1 / 43

physics

Final Exam Review (Day 1). physics. Energy Lecture Review Kinetic & Potential Energy Net Work ( W net = F net D x = F net cos q ) Work-Kinetic Energy Theorem ( W = D KE ) Conservation of Energy ( KE i + PE i = KE f + PE f ) Problem Review (Group Problems) Energy Net Forces.

ingrid
Download Presentation

physics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Final Exam Review (Day 1) physics

  2. Energy Lecture Review • Kinetic & Potential Energy • Net Work (Wnet = FnetDx = Fnetcosq) • Work-Kinetic Energy Theorem (W = DKE) • Conservation of Energy (KEi + PEi = KEf + PEf) • Problem Review (Group Problems) • Energy • Net Forces Final Exam review: day 1 overview

  3. Potential Energy • Stored Energy In Which An Object May Use To Perform Work On Another Object • Gravitational Potential (PE = mgh) • Elastic Potential (PE = ½ kx2) • Chemical Potential (PE = nCvDT) • Kinetic Energy • Energy Achieved By An Object In Motion • Kinetic Energy (KE = ½ mv2) Energy

  4. Work Done On An Object With Multiple Forces Is The Net Work For A Net Force Applied At An Angle, The Net Work Is The Product Of The Component Of Force In The Direction Of Motion & Its Displacement Wnet = FnetDx = Fnet cos q Net Work

  5. Work Done On (Net External Force) An Object Results In A Change In The Object’s Kinetic Energy • Work = Force x Displacement or W = FDx • Work-KE Theorem Is Equivelent To The Following Expression: W = KEf – KEi = ½ mvf2 – ½ mvi2 Work-KE Theorem

  6. Total Mechanical Energy (ME = KE + PE) Of An Object Remains Constant As The Object Moves • Assume No Net Work Done By External Forces (e.g. Friction, Air Resistance, Pressure, etc.) MEi = MEf KEi + PEi = KEf + PEf Conservation of ENergy

  7. The Drawing Shows A Boat Being Pulled By Two Locomotives Through A Canal Of Length 2.00 km. The Tension In Each Cable Is 5.00 x 103 N, And q = 20.0o. What Is The Net Work Done On The Boat By The Two Locomotives? Problem #1: work

  8. A 0.075 kg Arrow Is Fired Horizontally. The Bowstring Exerts An Average Force Of 65 N On The Arrow Over A Distance Of 0.90 m. With What Speed Does The Arrow Leave The Bow? Problem #2: Work & Energy

  9. A Cyclist Approaches The Bottom Of A gradual Hill AT A Speed Of 11 m/s. The Hill Is 5.0 m High, And The Cyclist Estimates That She Is Going Fast Enough To Coast Up And Over It Without Peddling. Ignoring Air Resistance And Friction, Find The Speed AT Which The Cyclist Crests The Hill. Problem #3: energy

  10. Final Exam Review (Day 2) physics

  11. Momentum Lecture Review • Impulse (I = Fnett = Dp = pf – pi) • Conservation Of Momentum (Pi = Pf) • Collisions (Elastic & Inelastic) • Circular Motion (ac = vt2/R) • Gravitation (Fg = Gm1m2/d2) • Problem Review (Group Problems) • Momentum • Collisions • Circular Motion Final Exam review: day 2 overview

  12. When A Large Force Acts on An Object For A Sufficient Amount Of Time, Their Product Is The Impulse Of The Force I = Fnett • The Object Will Thusly Experience A Change In Velocity (Or Change In Momentum) As Shown From Newton’s 2nd Law: I = Fnett = Dp = pf – pi Impulse

  13. The Total Linear Momentum Of An Isolated System Remains Constant • An Isolated System Is One For Which The Vector Sum Of The External Forces Acting On The System Is Zero Pi = Pf mv1i + mv2i = mv1f + mv2f momentum

  14. collisions • Elastic Collision Is One In Which The Total KE Of The System After The Collision Is Equal To The Total KE Before The Collision • Inelastic Collision Is One In Which The Total KE Of The System After The Collision Is Not Equal To The Total KE Before The Collision

