 20.0 above the horizontal. The block
More informationPSI AP Physics B Circular Motion
PSI AP Physics B Circular Motion Multiple Choice 1. A ball is fastened to a string and is swung in a vertical circle. When the ball is at the highest point of the circle its velocity and acceleration directions
More informationThe negative root tells how high the mass will rebound if it is instantly glued to the spring. We want
8.38 (a) The mass moves down distance.0 m + x. Choose y = 0 at its lower point. K i + U gi + U si + E = K f + U gf + U sf 0 + mgy i + 0 + 0 = 0 + 0 + kx (.50 kg)9.80 m/s (.0 m + x) = (30 N/m) x 0 = (60
More informationKinematics. v (m/s) ii. Plot the velocity as a function of time on the following graph.
Kinematics 1993B1 (modified) A student stands in an elevator and records his acceleration as a function of time. The data are shown in the graph above. At time t = 0, the elevator is at displacement x
More informationWork and Energy continued
Chapter 6 Work and Energy continued 6.2 The Work-Energy Theorem and Kinetic Energy Chapters 1 5 Motion equations were been developed, that relate the concepts of velocity, speed, displacement, time, and
More informationWorksheet #05 Kinetic Energy-Work Theorem
Physics Summer 08 Worksheet #05 June. 8, 08. A 0-kg crate is pulled 5 m up along a frictionless incline as shown in the figure below. The crate starts at rest and has a final speed of 6.0 m/s. (a) Draw
More informationMechanics and Heat. Chapter 5: Work and Energy. Dr. Rashid Hamdan
Mechanics and Heat Chapter 5: Work and Energy Dr. Rashid Hamdan 5.1 Work Done by a Constant Force Work Done by a Constant Force A force is said to do work if, when acting on a body, there is a displacement
More informationChapter Six News! DO NOT FORGET We ARE doing Chapter 4 Sections 4 & 5
Chapter Six News! DO NOT FORGET We ARE doing Chapter 4 Sections 4 & 5 CH 4: Uniform Circular Motion The velocity vector is tangent to the path The change in velocity vector is due to the change in direction.
More informationChapter 7 Energy of a System
Chapter 7 Energy of a System Course Outline : Work Done by a Constant Force Work Done by avarying Force Kinetic Energy and thework-kinetic EnergyTheorem Power Potential Energy of a System (Will be discussed
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 information(A) 10 m (B) 20 m (C) 25 m (D) 30 m (E) 40 m
Work/nergy 1. student throws a ball upward where the initial potential energy is 0. t a height of 15 meters the ball has a potential energy of 60 joules and is moving upward with a kinetic energy of 40
More informationPhysics 111. Lecture 15 (Walker: 7.1-2) Work & Energy March 2, Wednesday - Midterm 1
Physics 111 Lecture 15 (Walker: 7.1-2) Work & Energy March 2, 2009 Wednesday - Midterm 1 Lecture 15 1/25 Work Done by a Constant Force The definition of work, when the force is parallel to the displacement:
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 information(a) Find, in terms of a, g and θ, an expression for v 2. (3) (b) Find, in terms of m, g and θ, an expression for T. (4)
1. A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is fixed at the point O. The particle is initially held with OP horizontal and the
More informationPhysics 111: Mechanics Lecture 9
Physics 111: Mechanics Lecture 9 Bin Chen NJIT Physics Department Circular Motion q 3.4 Motion in a Circle q 5.4 Dynamics of Circular Motion If it weren t for the spinning, all the galaxies would collapse
More informationPhys101 Lectures 9 and 10 Conservation of Mechanical Energy
Phys101 Lectures 9 and 10 Conservation of Mechanical Energy Key points: Conservative and Nonconservative Forces Potential Energy Generalized work-energy principle Mechanical Energy and Its Conservation
More information= 1 2 kx2 dw =! F! d! r = Fdr cosθ. T.E. initial. = T.E. Final. = P.E. final. + K.E. initial. + P.E. initial. K.E. initial =
Practice Template K.E. = 1 2 mv2 P.E. height = mgh P.E. spring = 1 2 kx2 dw =! F! d! r = Fdr cosθ Energy Conservation T.E. initial = T.E. Final (1) Isolated system P.E. initial (2) Energy added E added
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 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 informationAP Physics C. Momentum. Free Response Problems
AP Physics C Momentum Free Response Problems 1. A bullet of mass m moves at a velocity v 0 and collides with a stationary block of mass M and length L. The bullet emerges from the block with a velocity
More informationRutgers University Department of Physics & Astronomy. 01:750:271 Honors Physics I Fall Lecture 10. Home Page. Title Page. Page 1 of 37.
