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Work and Energy

Work and Energy. Day 1 of 6. Part 1 of 2: Introduction to Work. Roller coaster video. FlipCam Video of marble launched horizontally from the roller coaster loop on the board. Information: Radius of loop = 0.125 m Camera shoots at 30 frames / sec

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Work and Energy

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  1. Work and Energy

  2. Day 1 of 6 Part 1 of 2: Introduction to Work

  3. Roller coaster video

  4. FlipCam Video of marble launched horizontally from the roller coaster loop on the board • Information: • Radius of loop = 0.125 m • Camera shoots at 30 frames / sec • The following snapshots are successive frames. • Meterstick shown begins at the end of the ramp • Is the marble’s vx constant? • What is it? • Is the marble’s vy constant? • Is the marble’s KE and/or PEg constant? • Is the marble’s ME constant?

  5. #1 35° Vx = 0.11 / 0.03=3.7m/s

  6. #2 ΔX = 47cm-36cm= 11cm Δy1 =78.5cm-70cm=8.5cm Δy2 =79.5cm-70cm=9.5cm 40° Vx = 0.11 / 0.03=3.7m/s

  7. #3 • Here the marble hits the table and loses energy in the impact. • How would you find the ME lost in the impact? • What would you measure from these pics?

  8. #4 • ΔX = 6 cm = 0.06 m • Δt = 0.03 s • Vx = 2 m/s • Is this the whole story?? • What about the vyf? • Could Energy make this • Easier? (YES!!!)

  9. #5 • ΔX = 6 cm

  10. #6 • ΔX = 6 cm

  11. Spring Constant Video

  12. Finding the Spring Constant • Length of tube to the top of the spring (Hypotenuse) = • Bearing’s Initial Height (opposite side) for: • 15° = • 25° = • 35° = • 45° = • Bearing’s mass = • Find Initial Potential Energies (Ug)

  13. Finding the Spring Constant • Initial Potential Energies: PEg = mgh • 15° = • 25° = • 35° = • 45° = • At the top of the spring, PEgi = KEf • The KE will now convert into PEs as the spring stops the ball bearing! (PEgi = KEf = PEs) • Find the spring constant from each measurement

  14. Announcements

  15. Work • Work tells us how much a force or combination of forces changes the energy of a system. • Work is the bridge between force (a vector) and energy (a scalar). • W = F Dr cos  • F: force (N) • Dr : displacement (m) • : angle between force and displacement

  16. Units of Work • SI System: Joule (N m) • 1 Joule of work is done when 1 N acts on a body moving it a distance of 1 meter • British System: foot-pound • (not used in Physics B) • cgs System: erg (dyne-cm) • (not used in Physics B) • Atomic Level: electron-Volt (eV)

  17. Force and direction of motion both matter in defining work! • There is no work done by a force if it causes no displacement. • Forces can do positive, negative, or zero work. When an box is pushed on a flat floor, for example… • The normal force and gravity do no work, since they are perpendicular to the direction of motion. • The person pushing the box does positive work, since she is pushing in the direction of motion. • Friction does negative work, since it points opposite the direction of motion.

  18. Positive vs. Negative Work

  19. Conceptual Checkpoint • Question: If a man holds a 50 kg box at arms length for 2 hours as he stands still, how much work does he do on the box?

  20. Conceptual Checkpoint • Question: If a man holds a 50 kg box at arms length for 2 hours as he walks 1 km forward, how much work does he do on the box?

  21. Conceptual Checkpoint • Question: If a man lifts a 50 kg box 2.0 meters, how much work does he do on the box?

  22. Work and Energy • Work changes mechanical energy! • If an applied force does positive work on a system, it tries to increase mechanical energy. • If an applied force does negative work, it tries to decrease mechanical energy. • The two forms of mechanical energy are called potential and kinetic energy.

  23. Sample problem Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to a tree house 9.0 meters above the ground. • How much work does Jane do when she lifts Tarzan? • How much work does gravity do when Jane lifts Tarzan?

  24. Sample problem Joe pushes a 10-kg box and slides it across the floor at constant velocity of 3.0 m/s. The coefficient of kinetic friction between the box and floor is 0.50. • How much work does Joe do if he pushes the box for 15 meters? • How much work does friction do as Joe pushes the box?

