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Chapter 6: Conservation of Energy

Chapter 6: Conservation of Energy. CQ: 1, 3. Problems: 1, 3, 5, 11, 17, 25, 29, 32, 49, 51, 53, 69, 71. Law of Conservation of Energy Work done by a constant force Kinetic energy Gravitational potential energy Work by variable force, Hooke’s law Elastic potential energy. Energy & Work.

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Chapter 6: Conservation of Energy

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  1. Chapter 6: Conservation of Energy • CQ: 1, 3. Problems: 1, 3, 5, 11, 17, 25, 29, 32, 49, 51, 53, 69, 71. • Law of Conservation of Energy • Work done by a constant force • Kinetic energy • Gravitational potential energy • Work by variable force, Hooke’s law • Elastic potential energy

  2. Energy & Work Energy is the capacity to do work, Energy is position & speed dependent Unit: joule = newton·meter (J = N·m) Work = force x distance (Fd) when force is in direction of motion (or opposite to motion) Ex. 50N pushes distance of 4 meters. W = (50N)(4m) = 200 J / 2

  3. Work & Force • Work is energy transferred by part of force in line of motion • Ex. Force 60° above path of motion

  4. Machines change an applied force by increasing it, decreasing it, or changing its direction. Types: inclined plane, screw, wedge pulley, wheel lever 4

  5. levers Work input Fd = Work output Fd Ex. Your hand moves 100m, causes car to rise 0.10m. The force amplification factor is, F d __ __ = F d

  6. inclined plane Weight x height change = Force x distance along plane Force along ramp less than Weight Ramp distance greater than height change ADA Standards: Ramp must be at least 12x longer than vertical rise Ex. A 1ft vertical rise requires 12ft of ramp. 6

  7. ADA Ramp 7

  8. Energy of Motion Called Kinetic Energy (KE) KE = ½(mass)(velocity)2 = ½mv2. Ex. 2000kg car moving at 2m/s. KE = ½ (2000)(2)2 = 4000J. Position Dependent Energies are called Potential Energies “PE” or “U” / 8

  9. Gravitational Potential Energy • U = weight x height (mgh) • 1kg at 1m height: • U = (1kg)(9.8N/kg)(1m) = 9.8J • Energy released in falling • /

  10. Hooke’s Law • The restoring force an object exerts is proportional to the amount it has been deformed (F = -kx)

  11. Elastic Potential Energy • PE-elastic = average force x distance • 10N compresses a spring 1m. U = (avg. force, 5 N)(1m) = 5 J • k = spring constant = force/distance • U = (½kx)x = ½kx2

  12. Work Energy Theorem Let direction of motion be +x

  13. Power Power is the rate work is performed Power = work/time = Force x velocity Unit: watt = joule/second = J/s Other Unit: horsepower 1 horsepower = 746 watts / 13

  14. Energy & Power Energy = power x time Ex. A toy car has 1000 J of energy at full charge. How long can it run at 100 watts? At 10 watts? Time = Energy/power = 1000J/100watts = 10 seconds = 1000J/10watts = 100 seconds/ 14

  15. Conservation of Energy • E = K + U = constant • Ex. Falling: Kinetic ↑ as Potential ↓ • /

  16. K E Ug

  17. Energy Summary • PE-gravitational = mgy • PE-elastic = ½kx2 • KE = ½mv2 • Mechanical Energy ME = KEs + PEs • E = ME + Thermal Energy

  18. 1 2 3

  19. y y Energy and speed are same at height y Accelerations are not same

  20. d Ex. Sled slides to a stop

  21. 1 5 2 4 3 A 2.00kg ball is dropped from rest from a height of 1.0m above the floor. The ball rebounds to a height of 0.500m. A movie-frame type diagram of the motion is shown below.

  22. Terminology • E: total energy of a system • E-mech = total energy minus the thermal energy • E-mech = E – Utherm. • Mech. Energy conserved in a frictionless system

  23. Power: The time rate of doing work. SI Unit: watt, W = J/s Example: How much average power is needed to accelerate a 2000kg car from rest to 20m/s in 5.0s? work = DKE

  24. Another equation for Power: Ex: A car drives at 20m/s and experiences air-drag of 400N. The engine must use (400N)(20m/s) = 8,000 watts of engine power to overcome this force. 8,000 watts = 10.7 hp.

  25. Work Example Moving down an inclined plane mgsinq mg h mg d q (mgsinq)d = mg(dsinq) = mgh

  26. Summary • Energy: Kinetic + Potential + Thermal, is conserved. • Mech. Energy: Kinetic + Potential, conserved in frictionless systems • Work is energy transfer (+ or -) • Power is rate of energy transfer

  27. Vehicle Efficiency 1 gallon gasoline has 138,000,000 J Engines only get a fraction of this: Ex. A 25% efficient car gets (0.25)(138,000,000 J) = 34,500,000J out of 1 gallon. A 20% efficient car gets 27,600,000J. 27

  28. Vehicle Frictional Work = Total Frictional Force x distance Ex. 400N friction for 1600 meters (1 mile) Work = (400N)(1600m) = 640,000J for one mile traveled /

  29. Mpg

  30. Ex. Mpg 20% Efficiency, f = 400N Engine gets 27,600,000 J/gal Frictional Work/Mile = 640,000J/mile = 43 mpg (at constant speed) 30

  31. Horsepower: 1 hp = 746 watts For the previous example:

  32. What size electric motor is needed to raise 2000lbs = 9000N of bricks at 10cm/s? Minimum Power: Pavg = Fvavg = (9000N)(0.1m/s) P = 900 W = 1.2 hp

  33. Similar to 5-45. • 3gram bullet moves at 200m/s and goes 10cm into a tree. What is the average force on the bullet? Tree? • Wnet on bullet = -Fd = change in K • Change in K = 0 – ½ (0.003kg)(200)(200) • -F(0.1m) = - 60Nm • F = 600N

  34. By energy conservation, the sum of all energies in each column is the same, = E1 = mg(1) = 19.6J Calculate v2: (use 1st and 2nd columns) mg(1) = ½ m(v2)2. g = ½ (v2)2. v2 = 4.43m/s Calculate PE-thermal: (use 1st and 5th columns) mg(1) = mg(1/2) + PE-thermal mg(1/2) = PE-thermal PE-thermal = 9.8J 37

  35. Calculate PE-elastic: (use 1st and 3rd columns) PE-elastic + PE-thermal = mg(1) PE-elastic + 9.8 = 19.6 PE-elastic = 9.8J Calculate v4: (use 1st and 4th columns) ½ m(v4)2 + PE-thermal = mg(1) ½ m(v4)2 + 9.8 = 19.6 ½ m(v4)2 = 9.8 (v4)2 = 2(9.8)/2 v4 = 3.13m/s 38

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