Rutgers University Department of Physics & Astronomy 01:750:271 Honors Physics I Fall 2015 Lecture 10 Page 1 of 37 Midterm I summary 100 90 80 70 60 50 40 30 20 39 43 56 28 11 5 3 0 1 Average: 82.00 Page
More informationChapters 10 & 11: Energy
Chapters 10 & 11: Energy Power: Sources of Energy Tidal Power SF Bay Tidal Power Project Main Ideas (Encyclopedia of Physics) Energy is an abstract quantity that an object is said to possess. It is not
More information( ) = ( ) W net = ΔKE = KE f KE i W F. F d x. KE = 1 2 mv2. Note: Work is the dot product of F and d. Work-Kinetic Energy Theorem
Work-Kinetic Energy Theorem KE = 1 2 mv2 W F change in the kinetic energy of an object F d x net work done on the particle ( ) = ( ) W net = ΔKE = KE f KE i Note: Work is the dot product of F and d W g
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 informationAP Physics Free Response Practice Oscillations
AP Physics Free Response Practice Oscillations 1975B7. A pendulum consists of a small object of mass m fastened to the end of an inextensible cord of length L. Initially, the pendulum is drawn aside through
More information14.9 Worked Examples. Example 14.2 Escape Velocity of Toro
14.9 Wored Examples Example 14. Escape Velocity of Toro The asteroid Toro, discovered in 1964, has a radius of about R = 5.0m and a mass of about m t =.0 10 15 g. Let s assume that Toro is a perfectly
More informationRotation. PHYS 101 Previous Exam Problems CHAPTER
PHYS 101 Previous Exam Problems CHAPTER 10 Rotation Rotational kinematics Rotational inertia (moment of inertia) Kinetic energy Torque Newton s 2 nd law Work, power & energy conservation 1. Assume that
More informationConservative vs. Non-conservative forces Gravitational Potential Energy. Conservation of Mechanical energy
Next topic Conservative vs. Non-conservative forces Gravitational Potential Energy Mechanical Energy Conservation of Mechanical energy Work done by non-conservative forces and changes in mechanical energy
More informationfrictionless horizontal surface. The bullet penetrates the block and emerges with a velocity of o
AP Physics Free Response Practice Momentum and Impulse 1976B2. A bullet of mass m and velocity v o is fired toward a block of mass 4m. The block is initially at rest on a v frictionless horizontal surface.
More informationCommon Exam 3, Friday, April 13, :30 9:45 A.M. at KUPF 205 Chaps. 6, 7, 8. HW #8 and HW #9: Due tomorrow, April 6 th (Fri)
Common Exam 3, Friday, April 13, 2007 8:30 9:45 A.M. at KUPF 205 Chaps. 6, 7, 8 Bring calculators (Arrive by 8:15) HW #8 and HW #9: Due tomorrow, April 6 th (Fri) Today. Chapter 8 Hints for HW #9 Quiz
More informationPhysics 1A, Summer 2011, Summer Session 1 Quiz 3, Version A 1
Physics 1A, Summer 2011, Summer Session 1 Quiz 3, Version A 1 Closed book and closed notes. No work needs to be shown. 1. Three rocks are thrown with identical speeds from the top of the same building.
More informationLecture 18. Newton s Laws
Agenda: l Review for exam Lecture 18 l Assignment: For Monday, Read chapter 14 Physics 207: Lecture 18, Pg 1 Newton s Laws Three blocks are connected on the table as shown. The table has a coefficient
More information5. A car moves with a constant speed in a clockwise direction around a circular path of radius r, as represented in the diagram above.
1. The magnitude of the gravitational force between two objects is 20. Newtons. If the mass of each object were doubled, the magnitude of the gravitational force between the objects would be A) 5.0 N B)
More informationPhysics 116A, Section 2, Second Exam A, February 26, Name (Please print)
Physics 116A, Section 2, Second Exam A, February 26, 2008 Name (Please print) Mulitiple choice questions are worth 3 points each. Mark your answers in the space provided at the right, and on the OPSCAN
More informationPhysics 116A, Section 2, Second Exam Version B, February 26, Name (Please print)
Physics 116A, Section 2, Second Exam Version B, February 26, 2008 Name (Please print) Mulitiple choice questions are worth 3 points each. Mark your answers in the space provided at the right, and on the
More informationW = F x W = Fx cosθ W = Fx. Work
Ch 7 Energy & Work Work Work is a quantity that is useful in describing how objects interact with other objects. Work done by an agent exerting a constant force on an object is the product of the component
More informationPHYS Summer Professor Caillault Homework Solutions. Chapter 8
PHYS 1111 - Summer 007 - Professor Caillault Homework Solutions Chapter 8 5. Picture the Problem The physical situation is depicted at right. for the Strategy Use W sp = 1 k x i x f work done by the spring.
