We divide the total displacement into a number N of small intervals of equal width Ax. This interval so small that the F(x)is approximately constant. Then in the interval x, to x,+dx, the work AW=FAx and similar△W,=F2△x The total work is W≈△+△W,+.=E1△x+F,△x+ or W≈>F△x (11-1 n=」 ch 11 Energy I: Work and kinetic energyCh.11 Energy I: Work and kinetic energy We divide the total displacement into a number N of small intervals of equal width . This interval so small that the F(x) is approximately constant. Then in the interval to +dx , the work and similar ……The total work is or (11-12)  =  W F x 1 1 x  =  W F x 2 21 x 1 x =   N n n W F x 1 ... ... W  W1 +W2 + = F1 x + F2 x +