(1 ) Divide (g) the path into a number of intervals of△r1,△r2,△r3…,△r…, the element work(元功) △ Wi done by e,over△ ri is given by(3-2): △=F2A=F2C0s日2△
i i i i i i W = F r = F cos r (1) Divide（分割） the path into a number of intervals of △r1, △r2, △ r3…, △ri, …, the element work（元功） △Wi done by over △ri is given by (3-2): Fi i r Fi i
(2) The total work(总功) over the path is W≈)AW;=)F →0 Reducing Ari to infinite small, max we have △r: W=∑F,63 limar:→0 How is the sum accomplished(完成)?
i r Fi i → = i lim r i i W F r (3-3) Reducing △ri,to infinite small, we have = i i i W W F r (2) The total work（总功）over the path is 0 i max r How is the sum accomplished(完成）?
(3) According to definition of(曲线积分定义) curvilinear integral, Eq 3-3) becomes W=C F dr= Fcose ds (3-4) where ds is the magnitude of dr Fcos e
⑶ According to definition of ( 曲线积分定义）curvilinear integral, Eq.(3-3) becomes = = B A B A W F dr F cos ds (3-4) where ds is the magnitude of r . d F cos
Special cases O if F= a constant vector over the entire path, (3-4) becomes W=F[dr=F(i-Ta) B52 =F·ab=F·s(3-5) where S=ab 2 In the case of one dimension W=F dr= F(x)dx(3-6)
Special cases: F ab F s W F r F r r b a b a = = = = − d ( ) (3-5) ② In the case of one dimension = = 2 1 d d x x b a W F r F(x) x (3-6) ① if F = a constant vector over the entire path, (3-4) becomes where S = ab a b F Constant vector mg B52
3 Work done by resultant force(合力的功) If F=f+F+F+., we can compute its work as follows W:=J F: dr 亚+E+[F+(67) which is equal to the sum of work done by each force F ￥ 罗安定团结
3. Work done by resultant force ( 合力的功） If , we can compute its work as follows F = F + F + F + = + + + = + + + = 1 2 3 1 2 3 d d d d W W W F r F r F r W F r b a b a b a b a (3-7) which is equal to the sum of work done by each force . Fi 安定团结