Work Contents Definition Tricky Bits Whiteboards Work Transfer

Work Contents: • Definition • Tricky Bits • Whiteboards

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

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

Whiteboards: Work 1|2|3

Fred O’Dadark exerts 13. 2 N on a rope that makes a 32 o 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(32 o) = 139. 9 Nm = 139. 9 Joules = 140 J So a Nm = J (Joule) 140 J

Jane Linkfence does 132 J of work lifting a box 1. 56 m. What is the weight of the box? W = Fd cos = 0 o (assuming she lifts straight up) W = Fd, F = W/d = (132 Nm)/(1. 56 m) = 84. 6 N

Myron Wondergaim exerts 43 N of force, and does 108 J of work pushing a sofa how far? (2 hints) W = Fd cos = 0 o (assume) W = Fd, d = W/F = (108 Nm)/(43 N) = 2. 5 m

Work - Transfer of energy Work = (Force)(Distance) W = Fd F = mg (if you are lifting) F = μmg (if you are dragging)

W = Fd F = mg (lifting) OR F = μmg (dragging) Art Zenkraft does 342 J of work lifting a 12. 0 kg mass how far? (2. 91 m) (J) W (N) F (m) d (kg) m ( ) μ

W = Fd F = mg (lifting) OR F = μmg (dragging) Shirley Nott does 552 J of work sliding a 45. 0 kg mass a distance of 10. 4 m. What must be the coefficient of friction? (0. 120) (J) W (N) F (m) d (kg) m ( ) μ

Whiteboards: Work and Weight and Friction 1|2|3

Helena Handbasket lifts a 5. 2 kg box from the floor to a 1. 45 m tall shelf. What work does she do? F = mg F = (5. 2 kg)(9. 8 N/kg) = 50. 96 N = 73. 892 J 74 J

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

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

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

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 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 m= 4. 0 kg d = 1. 5 m

Work and Friction Dragging things W = Fd cos sometimes F = μmg = (. 26)(5. 0 kg)(9. 8 N/kg) = 12. 74 N W = Fd = (12. 74 N)(3. 5 m) = 44. 59 J d = 3. 5 m m= 5. 0 kg μ =. 26