  25. Sample problem A father pulls his child in a little red wagon with constant speed. If the father pulls with a force of 16 N for 10.0 m, and the handle of the wagon is inclined at an angle of 60o above the horizontal, how much work does the father do on the wagon?

  26. Day 1 part 2 Kinetic Energy

  27. Kinetic Energy • Energy due to motion • K = ½ m v2 • K: Kinetic Energy • m: mass in kg • v: speed in m/s • Unit: Joules

  28. Sample problem A 10.0 g bullet has a speed of 1.2 km/s. • What is the kinetic energy of the bullet? • What is the bullet’s kinetic energy if the speed is halved? • What is the bullet’s kinetic energy if the speed is doubled?

  29. The Work-Energy Theorem • The net work due to all forces equals the change in the kinetic energy of a system. • Wnet = DK • Wnet: work due to all forces acting on an object • DK: change in kinetic energy (Kf – Ki)

  30. Sample problem An 80-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. • What would the speed of the acorn have been if there had been no air resistance? • Did air resistance do positive, negative or zero work on the acorn? Why?

  31. Sample problem An 80-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. • How much work was done by air resistance? • What was the average force of air resistance?

  32. Day 2 of 6 Part 1 of 2: Work done by variable forces

  33. Announcements

  34. Constant force and work F(x) • The force shown is a constant force. • W = FDr can be used to calculate the work done by this force when it moves an object from xa to xb. • The area under the curve from xa to xb can also be used to calculate the work done by the force when it moves an object from xa to xb x xa xb

  35. xa xb Variable force and work F(x) • The force shown is a variable force. • W = FDr CANNOT be used to calculate the work done by this force! • The area under the curve from xa to xb can STILL be used to calculate the work done by the force when it moves an object from xa to xb x

  36. Springs • When a spring is stretched or compressed from its equilibrium position, it does negative work, since the spring pulls opposite the direction of motion. • Ws = - ½ k x2 • Ws: work done by spring (J) • k: force constant of spring (N/m) • x: displacement from equilibrium (m) • The force doing the stretching does positive work equal to the magnitude of the work done by the spring. • Wapp = - Ws = ½ k x2

  37. F(N) 200 100 m 0 0 1 2 3 4 5 x (m) -100 Fs x Fs m -200 Springs: stretching 0 Ws = negative area = - ½ kx2 Fs = -kx (Hooke’s Law)

  38. Sample problem A spring with force constant 2.5 x 104 N/m is initially at its equilibrium length. • How much work must you do to stretch the spring 0.050 m? • How much work must you do to compress it 0.050 m?

  39. Sample problem It takes 130 J of work to compress a certain spring 0.10 m. • What is the force constant of the spring? • To compress the spring an additional 0.10 m, does it take 130 J, more than 130 J, or less than 130 J? Verify your answer with a calculation.

  40. Sample Problem • How much work is done by the force shown when it acts on an object and pushes it from x = 0.25 m to x = 0.75 m?

  41. Sample Problem • How much work is done by the force shown when it acts on an object and pushes it from x = 2.0 m to x = 4.0 m?

  42. Day 2 Part 2 Power

  43. Power • Power is the rate of which work is done. • P = W/Dt • W: work in Joules • Dt: elapsed time in seconds • When we run upstairs, t is small so P is big. • When we walk upstairs, t is large so P is small.

  44. Unit of Power • SI unit for Power is the Watt. • 1 Watt = 1 Joule/s • Named after the Scottish engineer James Watt (1776-1819) who perfected the steam engine. • British system • horsepower • 1 hp = 746 W

  45. How We Buy Energy… • The kilowatt-hour is a commonly used unit by the electrical power company. • Power companies charge you by the kilowatt-hour (kWh), but this not power, it is really energy consumed. • 1 kW = 1000 W • 1 h = 3600 s • 1 kWh = 1000J/s • 3600s = 3.6 x 106J

  46. Sample problem A record was set for stair climbing when a man ran up the 1600 steps of the Empire State Building in 10 minutes and 59 seconds. If the height gain of each step was 0.20 m, and the man’s mass was 70.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.

  47. Sample problem Calculate the power output of a 1.0 g fly as it walks straight up a window pane at 2.5 cm/s.

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