1 / 15

Work Contents: Definition Tricky Bits Whiteboards

Work Contents: Definition Tricky Bits Whiteboards. F. Fcos . . d. Work - Transfer of energy. Work = (Force)(Distance) W = Fd cos . TOC. F. Fcos . . d. W = Fd cos . Tricky bits : 0 o cos  = 1 180 o cos  = -1 (examples of zero, positive, negative work). TOC.

lenacunningham
Télécharger la présentation

Work Contents: Definition Tricky Bits Whiteboards

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Work • Contents: • Definition • Tricky Bits • Whiteboards

  2. F Fcos  d Work - Transfer of energy Work = (Force)(Distance) W = Fd cos TOC

  3. F Fcos  d W = Fd cos Tricky bits : 0o cos = 1 180ocos = -1 (examples of zero, positive, negative work) TOC

  4. Whiteboards: Work 1 | 2 | 3 TOC

  5. Fred O’Dadark exerts 13.2 N on a rope that makes a 32o angle with the ground, sliding a sled 12.5 m along the ground. What work did he do? W = Fd cos = (13.2 N)(12.5 m) cos(32o) = 139.9 Nm = 139.9 Joules = 140 J So a Nm = J (Joule) W 140 J

  6. Jane Linkfence does 132 J of work lifting a box 1.56 m. What is the weight of the box? W = Fd cos  = 0o (assuming she lifts straight up) W = Fd, F = W/d = (132 Nm)/(1.56 m) = 84.6 N W 84.6 N

  7. Bobbie Sachs exerts 43 N of force, and does 108 J of work pushing a sofa how far? (2 hints) W = Fd cos  = 0o (assume) W = Fd, d = W/F = (108 Nm)/(43 N) = 2.5 m W 2.5 m

  8. m= 4.0 kg d = 1.5 m Work and Weight Lifting things W = Fd cos sometimes F = mg (weight) F = mg = (4.0 kg)(9.8 N/kg) = 39.2 N W = Fd = (39.2 N)(1.5 m) = 58.8 J TOC

  9. Work and Friction Dragging things W = Fd cos sometimes F = μmg F = μmg = (.26)(5.0 kg)(9.8 N/kg) = 12.74 N W = Fd = (12.74 N)(3.5 m) = 44.59 J μ = .26 d = 3.5 m m= 5.0 kg TOC

  10. Whiteboards: Work and Weight 1 | 2 | 3 TOC

  11. Helena Handbasket brings a 5.2 kg box down from a 1.45 m tall shelf. What work does she do? (3 hints) F = mg W = Fd cos  = 180o (tricky tricky) F = (5.2 kg)(9.8 N/kg) = 50.96 N W = Fd cos(180o) = (50.96 N)(1.45 m) cos(180o) = -73.892 J = -74 J (The box does work on her. gives her energy) W -74 J

  12. Paul E. Wannacracker does 2375 J of work lifting what mass a height of 1.18 m? F = mg W = Fd cos = Fd 2375 J = F(1.18 m), F = 2012.712 N F = mg 2012.712 N = m(9.8 N/kg), m = 205.4 kg W 205.4 kg

  13. Tubi O’ Notubi does 137 J of work lifting a 5.25 kg mass to what height? F = mg F = (5.25 kg)(9.8 N/kg) = 51.45 N W = Fd cos = Fd 137 J = (51.45 N)d, d = 2.66 m W 2.66 m

  14. Hugh Jass drags a 125 kg sled with a coefficient of kinetic friction of .15 a distance of 34 m. What work does he do? F = μmg F = (.15)(125)(9.8) = 183.75 N W = Fd = (183.75 N)(34 m) = 6247.5 J ≈ 6250 J W 6250 J

  15. Seymour Butz does 1200 J of work dragging a 32 kg box with a coefficient of kinetic friction of .21 how far? F = μmg F = (.21)(32)(9.8) = 65.856 N W = Fd 1200 J = (65.856 N)d, d ≈ 18.2 N W 18.2 N

More Related