Work Contents: Definition Tricky Bits Whiteboards

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

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

Whiteboards: Work 1 | 2 | 3 TOC

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

Jane Linkfence does 132 J of work lifting a box 1. 56 m 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

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

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

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

Whiteboards: Work and Weight 1 | 2 | 3 TOC

Helena Handbasket brings a 5. 2 kg box down from a 1. 45 m tall shelf 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

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 2012.712 N = m(9.8 N/kg), m = 205.4 kg W 205.4 kg

Tubi O’ Notubi does 137 J of work lifting a 5 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

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

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 W 18.2 m