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Chapter 6-powerpoint-1224302877703008-9

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Chapter 6-powerpoint-1224302877703008-9

  1. 1. Chapter 6 Work and Energy
  2. 2. 6.1 Work Done by a Constant Force FsW or FdW = = ( )Jjoule1mN1 =⋅
  3. 3. 6.1 Work Done by a Constant Force ( )sFW θcos= 1180cos 090cos 10cos −= = =   
  4. 4. 6.1 Work Done by a Constant Force Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m. ( ) ( )[ ]( ) J2170 m0.750.50cosN0.45cos = ==  sFW θ
  5. 5. 6.1 Work Done by a Constant Force ( ) FssFW == 0cos ( ) FssFW −== 180cos
  6. 6. 6.1 Work Done by a Constant Force Example 3 Accelerating a Crate The truck is accelerating at a rate of +1.50 m/s2 . The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it?
  7. 7. 6.1 Work Done by a Constant Force The angle between the displacement and the friction force is 0 degrees. ( )[ ]( ) J102.1m650cosN180 4 ×==W ( )( ) N180sm5.1kg120 2 === mafs
  8. 8. 6.2 The Work-Energy Theorem and Kinetic Energy DEFINITION OF KINETIC ENERGY The kinetic energy KE of and object with mass m and speed v is given by 2 2 1 KE mv=
  9. 9. 6.2 The Work-Energy Theorem and Kinetic Energy THE WORK-ENERGY THEOREM When a net external force does work on and object, the kinetic energy of the object changes according to 2 2 12 f2 1 of KEKE omvmvW −=−=
  10. 10. 6.2 The Work-Energy Theorem and Kinetic Energy Example 4 Deep Space 1 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If the .056 N force acts on the probe through a displacement of 2.42×109 m, what is its final speed?
  11. 11. 6.2 The Work-Energy Theorem and Kinetic Energy 2 2 12 f2 1 W omvmv −= ( )[ ]sF θcosW ∑=
  12. 12. 6.2 The Work-Energy Theorem and Kinetic Energy ( ) ( ) ( ) ( )( )2 2 12 f2 19 sm275kg474kg474m1042.20cosN056. −=× v ( )[ ] 2 2 12 f2 1 cosF omvmvs −=∑ θ sm805=fv
  13. 13. 6.2 The Work-Energy Theorem and Kinetic Energy In this case the net force is kfmgF −=∑  25sin
  14. 14. 6.3 Gravitational Potential Energy ( )sFW θcos= ( )fo hhmgW −=gravity
  15. 15. 6.3 Gravitational Potential Energy ( )fo hhmgW −=gravity
  16. 16. 6.3 Gravitational Potential Energy Example 7 A Gymnast on a Trampoline The gymnast leaves the trampoline at an initial height of 1.20 m and reaches a maximum height of 4.80 m before falling back down. What was the initial speed of the gymnast?
  17. 17. 6.3 Gravitational Potential Energy 2 2 12 f2 1 W omvmv −= ( )fo hhmgW −=gravity ( ) 2 2 1 ofo mvhhmg −=− ( )foo hhgv −−= 2 ( )( ) sm40.8m80.4m20.1sm80.92 2 =−−=ov
  18. 18. 6.3 Gravitational Potential Energy fo mghmghW −=gravity DEFINITION OF GRAVITATIONAL POTENTIAL ENERGY The gravitational potential energy PE is the energy that an object of mass m has because of its position relative to the surface of the earth. mgh=PE ( )Jjoule1mN1 =⋅
  19. 19. 6.4 Conservative Versus Nonconservative Forces THE WORK-ENERGY THEOREM PEKE ∆+∆=ncW
  20. 20. 6.5 The Conservation of Mechanical Energy ( ) ( )ofof PEPEKEKEPEKE −+−=∆+∆=ncW ( ) ( )ooff PEKEPEKE +++=ncW of EE −=ncW If the net work on an object by nonconservative forces is zero, then its energy does not change: of EE =
  21. 21. 6.5 The Conservation of Mechanical Energy THE PRINCIPLE OF CONSERVATION OF MECHANICAL ENERGY The total mechanical energy (E = KE + PE) of an object remains constant as the object moves.
  22. 22. 6.5 The Conservation of Mechanical Energy
  23. 23. 6.5 The Conservation of Mechanical Energy Example 8 A Daredevil Motorcyclist A motorcyclist is trying to leap across the canyon by driving horizontally off a cliff 38.0 m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.
  24. 24. 6.5 The Conservation of Mechanical Energy of EE = 2 2 12 2 1 ooff mvmghmvmgh +=+ 2 2 12 2 1 ooff vghvgh +=+
  25. 25. 6.5 The Conservation of Mechanical Energy 2 2 12 2 1 ooff vghvgh +=+ ( ) 2 2 ofof vhhgv +−= ( )( ) ( ) sm2.46sm0.38m0.35sm8.92 22 =+=fv
  26. 26. 6.6 Nonconservative Forces and the Work-Energy Theorem THE WORK-ENERGY THEOREM of EE −=ncW ( ) ( )2 2 12 2 1 ooffnc mvmghmvmghW +−+=
  27. 27. 6.6 Nonconservative Forces and the Work-Energy Theorem Example 11 Fireworks Assuming that the nonconservative force generated by the burning propellant does 425 J of work, what is the final speed of the rocket. Ignore air resistance. ( ) ( )2 2 1 2 2 1 oo ffnc mvmgh mvmghW + −+=
  28. 28. 6.6 Nonconservative Forces and the Work-Energy Theorem 2 2 12 2 1 ofofnc mvmvmghmghW −+−= ( )( )( ) ( ) 2 2 1 2 kg20.0 m0.29sm80.9kg20.0J425 fv+ = ( ) 2 2 1 fofnc mvhhmgW +−= sm61=fv
  29. 29. 6.7 Power DEFINITION OF AVERAGE POWER Average power is the rate at which work is done, and it is obtained by dividing the work by the time required to perform the work. t W P == Time Work (W)wattsjoule =
  30. 30. 6.7 Power Time energyinChange =P watts745.7secondpoundsfoot550horsepower1 =⋅= vFP =
  31. 31. 6.7 Power
  32. 32. 6.8 Other Forms of Energy and the Conservation of Energy THE PRINCIPLE OF CONSERVATION OF ENERGY Energy can neither be created not destroyed, but can only be converted from one form to another.

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