  15. Uniform Circular Motion Is The Motion Of An Object Traveling At A Constant (Uniform) Speed On A Circular Path • Even Though An Object In Circular Motion Has A Constant Velocity (VT) It Experiences An Acceleration Toward The Center Of The Circular Path ac = vt2/R Circular motion

  16. Objects With Mass Are Seemingly Attracted To Each Other Through The Gravitational Force Fg = Gm1m2/d2 G Is The Gravitational Constant And Has An Experimentally Determined Value of: G = 6.67 x 10-11 Nm/kg2 gravitation

  17. Problem #1: Impulse • During A Storm, Rain Comes Straight Down With A Velocity Of vo = -15 m/s And Comes To Rest After Impacting The Car. If The Rain Drops Have A Mass Of 0.060 kg & It Takes 1.0s To Come To A Rest, What Is The Force Exerted By The Rain On The Car Roof? • If Hail Fell Instead Of Rain, Would The Force On The Roof Be Smaller Than, Equal To , Or Greater Than That Of The Raindrop?

  18. A Ball Of Mass m1 = 0.250 kg And Velocity v1i = +5.0 m/s Collides Head-on With A Ball Of Mass m2 = 0.800 kg That Is Initially At Rest. No External Forces Act On The Balls. If The Collision Is Elastic, What Are The Velocities Of The Balls After The Collision? Problem #2: momentum

  19. How Long Does It Take A Plane, Traveling At A Constant Speed Of 110 m/s, To Fly Once Around A Circle Whose Radius Is 2850 m? Problem #3: Circular motion

  20. Final Exam Review (Day 3) physics

  21. Kinematics Lecture Review • Displacement, Velocity & Acceleration • Speed Versus Velocity • Graphical Models Of Motion • Kinematic Equations • Freefall • 2-D Kinematics • Horizontal Projectile Motion • Projectile Motion @ Angle • Problem Review (Group Problems) • 1-D Kinematics • 2-D Kinematics Final Exam review: day 3 overview

  22. Displacement Is A Vector That Points From An Object’s Initial Position Toward Its Final Position (Shortest Distance Between Two Points) Dx = xf - xi • Velocity Is Defined As The Displacement (Change In Position) Of An Object Divided By The Change In Time v = Dx/Dt • Acceleration Is The Change In Velocity Of An Object Over The Change In Time a = Dv/Dt Displacement, Velocity & Acceleration

  23. What Is Speed? • Speed Is An Average Velocity • If I Asked How Fast You Drove From Home To School Today What Would You Say? • What About Velocity Then? • Velocity Is Instantaneous, It Changes From One Second To The Next • Watch Your Speedometer Next Time You’re Driving! Speed versus velocity

  24. Position Versus Time (Dx versus Dt) Velocity Versus Time (Dv versus Dt) Graphical models of motion Slope = Dx/Dt = Velocity Dx Dt Dv Slope = Dv/Dt = Acceleration Dt

  25. Kinematics is the study of objects’ motion at constant acceleration There are 4 kinematic equations Use these to solve all 1-D and 2-D motion problems! Kinematic equations

  26. Kinematic equations

  27. FREEFALL • Freefalling Bodies Move Freely Under The Influence Of Gravity ONLY! • The Acceleration Of An Object In Freefall Is ALWAYS The Acceleration Of Gravity ag = -9.81 m/s2 • Use Kinematic Equations To Solve Freefall Problems • What Is The Acceleration Of An Object Thrown Upwards? • What Is The Velocity Of An Object At The Peak Of Its Motion?

