Definition of work Ingredients: Force(F), displacement (Ar) Work W. of a constant force F acting through a displacement A r is F W=F△rcos0 0/△r Physics 121: Lecture 12, Pg 6
Physics 121: Lecture 12, Pg 6 Definition of Work: Ingredients: Force ( F ), displacement ( r ) Work, W, of a constant force F acting through a displacement r is: W = F r cos F r Fr
Definition of work Only the component of Falong the displacement is doing work EXample: Train on a track Fcos 0 Physics 121: Lecture 12, Pg 7
Physics 121: Lecture 12, Pg 7 Definition of Work... Only the component of F along the displacement is doing work. Example: Train on a track. r F F cos
Lecture 12 ACT 1 Work a box is pulled up a rough(u>0)incline by a rope-pulley weight arrangement as shown below How many forces are doing work on the box (a)2 (b)3 (C) Physics 121: Lecture 12, Pg 8
Physics 121: Lecture 12, Pg 8 Lecture 12, ACT 1 Work A box is pulled up a rough (m > 0) incline by a rope-pulleyweight arrangement as shown below. How many forces are doing work on the box ? (a) 2 (b) 3 (c) 4
Work: 1-D EXample (constant force) A force F= 10N pushes a box across a frictionless floor for a distance ax= 5m F △X Work done by on box F=F.△X=F△x (since F is parallel to AX) WF=(10N)x(5m)=50N-m Physics 121: Lecture 12, Pg 9
Physics 121: Lecture 12, Pg 9 Work: 1-D Example (constant force) A force F = 10N pushes a box across a frictionless floor for a distance x = 5m. x F Work done by F on box : WF = F ·x = F x (since F is parallel to x) WF = (10 N)x(5m) = 50 N-m
Units. Force x Distance= Work Newton x Meter Joule [M[L/2[][ML]2/[2 mks gs other N-m(Joule) Dyne-cm(erg) BTU 1054J 107J calorie = 4.184 J foot-Ib =1.356 J e∨ 1.6×10-19J Physics 121: Lecture 12, Pg 10
Physics 121: Lecture 12, Pg 10 Units: N-m (Joule) Dyne-cm (erg) = 10-7 J BTU = 1054 J calorie = 4.184 J foot-lb = 1.356 J eV = 1.6x10-19 J mks cgs other Force x Distance = Work Newton x [M][L] / [T]2 Meter = Joule [L] [M][L]2 / [T]2