More informationPHYS 101 Previous Exam Problems. Force & Motion I
PHYS 101 Previous Exam Problems CHAPTER 5 Force & Motion I Newton s Laws Vertical motion Horizontal motion Mixed forces Contact forces Inclines General problems 1. A 5.0-kg block is lowered with a downward
More informationWritten Homework problems. Spring (taken from Giancoli, 4 th edition)
Written Homework problems. Spring 014. (taken from Giancoli, 4 th edition) HW1. Ch1. 19, 47 19. Determine the conversion factor between (a) km / h and mi / h, (b) m / s and ft / s, and (c) km / h and m
More informationKINETIC ENERGY AND WORK
Chapter 7: KINETIC ENERGY AND WORK 1 Which of the following is NOT a correct unit for work? A erg B ft lb C watt D newton meter E joule 2 Which of the following groups does NOT contain a scalar quantity?
More informationSolutions Midterm Exam 1 October 3, surface. You push to the left on the right block with a constant force F.
Problem 1 (2.5 points) Two blocks of mass m 1 and m 3, connected by a rod of mass m 2, are sitting on a frictionless surface. You push to the left on the right block with a constant force F. What is the
More informationt = g = 10 m/s 2 = 2 s T = 2π g
Annotated Answers to the 1984 AP Physics C Mechanics Multiple Choice 1. D. Torque is the rotational analogue of force; F net = ma corresponds to τ net = Iα. 2. C. The horizontal speed does not affect the
More informationD) No, because of the way work is defined D) remains constant at zero. D) 0 J D) zero
CHAPTER 6 REVIEW NAME 1) Can work be done on a system if there is no motion? A) Yes, if an outside force is provided. B) Yes, since motion is only relative. C) No, since a system which is not moving has
More informationExtra Circular Motion Questions
Extra Circular Motion Questions Elissa is at an amusement park and is driving a go-cart around a challenging track. Not being the best driver in the world, Elissa spends the first 10 minutes of her go-cart
More informationName: Date: Period: AP Physics C Work HO11
Name: Date: Period: AP Physics C Work HO11 1.) Rat pushes a 25.0 kg crate a distance of 6.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction
More informationPhysics 2211 ABC Quiz #3 Solutions Spring 2017
Physics 2211 ABC Quiz #3 Solutions Spring 2017 I. (16 points) A block of mass m b is suspended vertically on a ideal cord that then passes through a frictionless hole and is attached to a sphere of mass
More informationThe content contained in all sections of chapter 6 of the textbook is included on the AP Physics B exam.
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 is always
More informationPotential Energy and Conservation of Energy Chap. 7 & 8
Level : AP Physics Potential Energy and Conservation of Energy Chap. 7 & 8 Potential Energy of a System see p.191 in the textbook - Potential energy is the energy associated with the arrangement of a system
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 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 informationPower: Sources of Energy
Chapter 5 Energy Power: Sources of Energy Tidal Power SF Bay Tidal Power Project Main Ideas (Encyclopedia of Physics) Energy is an abstract quantity that an object is said to possess. It is not something
More information2. To study circular motion, two students use the hand-held device shown above, which consists of a rod on which a spring scale is attached.
1. A ball of mass M attached to a string of length L moves in a circle in a vertical plane as shown above. At the top of the circular path, the tension in the string is twice the weight of the ball. At
More informationCh 8 Conservation of Energy
Ch 8 Conservation of Energy Cons. of Energy It has been determined, through experimentation, that the total mechanical energy of a system remains constant in any isolated system of objects that interact
More informationRELEASED. Go to next page. 2. The graph shows the acceleration of a car over time.
1. n object is launched across a room. How can a student determine the average horizontal velocity of the object using a meter stick and a calculator? The student can calculate the object s initial potential
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 informationRegents Physics. Physics Midterm Review - Multiple Choice Problems
Name Physics Midterm Review - Multiple Choice Problems Regents Physics 1. A car traveling on a straight road at 15.0 meters per second accelerates uniformly to a speed of 21.0 meters per second in 12.0
More informationUnit 4 Work, Power & Conservation of Energy Workbook
Name: Per: AP Physics C Semester 1 - Mechanics Unit 4 Work, Power & Conservation of Energy Workbook Unit 4 - Work, Power, & Conservation of Energy Supplements to Text Readings from Fundamentals of Physics
More informationChapter 8. Potential Energy
Chapter 8 Potential Energy CHAPTER OUTLINE 8. Potential Energy of a System 8.2 The Isolated System Conservation of Mechanical Energy 8.3 Conservative and Nonconservative Forces 8.4 Changes in Mechanical
More informationExam 3 Practice Solutions
Exam 3 Practice Solutions Multiple Choice 1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest at the top of an inclined plane. If all three are released at
More informationChapter 8. Potential Energy and Energy Conservation
Chapter 8. Potential Energy and Energy Conservation Introduction In Ch 7 Work- Energy theorem. We learnt that total work done on an object translates to change in it s Kinetic Energy In Ch 8 we will consider
More information