  28. Projectile Motion • Horizontal Projectile Motion • Projectile Motion At An Angle • Problem Solving Method • First Commandment Of Physics (Keep X & Y Directions Separate) • Solve For Time In Y-Direction Using A Kinematic Equation • Use Vix = Dx/DtIn The X-Direction 2-d kinematics

  29. Problem #1: Graphical Models What Is The Acceleration At The Following Points: a). 0 to 5 s b). 5 to 15 s c). 15 to 20 s

  30. A Baseball Player Hits A Triple And Ends Up On Third Base. A Baseball “Diamond” Is A Square, Each Side Of Length 27.4 m. What is The Magnitude Of His Displacement? If The Same Baseball Player Rounds 2nd Base With A Velocity Of 10 m/s & Slides To A Stop At 3rd Base, What Is His Deceleration From 2nd To 3rd? Problem #2: 1-D kinematics

  31. A Quarterback Throws A Pass To A Receiver, Who Catches It At The Same Height As The Pass Is Thrown. The Initial Velocity Of The Ball Is 15.0 m/s, At An Angle Of 25.0o Above The Horizontal. What Is The Horizontal Component Of The Ball’s Velocity When The Receiver Catches It? Problem #3: 2-d kinematics

  32. PROBLEM #4: 2-D KINEMATICS Given The Last Problem, How Far Did The Receiver Have To Run Before Making The Catch?

  33. Final Exam Review (Day 4) physics

  34. Dynamics Lecture Review • Force Types (4 Major Forces) • Newton’s Laws Of Motion • 1st Law (Freebody Diagrams) • 2nd Law (SF = Fnet = ma) • 3rd Law (F12 = -F21) • Problem Review (Group Problems) • Net Force • Newton’s Second Law I • Newton’s Second Law II Final Exam review: day 4 overview

  35. An Object With Mass (Takes Up Space) In The Presence Of Gravity Has Weight. W = mg • Any Object In Contact With A Surface Experiences A Force Normal To The Surface Called The Normal Force (FN) • An Applied Force (Mechanical or Electrical) Is One That Is Applied To An Object Causing It To Move (FA) • The Resistive Force Acting On An Object In Contact With Another Surface Is The Force Due To Friction • Static Friction (Fs = msFN) • Kinetic Friction (Fk = mkFN) Force types

  36. Force types Fk FN W

  37. Draw The Freebody(The Object Separated From Its Surroundings) • Draw Relevant Forces Acting On The Freebody From The Center Of The Object • Does The Body Have Mass In The Presence Of Gravity? • Is The Object In Contact With A Surface? • If Yes, Is The Contact Surface Frictionless or Is Friction Present? • Is There An Applied Force? Freebody diagrams

  38. Second Law: The Net Force (Fnet = SF) Acting On An Object Will Cause The Object To Accelerate (Motion) SF = ma • Add Forces In Different Directions Vectorally Fnet2 = Fx2 + Fy2 (Magnitude) Tan (q) = Fy/Fx (Direction) Newton’s second law: Sf = ma

  39. First Law An Object In Motion Will Stay In Motion Until Acted Upon By An External Net Force An Object At Rest Will Remain At Rest Until Acted Upon By An External Net Force • Third Law Two Objects In Contact Will Apply Forces Equal In Magnitude But Opposite In Direction F12 = -F21 Newton’s 1st & 3rd Laws:

  40. A Person With A Blackbelt In Karate Has A Fist That Has A Mass Of 0.70 kg. Starting From Rest, This Fist Attains A Velocity Of 8.0 m/s In 0.15 s. What Is The Magnitude Of The Average Net Force Applied To The Fist To Achieve This Level Of Performance? Problem #1: net force

  41. Two Forces, F1 & F2, Act On The 5.0 kg Block Shown In The Drawing. The Magnitudes Of The Forces Are F1 = 45.0 N And F2 = 25.0 N. What Is The Horizontal Acceleration (Magnitude & Direction) Of The Block? Problem #2: vector nature of 2nd law

  42. A 92 kg Baseball Player Slides Into Second Base. The Coefficient Of Kinetic Friction Between The Player And The Ground Is mk = 0.61. a). What Is The Magnitude Of The Frictional Force? b). If The Player Comes To Rest After 1.2 s, What Is His Initial Speed? Problem #3: friction

  43. Given The Following Atwood Machine With Masses M and m, Determine The Acceleration Of Mass m. Problem #4: Tension m M

More Related