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Announcements. smartPhysics grades uploaded to Canvas Preflights for Units 1 – 6 (Preflight = Prelecture + Checkpoint) Homework for Units 1 – 4

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Announcements

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  1. Announcements • smartPhysics grades uploaded to Canvas • Preflights for Units 1 – 6 (Preflight = Prelecture + Checkpoint) • Homework for Units 1 – 4 • To compare SP and Canvas, multiply the SP score (in percent) by 3 points (for Prelecture), by 2 points (for Checkpoint), by 5 points (for Homework) • Please check these and to make sure they are accurate (the process is cumbersome, so there may be some transfer errors) • Exam 1 • Grades will be posted soon • You can pick them up tomorrow in your discussion sessions • Fill out the survey on Canvas for up to 4 bonus points!!! (You are not eligible if you didn’t take the exam.)

  2. Classical Mechanics Lecture 7: Work and Energy Today’s Concepts: Work & Kinetic Energy

  3. Question of the Day A block of mass m = 2 kg is on a ramp inclined at an angle 30° from the horizontal. The coefficient of kinetic friction between the block and ramp is μk=0.12. A force F is applied to the block along the incline as shown, and the block moves up the ramp at constant velocity v = 2 m/s. How much work does the force F do in 0.6 seconds?

  4. Work-Kinetic Energy Theorem The total (net) work done by the all the forces acting on an object that moves between positions r1andr2is equal to the change in the object’s kinetic energy between those points:

  5. The Dot Product

  6. Work-Kinetic Energy Theorem: 1-D Example F d car = Fd In this case since cos(0) = 1 This is probably what you remember from High School. If the force is constant and the directions aren’t changing then this is very simple to evaluate:

  7. ACT: A car and a van F d Analysis car F d van A) They will have the same velocity B) They will have the same kinetic energy C) They will have the same momentum A lighter car and a heavier van, each initially at rest, are pushed with the same constant force F. After both vehicles travel a distance d, which of the following statements is true? (Ignore friction and air resistance.)

  8. ACT! A block of mass m = 2 kg is on a ramp inclined at an angle 30° from the horizontal. The coefficient of kinetic friction between the block and ramp is μk=0.12. A force F is applied to the block along the incline as shown, and the block moves up the ramp at constant velocity v = 2 m/s. What is the magnitude of F? • F = 0 • F = mg • F = μkmg • F is not constant • None of the above Constant speed:

  9. Question of the Day A block of mass m = 2 kg is on a ramp inclined at an angle 30° from the horizontal. The coefficient of kinetic friction between the block and ramp is μk=0.12. A force F is applied to the block along the incline as shown, and the block moves up the ramp at constant velocity v = 2 m/s. How much work does the force F do in 0.6 seconds?

  10. Question of the Day A block of mass m = 2 kg is on a ramp inclined at an angle 30° from the horizontal. The coefficient of kinetic friction between the block and ramp is μk=0.12. A force F is applied to the block along the incline as shown, and the block moves up the ramp at constant velocity v = 2 m/s. How much work does the force F do in 0.6 seconds? 30.03 J 22.82 J 18.94 J 14.22 J -24.43 J

  11. Work-Kinetic Energy Theorem You can just add up the work done by each force If there are several forces acting then Wnetis the work done by the net (total) force:

  12. ACT! A block of mass m = 2 kg is on a ramp inclined at an angle 30° from the horizontal. The coefficient of kinetic friction between the block and ramp is μk=0.12. A force F is applied to the block along the incline as shown, and the block moves up the ramp at constant velocity v = 2 m/s. What is the NET work done on the block as it moves a distance D up the ramp? • Wnet = FD • Wnet = FD - fkD • Wnet = FD - fkD - mgD • Wnet = 0 Constant speed:

  13. Checkpoint Review: Driving Up a Hill v A) WTOT> 0 B) WTOT= 0 C) WTOT< 0 • The car is going up the hill which indicates there is work done on the car. • Velocity isn't changing, therefore the Work (since this equals the change in kinetic energy) is zero. • We know that the W is equal to the integral of the net force multiplied by the total displacement. Since we are only concerned with the total displacement in the vertical direction. W = -mg(h) which will come out negative. If the car was driving downhill then W = -mg(-h) = positive. The total work will be negative. A car drives up a hill with constant speed. Which statement best describes the total work WTOT done on the car by all forces as it moves up the hill?

  14. ACT: Car & Hill v A) The normal force. B) The other component of gravity. C) Friction. A car drives up a hill with constant speed, so the total work done on it must be zero (checkpoint). Since gravity has a component that acts downhill, it will do negative work on the car. What force acts up the hill to yield zero total work?

  15. ACT: Car & Hill A) The normal force. B) The other component of gravity. C) Friction. If net force is zero, then no net work! If net force is zero, then no ΔK! A car drives up a hill with constant speed, so the total work done on it must be zero (checkpoint). Since gravity has a component that acts downhill, it will do negative work on the car. What force acts up the hill to yield zero total work?

  16. Work-Kinetic Energy Theorem The total (net) work done by the all the forces acting on an object that moves between positions r1andr2is equal to the change in the object’s kinetic energy between those points:

  17. Calculating the work in more than 1D Derivation – not so important Concept – very important A force pushing over some distancewill change the kinetic energy.

  18. Work done by gravity near the Earth’s surface path mg

  19. dl1 dy1 dx1 mg Work done by gravity near the Earth’s surface dlN dl1 dl2 mg

  20. Work done by gravity near the Earth’s surface dlN Δy dl1 dl2 mg

  21. ACT: Three falling objects H Free Fall Frictionless ramp String A) vf>vr>vsB)vf>vs>vrC)vf=vs=vr Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight down, one slides down a frictionless ramp, and one swings on the end of a string. What is the relationship between their speeds when they reach the bottom?

  22. Analysis Wg=mgH ΔK=1/2 mv22 H Free Fall Frictionless ramp String A) vf>vr>vsB) vf>vs>vrC)vf=vs=vr Only gravity will do work: Wg=ΔK

  23. ACT: Orbiting Satellite A satellite orbits earth at constant speed along a circular orbit. • Analysis 1: • Satellite maintains constant speed • Thus ΔK = 0 • Thus Wtot = 0 • Analysis 2: • Gravity is the only force on satellite • Gravity acts radially • Motion is tangential • Thus Fdl = 0 • Thus Wtot = 0 During one orbit, what is the Total Work done on the satellite? A) Positive B) Negative C) Zero

  24. Only Radial Motion Matters

  25. ACT: Work done by gravity In Case 1 we send an object from the surface of the earth to a height above the earth surface equal to one earth radius. In Case 2 we start the same object a height of one earth radius above the surface of the earth and we send it infinitely far away. In which case is the magnitude of the work done by the Earth’s gravity on the object biggest? A)Case 1 B) Case 2 C) They are the same Case 2 Case 1

  26. Same! ACT Solution Case 1: Case 2: RE Case 2 Case 1 2RE

  27. Work done by a Spring An integral is area under curve between two points: The area of a triangle is:

  28. ACT: Work by Spring, I A box attached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new location. D During this motion, the spring does: A) Positive Work B) Negative Work C) Zero work

  29. ACT: Work by Spring, II A box attached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new location. D During this motion, your hand does: A) Positive Work B) Negative Work C) Zero work

  30. ACT: Work by Spring, III A box attached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new location. D During this motion, the total work done on the box is: A) Positive B) Negative C) Zero

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