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Understanding Energy, Work, and Simple Machines in Physics

This chapter explores the fundamental concepts of energy, work, and simple machines. Energy is defined as the ability to produce change, categorized into kinetic and potential forms. The chapter discusses the equations that relate kinetic and potential energy to motion and height. It further delves into the work-energy theorem and practical calculations on work done in various scenarios. Additionally, the role of machines is analyzed, including mechanical advantage and efficiency, illustrating how they simplify tasks without reducing the work required.

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Understanding Energy, Work, and Simple Machines in Physics

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  1. Chapter 10 Energy, Work, & Simple Machines

  2. Energy • The ability to produce change

  3. Energy • The ability to do work

  4. Types of Energy • Kinetic • Potential

  5. Kinetic Energy (K) • The energy of motion

  6. Potential Energy (U) • Stored energy

  7. Kinetic Energy • vf2 = vi2 + 2ad • vf2 - vi2 = 2ad

  8. Kinetic Energy • a = F/m • vf2- vi2 = 2Fd/m

  9. Kinetic Energy ½ mvf2- ½ mvi2 = Fd

  10. Kinetic Energy K = ½ mv2

  11. Potential Energy U =mgh

  12. Work (W) • The process of changing the energy of a system

  13. Work • The product of force times displacement

  14. Work • W = Fd

  15. Work-Energy Theorem • W = DK

  16. Calculate the work required to lift a 50.0 kg box to a height of 2.0 m:

  17. Calculate the work done when a 250 N force is applied to move a cart 40.0 km:

  18. Calculate the work required to push a 500.0 kg box 250 m at a constant velocity.m = 0.20 between the box & the floor.

  19. Constant force at an Angle Direction of applied force a Direction of movement

  20. Constant force at an Angle W = F(cos a)d

  21. Calculate the work done when mowing the lawn when a boy applied a 50.0 N force at a 37o from horizontal for 2.0 km.

  22. Calculate the work done when a girl pulls a 4.0 kg box with a rope at a 37o from horizontal for 2.0 m. m = 2.5

  23. Power • The rate of doing work

  24. Power • P = W/t

  25. A 25 Mg elevator rises 125 m in 5.0 minutes. Calculate: F, W, & P

  26. A 10.0 Gg crate is accelerated by a cable up a 37o incline for 50.0 m in 2.5 hrs. m = 0.20 Calculate: FT, W, & P

  27. A 50.0 g box is accelerated up a 53o incline for 50.0 m at 250 cm/s2. m = 0.20 Calculate: FA, vf,W, P, K, & U at the top of the ramp

  28. Machines • Devices used to ease force one has to apply to move an object by changing the magnitude and direction of the force.

  29. Machines • Machines do not reduce the work required, but do reduce the force required.

  30. Machines • The force applied is called the effort force (Fe).

  31. Machines • The force exerted by the machine is called the resistant force (Fr).

  32. Mechanical Advantage • The ratio of resistant force to effort force

  33. Mechanical Advantage Fr Fe MA =

  34. In an Ideal Situation • 100 % of the work input into a system would be transferred to output work, thus:

  35. Wo = Wi or Frdr = Fede or Fr/Fe= de/dr

  36. Ideal Mechanical Advantage de dr IMA =

  37. Efficiency • The ratio of output work to input work times 100 %

  38. Efficiency = Wo Wi X 100 %

  39. Efficiency = MA IMA X 100 %

  40. Simple Machines Lever Inclined plane Wedge Wheel & Axle Screw Pulley

  41. Lever Fr Fe de dr

  42. Fr Fe de dr IMA = de/dr = length de/length dr

  43. Inclined Plane de Fr Fe dr

  44. de Fr Fe dr a IMA = de/dr = length hyp/hyp sin a

  45. Wedge ½ Fr Fe ½ Fr

  46. ½ Fr a Fe ½ Fr IMA = de/dr = cot ½ a

  47. Screw Fe Fr

  48. Pulley Fe Fr

  49. IMA = the number of lines pulling up Fe Fr

  50. Wheel & Axle Fr Fe

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